October 7, 2013 EIGENVALUES AND EIGENVECTORS RODICA D. COSTIN Contents 1. Motivation 3 1.1. Diagonal matrices 3 1.2. Example: solving linear di↵erential equations 3 2. Eigenvalues and eigenvectors: definition and calculation 4 2.1. Definitions 4 2.2. The characteristic equation 5 2.3. Geometric interpretation of eigenvalues and eigenvectors 6 2.4. Digression: the fundamental theorem of algebra 6 2.5. Diagonal matrices 8 2.6. Similar matrices have the same eigenvalues 8 2.7. Projections 9 2.8. Trace, determinant and eigenvalues 9 2.9. The eigenvalue zero 10 2.10. Eigenvectors corresponding to di↵erent eigenvalues are independent 10 2.11. Diagonalization of matrices with linearly independent eigenvectors 11 2.12. Eigenspaces 13 2.13. Real matrices with complex eigenvalues; decomplexification 14 2.14. Jordan normal form 16 2.15. Block matrices 20 3. Solutions of linear di↵erential equations with constant coefficients 22 3.1. The case when M is diagonalizable. 22 3.2. Non-diagonalizable matrix 26 3.3. Fundamental facts on linear di↵erential systems 29 3.4. Eigenvalues and eigenvectors of the exponential of a matrix 30 3.5. Higher order linear di↵erential equations; companion matrix 31 3.6. Stability in di↵erential equations 34 4. Di↵erence equations (Discrete dynamical systems) 40 4.1. Linear di↵erence equations with constant coefficients 40 4.2. Solutions of linear di↵erence equations 40 4.3. Stability 41 4.4. Example: Fibonacci numbers 42 4.5. Positive matrices 43 4.6. Markov chains 43 5. More functional calculus 45 1
Transcript
October 7, 2013
EIGENVALUES AND EIGENVECTORS
RODICA D. COSTIN
Contents
1. Motivation 31.1. Diagonal matrices 31.2. Example: solving linear di↵erential equations 32. Eigenvalues and eigenvectors: definition and calculation 42.1. Definitions 42.2. The characteristic equation 52.3. Geometric interpretation of eigenvalues and eigenvectors 62.4. Digression: the fundamental theorem of algebra 62.5. Diagonal matrices 82.6. Similar matrices have the same eigenvalues 82.7. Projections 92.8. Trace, determinant and eigenvalues 92.9. The eigenvalue zero 102.10. Eigenvectors corresponding to di↵erent eigenvalues are
independent 102.11. Diagonalization of matrices with linearly independent
eigenvectors 112.12. Eigenspaces 132.13. Real matrices with complex eigenvalues; decomplexification 142.14. Jordan normal form 162.15. Block matrices 203. Solutions of linear di↵erential equations with constant coe�cients 223.1. The case when M is diagonalizable. 223.2. Non-diagonalizable matrix 263.3. Fundamental facts on linear di↵erential systems 293.4. Eigenvalues and eigenvectors of the exponential of a matrix 303.5. Higher order linear di↵erential equations; companion matrix 313.6. Stability in di↵erential equations 344. Di↵erence equations (Discrete dynamical systems) 404.1. Linear di↵erence equations with constant coe�cients 404.2. Solutions of linear di↵erence equations 404.3. Stability 414.4. Example: Fibonacci numbers 424.5. Positive matrices 434.6. Markov chains 435. More functional calculus 45
1
2 RODICA D. COSTIN
5.1. Discrete equations versus di↵erential equations 455.2. Functional calculus for digonalizable matrices 465.3. Commuting matrices 49
EIGENVALUES AND EIGENVECTORS 3
1. Motivation
1.1. Diagonal matrices. Perhaps the simplest type of linear transforma-tions are those whose matrix is diagonal (in some basis). Consider for ex-ample the matrices
(1) M =
a1 00 a2
�, N =
b1 00 b2
�
It can be easily checked that
↵M + �N =
↵a1 + �b1 0
0 ↵a2 + �b2
�
and
M�1 =
1a1
00 1
a2
�, Mk =
ak1 00 ak2
�, MN =
a1b1 00 a2b2
�
Diagonal matrices behave like the bunch of numbers on their diagonal!The linear transformation consisting of multiplication by the matrix M
in (1) dialates a1 times vectors in the direction of e1 and a2 times vectorsin the direction of e2.
In this chapter we will see that most linear transformations do have diag-onal matrices in a special basis, whose elements are called the eigenvectorsof the transformation. We will learn how to find these bases. Along thespecial directions of the eigenvectors the transformation just dialation by afactor, called eigenvalue.
1.2. Example: solving linear di↵erential equations. Consider the sim-ple equation
du
dt= �u
which is linear, homogeneous, with constant coe�cients, and unknown func-tion u(t) 2 R (or in C). Its general solution is, as it is well known,u(t) = Ce�t.
Consider now a similar equation, but where the unknown u(t) is a vectorvalued function:
(2)du
dt= Mu, u(t) 2 Rn, M is an n⇥ n constant matrix
Inspired by the one dimensional case we look for exponential solutions.Substituting in (2) u(t) = e�tv (where � is a number and v is a constantvector, both be determined) and dividing by e�t, we obtain that the scalar� and the vector v must satisfy
(3) �v = Mv
or
(4) (M � �I)v = 0
4 RODICA D. COSTIN
If the null space of the matrix M � �I is zero, then the only solution of(4) is v = 0 which gives the (trivial!) solution u(t) ⌘ 0.
If however, we can find special values of � for which N (M � �I) is notnull, then we found a nontrivial solution of (2). Such values of � are calledeigenvalues of the matrix M , and vectors v 2 N (M � �I), v 6= 0, arecalled eigenvectors corresponding to the eigenvalue �.
Of course, the necessary and su�cient condition for N (M � �I) 6= {0} isthat
(5) det(M � �I) = 0
Example. Let us calculate the exponential solutions for
(6) M =
"�1 �3
0 2
#
Looking for eigenvalues of M we solve equation (5), which for (6) is
det
"�1 �3
0 2
#� �
"1 0
0 1
#!=
������1� � �3
0 2� �
����� = (�1� �) (2� �)
with solutions �1 = �1 and �2 = 2.We next determine an eigenvector corresponding to the eigenvalue � =
�1 = �1: looking for a nozero vector v1 such that (M��1I)v1 = 0 we solve"
0 �3
0 3
# "x1
x2
#=
"0
0
#
giving x2 = 0 and x1 arbitrary; therefore the first eigenvector is any scalarmultiple of v1 = (0, 1)T .
Similarly, for the eigenvalue � = �2 = 2 we solve (M � �2I)v2 = 0:"
�3 �3
0 0
# "y1
y2
#=
"0
0
#
which gives y2 = �y1 and y1 arbitrary, and the second eigenvector is anyscalar multiple of v2 = (1,�1)T .
We found two particular solutions of (2), (6), namely u1(t) = e�t(0, 1)T
and u2(t) = e2t(1,�1)T . These are functions belonging to the null spaceof the linear operator Lu = du
dx
�Mu, therefore any linear combination ofthese two solutions also belongs to the null space: any C1u1(t) +C2u2(t) isalso a solution, for and constants C1, C2.
A bit later we will show that these are all the solutions.
2. Eigenvalues and eigenvectors: definition and calculation
2.1. Definitions. Denote the set of n⇥n (square) matrices with entries inF (= R or C)
M
n
(F ) = {M |M = [Mij
]i,j=1,...n, M
ij
2 F}
EIGENVALUES AND EIGENVECTORS 5
A matrixM 2 M
n
(F ) defines an endomorphism the vector space Fn(overthe scalars F ) by usual multiplication x 7! Mx.
Note that a matrix with real entries can also act on Cn, since for anyx 2 Cn also Mx 2 Cn. But a matrix with complex non real entries cannotact on Rn, since for x 2 Rn the image Mx may not belong to Rn (whilecertainly Mx 2 Cn).
Definition 1. Let M be an n⇥n matrix acting on the vector space V = Fn.A scalar � 2 F is an eigenvalue of M if for some nonzero vector v 2 V ,
v 6= 0 we have
(7) Mv = �v
The vector v is called eigenvector corresponding to the eigenvalue �.
Of course, if v is an eigenvector corresponding to �, then so is any scalarmultiple cv (for c 6= 0).
2.2. The characteristic equation. Equation (7) can be rewritten asMv��v = 0, or (M � �I)v = 0, which means that the nonzero vector v belongsto the null space of the matrix M � �I, and in particular this matrix is notinvertible. Using the theory of matrices, we know that this is equivalent to
det(M � �I) = 0
The determinant has the form
det(M � �I) =
���������
M11 � � M12 . . . M1n
M21 M22 � � . . . M2n...
......
Mn1 M
n2 . . . Mnn
� �
���������
This is a polynomial in �, having degree n. To understand why this isthe case, consider first n = 2 and n = 3.
For n = 2 the characteristic polynomial is����M11 � � M12
M21 M22 � �
���� = (M11 � �) (M22 � �)�M12M21
= �2� (M11 +M22)�+ (M11M22 �M12M21)
which is a quadratic polynomial in �; the dominant coe�cient is 1.For n = 3 the characteristic polynomial is
������
M11 � � M12 M13
M21 M22 � � M23
M13 M23 M33 � �
������
and expanding along say, row 1,
= (�1)1+1 (M11��)
����M22 � � M23
M23 M33 � �
����+(�1)1+2M12
����M21 M23
M13 M33 � �
����
+(�1)1+3M13
����M21 M22 � �M13 M23
����
6 RODICA D. COSTIN
= ��3 + (M11 +M22 +M33)�2 + . . .
which is a cubic polynomial in �; the dominant coe�cient is �1.It is easy to show by induction that det(M � �I) is polynomial in �,
having degree n, and that the coe�cient of �n is (�1)n.
Definition 2. The polynomial det(M � �I) is called the characteristicpolynomial of the matrix M , and the equation det(M � �I) = 0 is calledthe characteristic equation of M .
Remark. Some authors refer to the characteristic polynomial as det(�I �M); the two polynomial are either equal or a �1 multiple of each other,since det(�I �M) = (�1)n det(M � �I).
2.3. Geometric interpretation of eigenvalues and eigenvectors. LetM be an n ⇥ n matrix, and T : Rn
! Rn defined by T (x) = Mx be thecorresponding linear transformation.
If v is an eigenvector corresponding to an eigenvalue � of M : Mv = �v,then T expands or contracts v (and any vector in its direction) � times (andit does not change its direction!).
If the eigenvalue/vector are not real, a similar fact is true, only thatmultiplication by a complex (not real) scalar cannot be easily called anexpansion or a contraction (there is no ordering in complex numbers), seethe example of rotations, §2.13.1.
The special directions of the eigenvectors are called principal axes ofthe linear transformation (or of the matrix).
2.4. Digression: the fundamental theorem of algebra.
2.4.1. Polynomials of degree two: roots and factorization. Consider polyno-mials of degree two, with real coe�cients: p(x) = ax2 + bx + c. It is well
known that p(x) has real solutions x1,2 =�b±
p
b2 � 4ac
2aif b2 � 4ac � 0
(where x1 = x2 when b2 � 4ac = 0), and p(x) has no real solutions ifb2 � 4ac < 0.
When the solutions are real, then the polynomial factors as
ax2 + bx+ c = a(x� x1)(x� x2)
In particular, if x1 = x2 then p(x) = a(x� x1)2 and x1 is called a doubleroot; x1 is said to have multiplicity two. It is convenient to say that also inthis case p(x) has two roots.
If, on the other hand, if p(x) has no real roots, then p cannot be factoredwithin real numbers, and it is called irreducible (over the real numbers).
2.4.2. Complex numbers and factorization of polynomials of degree two. Ifp(x) = ax2+bx+c is irreducible this means that b2�4ac < 0 and we cannottake the square root of this quantity in order to calculate the two roots of
EIGENVALUES AND EIGENVECTORS 7
p(x). However, writing b2 � 4ac = (�1) (�b2 + 4ac) and introducing thesymbol i for
p
�1 we can write the zeroes of p(x) as
x1,2 =�b± i
p
�b2 + 4ac
2a=
�b
2a± i
p
�b2 + 4ac
2a2 R+ iR = C
Considering the two roots x1, x2 complex, we can still factor ax2+bx+c =a(x � x1)(x � x2), only now the factors have complex coe�cients. Withincomplex numbers every polynomial of degree two is reducible!
Note that the two roots of a quadratic polynomial with real coe�cientsare complex conjugate: if a, b, c 2 R and x1,2 62 R then x2 = x1.
2.4.3. The fundamental theorem of algebra. It is absolutely remarkable thatany polynomial can be completely factored using complex numbers:
Theorem 3. The fundamental theorem of algebraAny polynomial p(x) = a
n
xn+an�1x
n�1+ . . .+a0 with coe�cients aj
2 Ccan be factored
(8) an
xn + an�1x
n�1 + . . .+ a0 = an
(x� x1)(x� x2) . . . (x� xn
)
for a unique set of complex numbers x1, x2, . . . , xn (not necessarily distinct),called the roots of the polynomial p(x).
Remark. With probability one, the zeroes x1, . . . , xn of polynomials p(x)are distinct. Indeed, if x1 is a double root (or has higher multiplicity) thenboth relations p(x1) = 0 and p0(x1) = 0 must hold. This means that there isa relation between the coe�cients a0, . . . an of p(x) (the multiplet (a0, . . . an)belongs to an n dimensional surface in Cn+1).
2.4.4. Factorization within real numbers. If we want to restrict ourselvesonly within real numbers then we can factor any polynomial into factors ofdegree one or two:
Theorem 4. Factorization within real numbersAny polynomial of degree n with real coe�cients can be factored into fac-
tors of degree one or two with real coe�cients.
Theorem4 is an easy consequence of the (deep) Theorem3. Indeed, first,factor the polynomial in complex numbers (8). Then note that the zeroesx1, x2, . . . , xn come in pairs of complex conjugate numbers, since if z satisfiesp(z) = 0, then also its complex conjugate z satisfies p(z) = 0. Then eachpair of factors (x� z)(x� z) must be replaced in (8) by its expanded value:
(x� z)(x� z) = x2 � (z + z)x+ |z|2
which is an irreducible polynomial of degree 2, with real coe�cients. 2
8 RODICA D. COSTIN
2.5. Diagonal matrices. Let D be a diagonal matrix:
(9) D =
2
6664
d1 0 . . . 00 d2 . . . 0...
......
0 0 . . . dn
3
7775
To find its eigenvalues, calculate
det(D��I) =
���������
d1 � � 0 . . . 00 d2 � � . . . 0...
......
0 0 . . . dn
� �
���������
= (d1��1)(d2��2) . . . (dn��n
)
The eigenvalues are precisely the diagonal elements, and the eigenvectorcorresponding to d
j
is ej
(as it is easy to check). The principal axes ofdiagonal matrices the coordinate axes. Vectors in the direction of one ofthese axes preserve their direction and are stretched or compressed: if x =ce
k
then Dx = dk
x.Diagonal matrices are easy to work with: what was noted for the 2 ⇥ 2
matrices in §1 is true in general, and one can easily check that any powerDk is the diagonal matrix having dk
j
on the diagonal.If p(x) is a polynomial
p(t) = ak
tk + ak�1t
k�1 + . . .+ a1t+ a0
then for any square matrix M one can define p(M) as
(10) p(M) = ak
Mk + ak�1M
k�1 + . . .+ a1M + a0I
If D is a diagonal matrix (9) then p(D) is the diagonal matrix havingp(d
j
) on the diagonal. (Check!)Diagonal matrices can be viewed as the collection of their eigenvalues!
Exercise. Show that the eigenvalues of an upper (or lower) triangularmatrix are the elements on the diagonal.
2.6. Similar matrices have the same eigenvalues. It is very easy towork with diagonal matrices and a natural question arises: which lineartransformations have a diagonal matrix in a well chosen basis? This is themain topic we will be exploring for many sections to come.
Recall that if the matrix M represents the linear transformation L : V !
V in some basis B
V
of V , and the matrix M̃ represents the same lineartransformation L, only in a di↵erent basis B̃
V
, then the two matrices aresimilar: M̃ = S�1MS (where S the the matrix of change of basis).
Eigenvalues are associated to the linear transformation (rather than itsmatrix representation):
EIGENVALUES AND EIGENVECTORS 9
Proposition 5. Two similar matrices have the same eigenvalues: if M, M̃, Sare n⇥ n matrices, and M̃ = S�1MS then the eigenvalues of M and of M̃are the same.
This is very easy to check, since
det(M̃ � �I) = det(S�1MS � �I) = det⇥S�1(M � �I)S
⇤
= detS�1 det(M � �I) detS = det(M � �I)
so M and M̃ have the same characteristic equation. 2
2.7. Projections. Recall that projections do satisfy P 2 = P (we saw thisfor projections in dimension two, and we will prove it in general).
Proposition 6. Let P be a square matrix satisfying P 2 = P . Then theeigenvalues of P can only be 0 or 1.
Proof. Let � be an eigenvalue; this means that there is a nonzero vectorv so that Pv = �v. Applying P to both sides of the equality we obtainP 2v = P (�v) = �Pv = �2v. Using the fact that P 2v = Pv = �v it followsthat �v = �2v so (� � �2)v = 0 and since v 6= 0 then � � �2 = 0 so� 2 {0, 1}. 2
Example. Consider the projection of R3 onto the x1x2 plane. Its matrix
P =
2
41 0 00 1 00 0 0
3
5
is diagonal, with eigenvalues 1, 1, 0.
2.8. Trace, determinant and eigenvalues.
Definition 7. Let M be an n ⇥ n matrix, M = [Mij
]i,j=1,...,n. The trace
of M is the sum of its elements on the principal diagonal:
TrM =nX
j=1
Mjj
The following theorem shows that what we noticed in §2.2 for n = 2 istrue for any n:
Theorem 8. Let M be an n⇥n matrix, an let �1, . . . ,�n
be its n eigenvalues(complex, not necessarily distinct). Then
(11) detM = �1�2 . . .�n
and
(12) TrM = �1 + �2 + . . .+ �n
In particular, the traces of similar matrices are equal, and so are theirdeterminants.
10 RODICA D. COSTIN
Sketch of the proof.The coe�cients of any polynomial can be written in terms of its roots1,
and, in particular, it is not hard to see that
(13) p(x) ⌘ (x� �1)(x� �2) . . . (x� �n
)
= xn � (�1 + �2 + . . .+ �n
)xn�1 + . . .+ (�1)n(�1�2 . . .�n
)
In particular, p(0) = (�1)n�1�2 . . .�n
.The characteristic polynomial factors as
det(M � �I) = (�1)n(�� �1) . . . (�� �n
) ⌘ (�1)np(�)
(recall that the dominant coe�cient of the characteristic polynomial is (�1)n)and (11) follows.
To show (12) we expand the determinant det(M � �I) using minors andcofactors keeping track of the coe�cient of �n�1. As seen on the examplesin §2.2, only the first term in the expansion contains the power �n�1, andcontinuing to expand to lower and lower dimensional determinants, we seethat the only term containing �n�1 is
(M11 � �)(M22 � �) . . . (Mnn
� �)
= (�1)n�n
� (�1)n(M11 +M22 + . . .+Mnn
)�n�1 + lower powers of �
which compared to (13) gives (12). 2
2.9. The eigenvalue zero. As an immediate consequence of Theorem8,we can recognize invertible matrices by looking at their eigenvalues:
Corollary 9. A matrix M is invertible if and only if all its eigenvalues arenonzero.
Note that a matrix M has an eigenvalue equal to zero if and only if itsnull space N (M) is nontrivial. Moreover, the matrix M has dimN (M)eigenvectors linearly independent which correspond to the eigenvalue zero.
2.10. Eigenvectors corresponding to di↵erent eigenvalues are inde-pendent.
Theorem 10. Let M be an n⇥ n matrix.Let �1, . . .�
k
a set of distinct eigenvalues of M and v1, . . . ,vk
be corre-sponding eigenvectors.
Then the set v1, . . . ,vk
is linearly independent.In particular, if M has entries in F = R or C, and all the eigenvalues
of M are in F and are distinct, then the set of corresponding eigenvectorsform a basis for Fn.
1These are called Vieta’s formulas.
EIGENVALUES AND EIGENVECTORS 11
Proof.Assume, to obtain a contradiction, that the eigenvectors are linearly de-
pendent: there are c1, . . . , ck
2 F not all zero such that
(14) c1v1 + . . .+ ck
vk
= 0
Step I. We can assume that all cj
are not zero, otherwise we just removethose v
j
from (14) and we have a similar equation with a smaller k.If after this procedure we are left with k = 1, then this implies c1v1 = 0
which contradicts the fact that not all cj
are zero or the fact that v1 6= 0.Otherwise, for k � 2 we continue as follows.Step II. Then we can solve (14) for v
k
:
(15) vk
= c01v1 + . . .+ c0k�1vk�1
where c0j
= �cj
/ck
.Applying M to both sides of (15) we obtain
(16) �k
vk
= c01�1v1 + . . .+ c0k�1�k�1vk�1
Multiplying (15) by �k
and subtracting from (16) we obtain
(17) 0 = c01(�1 � �k
)v1 + . . .+ c0k�1(�k�1 � �
k
)vk�1
Note that all c0j
(�j
� �k
) are non-zero (since all c01 are non-zero, and�j
6= �k
).If k=2, then this implies v1 = 0 which is a contradiction.If k > 2 we go to Step I. with a lower k.The procedure decreases k, therefore it must end, and we have a contra-
diction. 2
2.11. Diagonalization of matrices with linearly independent eigen-vectors. Suppose that the M be an n⇥n matrix has n independent eigen-vectors v1, . . . ,vn
.Note that, by Theorem10, this is the case if we work in F = C and all
the eigenvalues are distinct (recall that this happens with probability one).Also this is the case if we work in F = R and all the eigenvalues are realand distinct.
Let S be the matrix with columns v1, . . . ,vn
:
S = [v1, . . . ,vn
]
which is invertible, since v1, . . . ,vn
are linearly independent.Since Mv
j
= �j
vj
then
(18) M [v1, . . . ,vn
] = [�1v1, . . . ,�n
vn
]
12 RODICA D. COSTIN
The left side of (18) equals MS. To identify the matrix on the right sideof (18) note that since Se
j
= vj
then S(�j
ej
) = �j
vj
so
[�1v1, . . . ,�n
vn
] = S[�1e1, . . . ,�n
en
] = S⇤, where ⇤ =
2
6664
�1 0 . . . 00 �2 . . . 0...
......
0 0 . . . �n
3
7775
Relation (18) is therefore
MS = S⇤, or S�1MS = ⇤ = diagonal
Note that the matrix S which diagonalizes a matrix is not unique. Forexample, we can replace any eigenvector by a scalar multiple of it. Also, wecan use di↵erent orders for the eigenvectors (this will result on a diagonalmatrix with the same values on the diagonal, but in di↵erent positions).
Example 1. Consider the matrix (6) for which we found the eigenvalues�1 = �1 and �2 = 2 and the corresponding eigenvectors v1 = (0, 1)T .v2 = (1,�1)T . Taking
S =
0 11 �1
�
we have
S�1MS =
�1 00 2
�
Not all matrices are diagonalizable, certainly those with distinct eigenval-ues are, and some matrices with multiple eigenvalues.
Example 2. The matrix
(19) N =
0 10 0
�
has eigenvalues �1 = �2 = 0 but only one (up to a scalar multiple) eigen-vector v1 = e1.
Multiple eigenvalues are not guaranteed to have an equal number of in-dependent eigenvectors!
N is not diagonalizable. Indeed, assume the contrary, to arrive at acontradiction. Suppose there exists an invertible matrix S so that S�1NS =⇤ where ⇤ is diagonal, hence it has the eigenvalues of N on its diagonal,and therefore it is the zero matrix: S�1NS = 0, which multiplied by S tothe left and S�1 to the right gives N = 0, which is a contradiction.
Some matrices with multiple eigenvalues may still be diagonalized; nextsection explores when this is the case.
EIGENVALUES AND EIGENVECTORS 13
2.12. Eigenspaces. Consider an n ⇥ n matrix M with entries in F , witheigenvalues �1, . . . ,�n
in F .
Definition 11. The set
V�j = {x 2 Fn
|Mx = �j
x}
is called the eigenspace of M associated to the eigenvalue �j
.
Exercise. Show that V�j is the null space of the transformation M � �I
and that V�j is a subspace of Fn.
Note that all the nonzero vectors in V�j are eigenvectors of M correspond-
ing to the eigenvalues �j
.
Definition 12. A subspace V is called an invariant subspace for M ifM(V ) ⇢ V (which means that if x 2 V then Mx 2 V ).
The following Remark gathers main features of eigenspaces; their proofis left to the reader.
Remark. 1. Each V�j is an invariant subspace for M .
2. V�j \ V
�l= {0} if �
j
6= �l
.3. Denote by �1, . . . ,�
k
the distinct eigenvalues of M and by rj
themultiplicity of the eigenvalue �
j
, for each j = 1, . . . , k; it is clear that
det(M � �I) =kY
j=1
(�j
� �)rj and r1 + . . .+ rk
= n
Then
dimV�j r
j
4. M is diagonalizable in Fn if and only if dimV�j = r
j
for all j = 1, . . . , kand then
V�1 � . . .� V
�k= Fn
Example. Consider the matrix
(20) M :=
2
642 0 0
1 0 �1
1 �2 1
3
75
Its characteristic polynomials is
det(M � �I) = ��3 + 3�2� 4 = � (�+ 1) (�� 2)2
so �1 = �1 and �2 = �3 = 2 is a double eigenvalue. The eigenspace V�1 is
one dimensional, spanned by an eigenvector, which, after a simple calcula-tion turns out to be v1 = (0, 1, 1)T . If the eigenspace V
�2 is two-dimensional(which is not guaranteed) then the matrix M is diagonalizable. A simple
14 RODICA D. COSTIN
calculation shows that there are two independent eigenvectors correspond-ing to the eigenvalue �2 = 2, for example v2 = (1, 0, 1)T and v3 = (2, 1, 0)T
(the null space of M � �2I is two-dimensional). Let
S = [v1,v2,v3] =
2
640 1 2
1 0 1
1 1 0
3
75
then
S�1MS =
2
64�1 0 0
0 2 0
0 0 2
3
75
2.13. Real matrices with complex eigenvalues; decomplexification.
2.13.1. Complex eigenvalues of real matrices. For an n⇥n matrix with realentries, if we want to have n guaranteed eigenvalues, then we have to acceptworking in Cn. Otherwise, if we want to restrict ourselves to working onlywith real vectors, then we have to accept that we may have fewer (real)eigenvalues, or perhaps none.
Complex eigenvalues of real matrices come in pairs: if � is an eigenvalue ofM , then so is its complex conjugate � (since the characteristic equation hasreal coe�cients). Also, if v is an eigenvector corresponding to the eigenvalue�, then v is eigenvector corresponding to the eigenvalue � (check!). The realand imaginary parts of v span a plane where the linear transformation actsby rotation, and a possible dilation. Simple examples are shown below.
Example 1: rotation in the xy-plane. Consider a rotation matrix
(21) R✓
=
cos ✓ � sin ✓sin ✓ cos ✓
�
To find its eigenvalues calculate
det(R✓
��I) =
����cos ✓ � � � sin ✓sin ✓ cos ✓ � �
���� = (cos ✓��)2+sin2 ✓ = �2�2 cos ✓+1
hence the solutions of the characteristic equations det(R✓
� �I) = 0 are�1,2 = cos ✓ ± i sin ✓ = e±i✓. It is easy to see that v1 = (i, 1)T is theeigenvector corresponding to �1 = ei✓ and v2 = (�i, 1)T is the eigenvectorcorresponding to �2 = e�i✓.
Example 2: complex eigenvalues in R3. Consider the matrix
M =
2
641� 1
2
p
3 �
52
p
3 012
p
3 1 + 12
p
3 0
0 0 �4
3
75
Its characteristic polynomial is
EIGENVALUES AND EIGENVECTORS 15
det(M � �I) = ��3� 2�2 + 4�� 16 = � (�+ 4)
��2
� 2�+ 4�
and its eigenvalues are: �1,2 = 1 ± ip
3 = 2e±i⇡/3 and �3 = �4, and cor-responding eigenvectors v1,2 = (�1 ± 2i, 1, 0)T , and v3 = e3. The matrixS = [v1,v2,v3] diagonalized the matrix: S�1MS is the diagonal matrix,having the eigenvalues on the diagonal, but all these are complex matrices.
To understand how the matrix acts on R3, we consider the real andimaginary parts of v1: let x1 = <v1 = 1
2(v1 + v2) = (�1, 1, 0)T andy1 = =v1 = 1
2i(v1 � v2) = (2, 0, 0)T . Since the eigenspaces are invari-ant under M , then so is Sp(x1,y1), over the complex and even over the realnumbers (since M has real elements). The span over the real numbers is thexy-plane, and it is invariant under M . The figure shows the image of theunit circle in the xy-plane under the matrix M : it is an ellipse.
Figure 1. The image of the unit circle in the xy-plane.
Along the direction of the third eigenvector (the z-axis) the matrix mul-tiples any c e3 by �4.
In the basis x1,y1,v3 the matrix of the linear transformation has itssimplest form: using SR = [x1,y1,v3] we obtain the matrix of the transfor-mation in this new basis as
S�1R MSR =
2
641
p
3 0
�
p
3 1 0
0 0 �4
3
75
and the upper 2⇥ 2 block represents the rotation and dilation 2R�⇡/3.
2.13.2. Decomplexification. Suppose the n⇥ n matrix M has real elements,eigenvalues �1, . . . ,�n
and n independent eigenvectors v1, . . . ,vn
. Then Mis diagonalizable: if S = [v1, . . . ,vn
] then S�1MS = ⇤ where ⇤ is a diagonalmatrix with �1, . . . ,�n
on its diagonal.
16 RODICA D. COSTIN
Suppose that some eigenvalues are not real. Then the matrices S and ⇤are not real either, and the diagonalization of M must be done in Cn.
Suppose that we want to work in Rn only. Recall that the nonreal eigen-values and eigenvectors of real matrices come in pairs of complex-conjugateones. In the complex diagonal form ⇤ one can replace diagonal 2⇥ 2 blocks
�j
00 �
j
�
by a 2⇥ 2 matrix which is not diagonal, but has real entries.To see how this is done, suppose �1 2 C \ R and �2 = �1, v2 = v1.
Splitting into real and imaginary parts, write �1 = ↵1+i�1 and v1 = x1+iy1.Then from M(x1 + iy1) = (↵1 + i�1)(x1 + iy1) identifying the real andimaginary parts, we obtain
Mx1 + iMy1 = (↵1x� �1y) + i(�1x+ ↵1y)
In the matrix S = [v1,v2, . . . ,vn
] composed of independent eigenvectorsof M , replace the first two columns v1,v2 = v1 by x1,y1 (which are vectorsin Rn): using the matrix S̃ = [x1,y1,v3, . . . ,vn
] instead of S we have MS̃ =S̃⇤̃ where
⇤̃ =
2
666664
↵1 �1 0 . . . 0��1 ↵1 0 . . . 00 0 �3 . . . 0...
......
...0 0 0 . . . �
m
3
777775
We can similarly replace any pair of complex conjugate eigenvalues with2⇥ 2 real blocks.
Exercise. Show that each 2⇥ 2 real block obtained through decomplex-ification has the form
↵ ��� ↵
�= ⇢R
✓
for a suitable ⇢ > 0 and R✓
rotation matrix (21).
2.14. Jordan normal form. We noted in §2.12 that a matrix is similar toa diagonal matrix if and only if the dimension of each eigenspace V
�j equalsthe order of multiplicity of the eigenvalue �
j
. Otherwise, there are fewerthan n independent eigenvectors; such a matrix is called defective.
2.14.1. Jordan blocks. Defective matrices can not be diagonalized, but wewill see that they are similar to block diagonal matrices, called Jordan nor-mal forms; these are upper triangular, have the eigenvalues on the diagonal,1 in selected placed above the diagonal, and zero in the rest. After that, insection §2.14.3 it is shown how to construct the transition matrix S, whichconjugates a defective matrix to its Jordan form; its columns are made ofgeneralized eigenvectors.
EIGENVALUES AND EIGENVECTORS 17
The Jordan blocks which appear on the diagonal of a Jordan normal formare as follows.
1⇥ 1 Jordan blocks are just [�].2⇥ 2 Jordan blocks have the form
(22) J2(�) =
� 10 �
�
For example, the matrix (19) is a Jordan block J2(0).3⇥ 3 Jordan blocks have the form
(23) J3(�) =
2
664
� 1 0
0 � 1
0 0 �
3
775
In general, a k ⇥ k Jordan block, Jk
(�), is a matrix having the samenumber, �, on the diagonal, 1 above the diagonal and 0 everywhere else.
Note that Jordan blocks Jk
(�) have the eigenvalue � with multiplicity k,and the dimension of the eigenspace is one.
Example of a matrix in Jordan normal form:2
66666666664
3 | 0 0 0 0� �
0 | 2 | 0 0 0� � �
0 0 | 2 1 0
0 0 | 0 2 1
0 0 | 0 0 2
3
77777777775
which is block-diagonal, having two 1⇥1 Jordan blocks and one 3⇥3 Jordanblock along its diagonal. The eigenvalue 3 is simple, while 2 has multiplicityfour. The eigenspace corresponding to 2 is two-dimensional (e2 and e3 areeigenvectors).
Note how Jordan blocks act on the vectors of the basis. For (22): J2(�)e1 =�e1, so e1 is an eigenvector. Also
(24) J2(�)e2 = e1 + �e2
which implies that (J2(�)� �I)2e2 = (J2(�)� �I)e1 = 0.Similarly, for (30): J3(�)e1 = �e1 so e1 is an eigenvector. Then
(25) J3(�)e2 = e1 + �e2
implying that (J3(�)� �I)2e2 = (J3(�)� �I)e1 = 0. Finally,
(26) J3(�)e3 = e2 + �e3
18 RODICA D. COSTIN
implying that (J3(�) � �I)3e3 = (J3(�) � �I)2e2 = 0. This illuminatesthe idea behind the notion of generalized eigenvectors defined in the nextsection.
2.14.2. The generalized eigenspace. Defective matrices are similar to a ma-trix which is block-diagonal, having Jordan blocks on its diagonal. An ap-propriate basis is formed using generalized eigenvectors:
Definition 13. A generalized eigenvector of M corresponding to theeigenvalue � is a vector x 6= 0 so that
(27) (M � �I)kx = 0
for some positive integer k.
Examples.1) Eigenvectors are generalized eigenvectors (take k = 1 in (27)).2) Vectors in the standard basis are generalized eigenvectors for Jordan
blocks.
Definition 14. The generalized eigenspace of M corresponding to theeigenvalue � is the subspace
E�
= {x | (M � �I)kx = 0 for some k 2 Z+}
Sometimes we want to refer to only at the distinct eigenvalues of a matrix,this set is called ”the spectrum”:
Definition 15. The spectrum �(M) of a matrix M is the set of its eigen-values.
Theorem 16. For any n⇥ n matrix M the following hold:(i) V
�
⇢ E�
;(ii) E
�
is a subspace;(iii) E
�
is an invariant subspace under M ;(iv) E
�1 \ E�2 = 0 for �1 6= �2.
(v) dimE�
=the multiplicity of �.(vi)The set of eigenvectors and generalized eigenvectors of M span the
whole space Cn:
�
�2�(M) E�
= Cn
The proofs of (i)-(iv) are simple exercises.The proofs of (v), (vi) are not included here.
EIGENVALUES AND EIGENVECTORS 19
2.14.3. How to find a basis for each E�
that can be used to conjugate amatrix to a Jordan normal form.
Example 1. The matrix
(28) M =
"1 + a �1
1 a� 1
#
is defective: it has eigenvalues a, a and only one independent eigenvector,(1, 1)T . It is therefore similar to J2(a). To find a basis x1,x2 in which thematrix takes this form, let x1 = (1, 1)T (the eigenvector); to find x2 wesolve (M � aI)x2 = x1 (as seen in (24) and in (25)). The solutions arex2 2 (1, 0)T +N (M � aI), and any vector in this space works, for examplex2 = (1, 0)T . For
(29) S = [x1,x2] =
"1 1
1 0
#
we have S�1MS = J2(a).
Example 2.The matrix
M =
2
641 �2 3
1 2 �1
0 �1 3
3
75
has eigenvalues 2, 2, 2 and only one independent eigenvector v1 = (1, 1, 1)T .Let x1 = v1 = (1, 1, 1)T . Solving (M � 2I)x2 = x1 we obtain x2 =
(1,�1, 0)T (plus any vector in N (M �2I) = V�1). Next solve (M �2I)x3 =
x2 which gives x3 = (0, 1, 1)T (plus any vector in the null space of M � 2I).For S = [x1,x2,x3] we have
(30) S�1MS =
2
664
2 1 0
0 2 1
0 0 2
3
775
In general, if � is an eigenvalue of M for which dimV�
is less than themultiplicity of �, we do the following. Choose a basis for V
�
. For eacheigenvector v in this basis set x1 = v and solve recursively
(31) (M � �I)xk+1 = x
k
, k = 1, 2, . . .
Note that each x1 satisfies (27) for k = 1, x2 satisfies (27) for k = 2, etc.At some step k1 the system (M � �I)x
k1+1 = xk1 will have no solution;
we found the generalized eigenvectors x1, . . . ,xk1 (which will give a k1 ⇥ k1
Jordan block). We then repeat the procedure for a di↵erent eigenvector inthe chosen basis for V
�
, and obtain a new set of generalized eigenvectors,corresponding to a new Jordan block.
Note: Jordan form is not unique.
20 RODICA D. COSTIN
2.14.4. Real Jordan normal form. If a real matrix has multiple complexeigenvalues and is defective, then its Jordan form can be replaced with anupper block diagonal matrix in a way similar to the diagonal case illus-trated in §2.13.2, by replacing the generalized eigenvectors with their realand imaginary parts.
For example, a real matrix which can be brought to the complex Jordannormal form 2
664
↵+ i� 1 0 00 ↵+ i� 0 00 0 ↵� i� 10 0 0 ↵� i�
3
775
can be conjugated (by a real matrix) to the real matrix2
664
↵ � 1 0�� ↵ 0 10 0 ↵ �0 0 �� ↵
3
775
2.15. Block matrices.
2.15.1. Multiplication of block matrices. It is sometimes convenient to workwith matrices split in blocks. We have already used this when we wrote
M [v1, . . . ,vn
] = [Mv1, . . . ,Mvn
]
More generally, if we have two matrices M, P with dimensions that allowfor multiplication (i.e. the number of columns of M equals the number ofrows of P ) and they are split into blocks:
M =
2
4M11 | M12
��� � ���
M21 | M22
3
5 , P =
2
4P11 | P12
��� � ���
P21 | P22
3
5
then
MP =
2
4M11P11 +M12P21 | M11P12 +M12P22
������ � ������
M21P11 +M22P21 | M21P12 +M22P22
3
5
if the number of columns of M11 equals the number of rows of P11.Exercise. Prove that the block multiplication formula is correct.
More generally, one may split the matrices M and P into many blocks, sothat the number of block-columns of M equal the number of block-rows ofP and so that all products M
jk
Pkl
make sense. Then MP can be calculatedusing blocks by a formula similar to that using matrix elements.
In particular, if M,P are block diagonal matrices, having the blocks Mjj
,Pjj
on the diagonal, then MP is a block diagonal matrix, having the blocksM
jj
Pjj
along the diagonal.
EIGENVALUES AND EIGENVECTORS 21
For example, if M is a matrix in Jordan normal form, then it is blockdiagonal, with Jordan blocks M
jj
along the diagonal. Then the matrixM2 is block diagonal, having M2
jj
along the diagonal, and all powers Mk
are block diagonal, having Mk
jj
along the diagonal. Furthermore, any linear
combination of these powers of M , say c1M+c2M2 is block diagonal, having
the corresponding c1Mjj
+ c2M2jj
along the diagonal.
2.15.2. Determinant of block matrices.
Proposition 17. Let M be a square matrix, having a triangular block form:
M =
A B0 D
�or M =
A 0C D
�
where A and D are square matrices, say A is k ⇥ k and D is l ⇥ l.Then detM = detA detD.Moreover, if a1, . . . , a
k
are the eigenvalues of A, and d1, . . . , dl
are theeigenvalues of D, then the eigenvalues of M are a1, . . . , a
k
, d1, . . . , dl
.
The proof is left to the reader as an exercise.2
For a more general 2⇥ 2 block matrix, with D invertible3
M =
A BC D
�
the identity
A BC D
� I 0
�D�1C I
�=
A�BD�1C B
0 D
�
together with Proposition 17 implies that
det
A BC D
�= det(A�BD�1C) detD
which, of course, equals det(AD �BD�1CD).For larger number of blocks, there are more complicated formulas.
2Hint: bring A, D to Jordan normal form, then M to an upper triangular form.3References: J.R. Silvester, Determinants of block matrices, Math. Gaz., 84(501)
(2000), pp. 460-467, and P.D. Powell, Calculating Determinants of Block Matrices,http://arxiv.org/pdf/1112.4379v1.pdf
22 RODICA D. COSTIN
3. Solutions of linear differential equations with constant
coefficients
In §1.2 we saw an example which motivated the notions of eigenvaluesand eigenvectors. General linear first order systems of di↵erential equationswith constant coe�cients can be solved in a quite similar way. Consider
(32)du
dt= Mu
where M is an m⇥m constant matrix and u in an m-dimensional vector.As in §1.2, it is easy to check that u(t) = e�tv is a solution of (32) if �
is an eigenvalue of M , and v is a corresponding eigenvector. The goal isto find the solution to any initial value problem: find the solution of (32)satisfying
(33) u(0) = u0
for any given vector u0.
3.1. The case when M is diagonalizable. Assume that M has m inde-pendent eigenvectors v1, . . . ,vm
, corresponding to the eigenvalues �1, . . . ,�m
.Then (32) has the solutions u
j
(t) = e�jtvj
for each j = 1, . . . ,m.These solutions are linearly independent. Indeed, assume that for some
constants c1, . . . , cm we have c1u1(t) + . . . + cm
um
(t) = 0 for all t. Then,in particular, for t = 0 it follows that c1v1 + . . .+ c
m
vm
= 0 which impliesthat all c
j
are zero (since v1, . . . ,vm
were assumed independent).
3.1.1. Fundamental matrix solution. Since equation (32) is linear, then anylinear combination of solutions is again a solution:
(34) u(t) = a1u1(t) + . . .+ am
um
(t)
= a1e�1tv1 + . . .+ a
m
e�mtvm
, aj
arbitrary constants
The matrix
(35) U(t) = [u1(t), . . . ,um
(t)]
is called a fundamental matrix solution. Formula (34) can be writtenmore compactly as
(36) u(t) = U(t)a, where a = (a1, . . . , am)T
The initial condition (33) determines the constants a, since (33) impliesU(0)a = u0. Noting that U(0) = [v1, . . . ,vm
] = S therefore a = S�1u0 andthe initial value problem (32), (33) has the solution
u(t) = U(t)S�1u0
General results in the theory of di↵erential equations (on existence anduniqueness of solutions to initial value problems) show that this is onlyone solution.
In conclusion:
EIGENVALUES AND EIGENVECTORS 23
Proposition 18. If the m⇥m constant matrix M has has m independenteigenvectors v1, . . . ,vm
, corresponding to the eigenvalues �1, . . . ,�m
, thenequation (32) has m linearly independent solutions u
j
(t) = e�jtvj
, j =1, . . . ,m and any solution of (32) is a linear combination of them.
Example. Solve the initial value problem
(37)dx
dt
= x� 2y, x(0) = ↵dy
dt
= �2x+ y, y(0) = �
Denoting u = (x, y)T , problem (37) is
(38)du
dt= Mu, where M =
"1 �2
�2 1
#, with u(0) =
↵�
�
Calculating the eigenvalues of M , we obtain �1 = �1, �2 = 3, and corre-sponding eigenvectors v1 = (1, 1)T , v2 = (�1, 1)T . There are two indepen-dent solutions of the di↵erential system:
u1(t) = e�t
11
�, u2(t) = e3t
�11
�
and a fundamental matrix solution is
(39) U(t) = [u1(t),u2(t)] =
"e�t
�e3t
e�t e3t
#
The general solution is a linear combination of the two independent solutions
u(t) = a1e�t
11
�+ a2e
3t
�11
�= U(t)
a1a2
�
This solution satisfies the initial condition if
a1
11
�+ a2
�11
�=
↵�
�
which is solved for a1, a2: from"1 �1
1 1
# a1a2
�=
↵�
�
it follows thata1a2
�=
"1 �1
1 1
#�1 ↵�
�=
"12
12
�
12
12
# ↵�
�=
↵+�
2�↵+�
2
�
therefore
(40) u(t) = ↵+�
2 e�t
11
�+ �↵+�
2 e3t�11
�
sox(t) = ↵+�
2 e�t
�
�↵+�
2 e3t
y(t) = ↵+�
2 e�t + �↵+�
2 e3t
24 RODICA D. COSTIN
3.1.2. The matrix eMt. It is often preferable to work with a matrix of in-dependent solutions U(t) rather than with a set of independent solutions.Note that the m⇥m matrix U(t) satisfies
(41)d
dtU(t) = M U(t)
In dimension one this equation reads du
dt
= �u having its general solutionu(t) = Ce�t. Let us check this fact based on the fact that the exponentialis the sum of its Taylor series:
ex = 1 +1
1!x+
1
2!x2 + . . .+
1
n!xn + . . . =
1X
n=0
1
n!xn
where the series converges for all x 2 C. Then
e�t = 1 +1
1!�t+
1
2!�2t2 + . . .+
1
n!�nxn + . . . =
1X
n=0
1
n!�ntn
and the series can be di↵erentiated term-by-term, giving
d
dte�t =
d
dt
1X
n=0
1
n!�ntn =
1X
n=0
1
n!�n
d
dttn =
1X
n=1
1
(n� 1)!�ntn�1 = �e�t
Perhaps one can define, similarly, the exponential of a matrix and obtainsolutions to (41)?
For any square matrix M , one can define polynomials, as in (10), and itis natural to define
(42) etM = 1 +1
1!tM +
1
2!t2M2 + . . .+
1
n!tnMn + . . . =
1X
n=0
1
n!tnMn
provided that the series converges. If, furthermore, the series can di↵erenti-ated term by term, then this matrix is a solution of (41) since(43)
d
dtetM =
d
dt
1X
n=0
1
n!tnMn =
1X
n=0
1
n!
d
dttnMn =
1X
n=1
n
n!tn�1Mn = MetM
Convergence and term-by-term di↵erentiation can be justified by diago-nalizing M .
Let v1, . . . ,vm
be independent eigenvectors corresponding to the eigen-values �1, . . . ,�m
of M , let S = [v1, . . . ,vm
]. Then M = S⇤S�1 with ⇤ thediagonal matrix with entries �1, . . . ,�m
.Note that
M2 =�S⇤S�1
�2= S⇤S�1S⇤S�1 = S⇤2S�1
thenM3 = M2M =
�S⇤2S�1
� �S⇤S�1
�= S⇤3S�1
and so on; for any powerMn = S⇤nS�1
EIGENVALUES AND EIGENVECTORS 25
Then the series (42) is
(44) etM =1X
n=0
1
n!tnMn =
1X
n=0
1
n!tnS⇤nS�1 = S
1X
n=0
1
n!tn⇤n
!S�1
For
(45) ⇤ =
2
6664
�1 0 . . . 00 �2 . . . 0...
......
0 0 . . . �m
3
7775
it is easy to see that
(46) ⇤n =
2
6664
�n
1 0 . . . 00 �n
2 . . . 0...
......
0 0 . . . �n
m
3
7775for n = 1, 2, 3 . . .
therefore
(47)1X
n=1
1
n!tn⇤n =
2
6664
P1n=1
1n! t
n�n
1 0 . . . 00
P1n=1
1n! t
n�n
2 . . . 0...
......
0 0 . . .P1
n=11n! t
n�n
m
3
7775
(48) =
2
6664
et�1 0 . . . 00 et�2 . . . 0...
......
0 0 . . . et�m
3
7775= et⇤
and (44) becomes
(49) etM = Set⇤S�1
which shows that the series defining the matrix etM converges and can bedi↵erentiated term-by-term (since these are true for each of the series in(47)). Therefore etM is a solution of the di↵erential equation (41).
Multiplying by an arbitrary constant vector b we obtain vector solutionsof (32) as
(50) u(t) = etMb, with b an arbitrary constant vector
Noting that u(0) = b it follows that the solution of the initial value problem(32), (33) is
u(t) = etMu0
Note: the fundamental matrix U(t) in (35) is linked to the fundamentalmatrix etM by
(51) U(t) = Set⇤ = etMS
26 RODICA D. COSTIN
Example. For the example (38) we have
S =
"1 �1
1 1
#, ⇤ =
"�1 0
0 3
#, et⇤ =
"e�t 0
0 e3t
#
and
u1(t) = e�t
11
�, u2(t) = e3t
�11
�
The fundamental matrix U(t) is given by (39).Using (49)
etM = Set⇤S�1 =
"12 e
�t + 12 e
3 t 12 e
�t
�
12 e
3 t
12 e
�t
�
12 e
3 t 12 e
�t + 12 e
3 t
#
and the solution to the initial value problem is
etM↵�
�=
" �12 e
�t + 12 e
3 t�↵+
�12 e
�t
�
12 e
3 t��
�12 e
�t
�
12 e
3 t�↵+
�12 e
�t + 12 e
3 t��
#
which, of course, is the same as (40).
3.2. Non-diagonalizable matrix. The exponential etM is defined simi-larly, only a Jordan normal form must be used instead of a diagonal form:writing S�1MS = J where S is a matrix formed of generalized eigenvectorsof M , and J is a Jordan normal form, then
(52) etM = SetJS�1
It only remains to check that the series defining the exponential of a Jordanform converges, and that it can be di↵erentiated term by term.
Also to be determined are m linearly independent solutions, since if M isnot diagonalizable, then there are fewer than m independent eigenvectors,hence fewer than m independent solutions of pure exponential type. Thiscan be done using the analogue of (51), namely by considering the matrix
(53) U(t) = SetJ = etMS
The columns of the matrix (53) are linearly independent solutions, and wewill see that among them are the purely exponential ones multipling theeigenvectors of M .
Since J is block diagonal (with Jordan blocks along its diagonal), then itsexponential will be block diagonal as well, with exponentials of each Jordanblock (see §2.15.1 for multiplication of block matrices).
3.2.1. Example: 2⇥ 2 blocks: for
(54) J =
"� 1
0 �
#
EIGENVALUES AND EIGENVECTORS 27
direct calculations give
(55) J2 =
"�2 2�
0 �2
#, J3 =
"�3 3�2
0 �3
#, . . . , Jk =
"�k k �k�1
0 �k
#, . . .
and then
(56) etJ =1X
k=0
1
k!tkJk =
"et� tet�
0 et�
#
For the equation (32) with the matrix M is similar to a 2 ⇥ 2 Jordanblock: S�1MS = J with J as in (54), and S = [x1,x2] a fundamentalmatrix solution is U(t) = SetJ = [et�x1, e
Denoting u = (x, y)T , the di↵erential system (59) is dudt
= Mu with M givenby (28), matrix for which we found that it has a double eigenvalue a andonly one independent eigenvector x1 = (1, 1)T .
Solution 1. For this matrix we already found an independent generalizedeigenvector x2 = (1, 0)T , so we can use formula (58) to write down thegeneral solution of (59).
Solution 2. We know one independent solution to the di↵erential system,namely u1(t) = eatx1. We look for a second independent solution as thesame exponential multiplying a polynomial in t, of degree 1: substitutingu(t) = eat(tb + c) in du
dt
= Mu we obtain that a(tb + c) + b = M(tb + c)holds for all t, therefore Mb = ab and (M � aI)c = b which means that bis an eigenvector of M (or b = 0), and c is a generalized eigenvector. Wehave re-obtained the formula (57).
By either method it is found that a fundamental matrix solution is
U(t) = [u1(t),u2(t)] = eat1 t+ 11 t
�
and the general solution has the form u(t) = U(t)c for an arbitrary constantvector c. We now determine c so that u(0) = (↵,�)T , so we solve
Higher powers can be calculated by induction; it is clear that
(61) Jk =
2
664
�k k�k�1 k(k�1)2 �k�2
0 �k k�k�1
0 0 �k
3
775
Then
(62) etJ =1X
k=0
1
k!tkJk =
2
664
et� tet� 12 t
2et�
0 et� tet�
0 0 et�
3
775
For M = SJS�1 with J as in (60) and S = [x1,x2,x3], a fundamentalmatrix solution for (32) is
SetJ = [x1e�t, (tx1 + x2)e
�t, (1
2t2x1 + tx2 + x3)e
�t]
EIGENVALUES AND EIGENVECTORS 29
3.2.3. In general, if an eigenvalue � has multiplicity r, but there are onlyk < r independent eigenvectors v1, . . . ,v
k
then, besides the k independentsolutions e�tv1, . . . , e
�tvk
there are other r� k independent solutions in theform e�tp(t) with p(t) polynomials in t of degree at most r� k, with vectorcoe�cients (which turn out to be generalized eigenvectors of M).
Then the solution of the initial value problem (32), (33) is
u(t) = etMu0
Combined with the results of uniqueness of the solution of the initial valueproblem (known from the general theory of ordinary di↵erential equations)it follows that:
Theorem 19. Any linear di↵erential equation u0 = Mu where M is an m⇥
m constant matrix, and u is an m-dimensional vector valued function hasm linearly independent solutions, and any solution is a linear combinationof these. In other words, the solutions of the equation form a linear spaceof dimension m.
3.3. Fundamental facts on linear di↵erential systems.
Theorem 20. Let M be an n⇥ n matrix (diagonalizable or not).(i) The matrix di↵erential problem
(63)d
dtU(t) = M U(t), U(0) = U0
has a unique solution, namely U(t) = eMtU0.(ii) Let W (t) = detU(t). Then
(64) W 0(t) = TrM W (t)
therefore
(65) W (t) = W (0) etTrM
(iii) If U0 is an invertible matrix, then the matrix U(t) is invertible forall t, called a fundamental matrix solution; the columns of U(t) forman independent set of solutions of the system
(66)du
dt= Mu
(iv) Let u1(t), . . . ,un
(t) be solutions of the system (66). If the vectorsu1(t), . . . ,un
(t) are linearly independent at some t then they are linearlyindependent at any t.
Proof.(i) Clearly U(t) = eMtU0 is a solution, and it is unique by the general
theory of di↵erential equations: (63) is a linear system of n2 di↵erentialequation in n2 unknowns.
(ii) Using (52) it follows that
W (t) = detU(t) = det(SetJS�1U0) = det etJ detU0 = etPn
j=1 �j detU0
30 RODICA D. COSTIN
= etTrM detU0 = etTrMW (0)
which is (65), implying (64).(iii), (iv) are immediate consequences of (65). 2
3.4. Eigenvalues and eigenvectors of the exponential of a matrix.It is not hard to show that
(eM )�1 = e�M , (eM )k = ekM , eM+cI = eceM
More generally, it can be shown that if MN = NM , then eMeN = eM+N .Warning: if the matrices do not commute, this may not be true!
Recall that if M is diagonalizable, in other words if M = S⇤S�1 where⇤ = diag(�1, . . . ,�n
) is a diagonal matrix, then eM = Se⇤S�1 where e⇤ =diag(e�1 , . . . , e
�
n
). If follows that the eigenvalues of eM are e�1 , . . . , e�
n
and thecolumns of S are eigenvectors of M , and also of eM .
If M is not diagonalizable, let J be its Jordan normal form. Recall thatif M = SJS�1 then eM = SeJS�1 where eJ is an upper triangular matrix,with diagonal elements still being exponentials of the eigenvalues of M . Thematrix eJ is not a Jordan normal form; however, generalized eigenvectors ofM are also of eM .
Exercise.1. Show that if Mx = 0 then eMx = x.2. Show that if v is an eigenvector of M corresponding to the eigenvalues
�, then v is also an eigenvector of eM corresponding to the eigenvalues e�.3. Show that if Mv = �v then eMv = e�[v + (M � �I)v].
Note that if (M � �I)2x = 0 then (eM � e�)2x = 0. Indeed, (eM �
In general, if x is a generalized eigenvector of M corresponding to theeigenvalues �, then x is also a generalized eigenvector of eM correspondingto the eigenvalues e�.
EIGENVALUES AND EIGENVECTORS 31
3.5. Higher order linear di↵erential equations; companion matrix.Consider scalar linear di↵erential equations, with constant coe�cients, oforder n:
(67) y(n) + an�1y
(n�1) + . . .+ a1y0 + a0y = 0
where y(t) is a scalar function and a1, . . . , an�1 are constants.Such equations can be transformed into systems of first order equations:
the substitution
(68) u1 = y, u2 = y0, . . . , un
= y(n�1)
transforms (67) into the system
(69) u0 = Mu, where M =
2
666664
0 1 0 . . . 00 0 1 . . . 0...
...0 0 0 . . . 1
�a0 �a1 �a2 . . . �an�1
3
777775
The matrix M is called the companion matrix to the di↵erential equation(67).
To find its eigenvalues an easy method is to search for � so that the linearsystem Mx = �x has a solution x 6= 0:
x2 = �x1, x3 = �x2, . . . , xn
= �xn�1, �a0x1 � a1x2 � . . .� a
n�1xn = �xn
which implies that
(70) �n + an�1�
n�1 + . . .+ a1�+ a0 = 0
which is the characteristic equation of M .Note that the characteristic equation (70) can also be obtained by search-
ing for solutions of (67) which are purely exponential: y(t) = e�t.
3.5.1. Linearly independent sets of functions. We are familiar with the no-tion of linear dependence or independence of functions belonging to a givenlinear space. In practice, functions arise from particular problems, or classesof problems, for example as solutions of equations and only a posteriori wefind a linear space to accommodates them. A natural definition of lineardependence or independence which can be used in most usual linear spaceof functions is:
Definition 21. A set of function f1, . . . , fn are called linearly dependenton an interval I if there are constants c1, . . . , cn, not all zero, so that
(71) c1f1(t) + . . .+ cn
fn
(t) = 0 for all t 2 I
A set of functions which are not linearly dependent on I are called linearlyindependent on I. This means that if, for some constants c1, . . . , cn relation(71) holds, then necessarily all c1, . . . , cn are zero.
If all functions f1, . . . , fn are enough many times di↵erentiable then thereis a simple way to check linear dependence or independence:
32 RODICA D. COSTIN
Theorem 22. Assume functions f1, . . . , fn are n� 1 times di↵erentiable onthe interval I. Consider their Wronskian
W [f1, . . . , fn](t) =
���������
f1(t) . . . fn
(t)f 01(t) . . . f 0
n
(t)...
...
f(n�1)1 (t) . . . f
(n�1)n
(t)
���������
(i) If the functions f1, . . . , fn are linearly dependent then
W [f1, . . . , fn](t) = 0 for all t 2 I
(ii) If there is some point t0 2 I so that W [f1, . . . , fn](t0) 6= 0 then thefunctions are linearly independent on I.
Indeed, to show (i), assume that (71) holds for some constants c1, . . . , cn,not all zero; then by di↵erentiation, we see that the columns of W (t) arelinearly dependent for each t, hence W (t) = 0 for all t.
Part (ii) is just the negation of (i). 2Example 1. To check if the functions 1, t2, et are linearly dependent we
calculate their Wronskian
W [1, t2, et] =
������
1 t2 et
0 2t et
0 2 et
������= 2 et (t� 1) is not identically 0
so they are linearly independent (even if the Wronskian happens to be zerofor t = 1).
Example 2. If the numbers �1, . . . ,�n
are all distinct then the functionset�1 , . . . , et�n are linearly independent.
Indeed, their Wronskian equals the product et�1 . . . et�n multiplied by aVandermonde determinant which equals
Qi<j
(�j
� �i
) which is never zeroif �1, . . . ,�n
are all distinct, or identically zero if two of �s are equal.
I what follows we will see that if the functions f1, . . . , fn happen to besolutions of the same linear di↵erential equation, then their Wronskian iseither identically zero, or never zero.
3.5.2. Linearly independent solutions of nth order linear di↵erential equa-tions. Using the results obtained for first order linear systems, and lookingjust at the first component u1(t) of the vector u(t) (since y(t) = u1(t)) wefind:
(i) if the characteristic equation (70) has n distinct solutions �1, . . . ,�n
then the general solution is a linear combination of purely exponential solu-tions
y(t) = a1e�1t + . . .+ a
n
e�nt
(ii) if �j
is a repeated eigenvalue of multiplicity rj
then there are rj
independent solutions of the type e�jtq(t) where q(t) are polynomials in t ofdegree at most r
j
, therefore they can be taken to be e�jt, te�jt, . . . , trj�1e�jt.
EIGENVALUES AND EIGENVECTORS 33
Example. Solve the di↵erential equation
y000 � 3y00 + 4y = 0
The characteristic equation, obtained by substituting y(t) = e�t, is �3�3�2+
4 = 0 which factored is (�� 2)2(�+ 1) = 0 giving the simple eigenvalue �1and the double eigenvalue 2. There are tree independent solutions y1(t) =e�t, y2(t) = e2t, y3(t) = te2t and any solution is a linear combination ofthese.
3.5.3. The Wronskian. Let y1(t), . . . , yn(t) be n solutions of the equation(67). The substitution (68) produces n solutions u1, . . .un
of the system(69). Let U(t) = [u1(t), . . .un
(t)] be the matrix solution of (69). Theorem20applied to the companion system yields:
Theorem 23. Let y1(t), . . . , yn(t) be solutions of equation (67).(i) Their Wronskian W (t) = W [y1, . . . , yn](t) satisfies
W (t) = e�tan�1W (0)
(ii) y1(t), . . . , yn(t) are linearly independent if and only if their Wronskianis not zero.
3.5.4. Decomplexification. Suppose equation (67) has real coe�cients, aj
2
R, but there are nonreal eigenvalues, say �1,2 = ↵1±i�1. Then there are twoindependent solutions y1,2(t) = et(↵1±i�1) = et↵1 [cos(t�1)± i sin(t�1)]. If realvalued solutions are needed (or desired), note that Sp(y1, y2) = Sp(y
c
, ys
)where
yc
(t) = et↵1 cos(t�1), ys
(t) = et↵1 sin(t�1)
and ys
, yc
are two independent solutions (real valued).Furthermore, any solution in Sp(y
c
, ys
) can be written as
(72) C1et↵1 cos(t�1) + C2e
t↵1 sin(t�1) = Aet↵1 sin(t�1 +B)
where
A =qC21 + C2
2
and B is the unique angle in [0, 2⇡) so that
cosB =C2p
C21 + C2
2
, sinB =C1p
C21 + C2
2
Example 1. Solve the equation of the harmonic oscillator
(73) x0 = �y, y0 = k2x where k 2 Rgiving both complex and real forms.
In this example it is quicker to solve by turning the system into a secondorder scalar equation: taking the derivative in the first equation we obtainx00 = �y0 and using the second equation it follows that x00 + k2x = 0, with
34 RODICA D. COSTIN
characteristic equation �2+k2 = 0 and eigenvalues �1,2 = ±ik. The generalsolution is x(t) = c1e
ikt+ c2e�ikt. Then y(t) = �x0(t) = ikc1e
ikt
� ikc2e�ikt.
In real form
(74) x(t) = A sin(kt+B), y(t) = �Ak cos(kt+B)
Example 2. Solve the di↵erential equation y(iv) + y = 0. Find four realvalued independent solutions.
The characteristic equation is �4+1 = 0 with solutions �k
= ei⇡(2k+1)/4, k =0, 1, 2, 3. The equation has four independent solutions y
k
(t) = exp(i⇡(2k +1)/4 t), k = 0, 1, 2, 3.
To identify the real and imaginary parts of the eigenvalues, note that
3.6.1. Stable versus unstable equilibrium points. A linear, first order systemof di↵erential equation
(75)du
dt= Mu
always has the zero solution: u(t) = 0 for all t. The point 0 is called anequilibrium point of the system (75). More generally,
Definition 24. An equilibrium point of a di↵erential equation u0 = f(u)is a point u0 for which the constant function u(t) = u0 is a solution, there-fore f(u0) = 0.
It is important in applications to know how solutions behave near anequilibrium point.
An equilibrium point u0 is called stable if any solutions which start closeenough to u0 remain close to u0 for all t > 0. (This definition can be mademore mathematically precise, but it will not be needed here, and it is besidesthe scope of these lectures.)
Definition 25. An equilibrium point u0 is called asymptotically stableif
limt!1
u(t) = u0 for any solution u(t)
It is clear that an asymptotically stable point is stable, but the converse isnot necessarily true. For example, the harmonic oscillator (73) has solutionsconfined to ellipses, since from (74) it follows that x2+y2/k2 = A2. Solutionsare close to the origin if A is small, and they go around the origin along anellipse, never going too far, and not going towards the origin: the origin isa stable, but not asymptotically stable equilibrium point.
An equilibrium point which is not stable is called unstable.
EIGENVALUES AND EIGENVECTORS 35
Suppose one is interested in the stability of an equilibrium point of anequation u0 = f(u). By a change of variables the equilibrium point can bemoved to u0 = 0, hence we assume f(0) = 0. It is natural to approximatethe equation by its linear part: f(u) ⇡ Mx, where the matrix M has the ele-ments M
ij
= @fi
/@xj
(0), and expect that the stability (or instability) of theequilibrium point of u0 = f(u) to be the same as for its linear approximationu0 = Mu.
This is true for asymptotically stable points, and for unstable points,under fairly general assumptions on f . But it is not necessarily true forstable, not asymptotically stable, points as in this case the neglected termsof the approximation may change the nature of the trajectories.
Understanding the stability for linear systems helps understand the sta-bility for many nonlinear equations.
3.6.2. Characterization of stability for linear systems. The nature of theequilibrium point u0 = 0 of linear di↵erential equations depends on theeigenvalues of the matrix M as follows.
We saw that solutions of a linear system (75) are linear combinations ofexponentials et�j where �
j
are the eigenvalues of the matrix M , and if M isnot diagonalizable, also of tket�j for 0 < k (multiplicity of �
j
)� 1.Recall that
limt!1
tket�j = 0 if and only if <�j
< 0
Therefore:(i) if all �
j
have negative real parts, then any solution u(t) of (75) convergeto zero: lim
t!1 u(t) = 0, and 0 is asymptotically stable.(ii) If all <�
j
0, and some real parts are zero, and eigenvalues with zeroreal part have the dimension of the eigenspace equal to the multiplicity ofthe eigenvalue4 then 0 is stable.
(iii) If any eigenvalue has a positive real part, then 0 is unstable.As examples, let us consider 2 by 2 systems with real coe�cients.Example 1: an asymptotically stable case, with all eigenvalues real.For
M =
"�5 �2
�1 �4
#, M = S⇤S�1, with ⇤ =
"�3 0
0 �6
#, S =
"1 2
�1 1
#
The figure shows the field plot (a representation of the linear transfor-mation x ! Mx of R2). The trajectories are tangent to the line field, andthey are going towards the origin. Solutions with initial conditions alongthe directions of the two eigenvectors of M are straight half-lines (two such
4This means that if <�j = 0 then there are no solutions q(t)et�j with nonconstantq(t).
36 RODICA D. COSTIN
solutions are shown in the picture); these are the solutions u(t) = e�jtcvj
.(Solutions with any other initial conditions are not straight lines.)
Figure 2. Asymptotically stable equilibrium point, nega-tive eigenvalues.
The point 0 is a hyperbolic equilibrium point.
EIGENVALUES AND EIGENVECTORS 37
Example 2: an asymptotically stable case, with nonreal eigenvalues.For
M =
"�1 �2
1 �3
#with ⇤ =
"�2 + i 0
0 �2� i
#, S =
"1 + i 1� i
1 1
#
The figure shows the field plot and two trajectories. All trajectories aregoing towards the origin, though rotating around it. The equilibrium point0 is hyperbolic.
Figure 3. Asymptotically stable equilibrium point, nonreal eigenvalues.
38 RODICA D. COSTIN
Example 3: an unstable case, with one negative eigenvalue, and a positiveone.
For
M =
"3 6
3 0
#with ⇤ =
"�3 0
0 6
#, S =
"1 2
�1 1
#
The figure shows the field plot. Note that there is a stable direction (inthe direction of the eigenvector corresponding to the negative eigenvalue),and an unstable one (the direction of the second eigenvector, correspondingto the positive eigenvalue). Any trajectory starting at a point not on thestable direction has infinite limit as t ! 1.
Figure 4. Unstable equilibrium point, one positive eigenvalues.
The equilibrium point 0 is a saddle point.
EIGENVALUES AND EIGENVECTORS 39
Example 4: an unstable point, with both positive eigenvalues.For
M =
"5 2
1 4
#with ⇤ =
"3 0
0 6
#, S =
"1 2
�1 1
#
The field plot is similar to that that of Example 1, only the arrows haveopposite directions; the trajectories go away from the origin. In fact thissystem is obtained from Example 1 by changing t into �t.
Example 5: the equilibrium point 0 is stable, not asymptotically stable.For
M =
"1 �2
1 �1
#with ⇤ =
"i 0
0 �i
#, S =
"1 + i 1� i
1 1
#
The trajectories rotate around the origin on ellipses, with axes determinedby the real part and the imaginary part of the eigenvectors.
Figure 5. Stable, not asymptotically stable, equilibrium point.
4.1. Linear di↵erence equations with constant coe�cients. A firstorder di↵erence equation, linear, homogeneous, with constant coe�cients,has the form
(76) xk+1 = Mx
k
where M is an n ⇥ n matrix, and xk
are n-dimensional vectors. Given aninitial condition x0 the solution of (76) is uniquely determined: x1 = Mx0,then we can determine x2 = Mx1, then x3 = Mx2, etc. Clearly the solutionof (76) with the initial condition x0 is
(77) xk
= Mkx0
A second order di↵erence equation, linear, homogeneous, with constantcoe�cients, has the form
(78) xk+2 = M1x
k+1 +M0xk
A solution of (78) is uniquely determined if we give two initial conditions,x0 and x1. Then we can find x2 = M1x1 +M0x0, then x3 = M1x2 +M0x1
etc.Second order di↵erence equations can be reduced to first order ones: let
yk
be the 2n dimensional vector
yk
=
xk
xk+1
�
Then yk
satisfies the recurrence
yk+1 = My
k
where M =
0 IM0 M1
�
which is of the type (76), and has a unique solution if y0 is given.More generally, a di↵erence equation of order p which is linear, homoge-
neous, with constant coe�cients, has the form
(79) xk+p
= Mp�1x
k+p�1 + . . .+M1xk+1 +M0x
k
which has a unique solution if the initial p values are specified x0,x1 . . . ,xp�1.The recurrence (79) can be reduced to a first order one for a vector of di-mension np.
To understand the solutions of the linear di↵erence equations it thensu�ces to study the first order ones, (76).
4.2. Solutions of linear di↵erence equations. Consider the equation(76). If M has n independent eigenvectors v1, . . . ,vn
(i.e. M is diagonaliz-able) let S = [v1, . . . ,vm
] and then M = S⇤S�1 with ⇤ the diagonal matrixwith entries �1, . . . ,�n
. The solution (77) can be written as
xk
= Mkx0 = S⇤kS�1x0
EIGENVALUES AND EIGENVECTORS 41
and denoting S�1x0 = b,
xk
= S⇤kb = b1�k
1v1 + . . .+ bn
�k
n
vn
hence solutions xk
are linear combinations of �k
j
multiples of the eigenvectorsvj
.
Example. Solve the recurrence relation zn+2 = 3z
n+1 � 2zn
if z0 = ↵,z1 = �.
This is a scalar di↵erence equation, and we could turn it into a first ordersystem. But, by analogy to higher order scalar di↵erential equations, it maybe easier to work directly with the scalar equation. We know that there aresolutions of the form z
n
= �n, and substituting this in the recurrence we get�n+2 = 3�n+1
= 2n. We can alwaysdiscard the value � = 0 since it corresponds to the trivial zero solution. Thegeneral solution is z
n
= c1 + 2nc2. The constants c1, c2 can be determinedfrom the initial conditions: z0 = c1 + c2 = ↵, z1 = c1 + 2c2 = �, thereforezn
= (2↵+ �) + (� � ↵)2n.
If M is not diagonalizable, just as in the case of di↵erential equations,then consider a matrix S so that S�1MS is in Jordan normal form.
Consider the example of a 2⇥2 Jordan block: M = SJS�1 with J given by(60). As in Let S = [y1,y2] where y1 is the eigenvector of M correspondingto the eigenvalue � and y2 is a generalized eigenvector. Using (55) we obtainthe general solution
xk
= [y1,y2]
"�k k �k�1
0 �k
# b1b2
�= b1�
ky1 + b2
⇣k�ky1 + �ky2
⌘
and the recurrence has two linearly independent solutions of the form �ky1
and q(k)�k where q(k) is a polynomial in k of degree one.In a similar way, for p⇥ p Jordan blocks there are p linearly independent
solutions of the form q(k)�k where q(k) are polynomials in k of degree atmost p� 1, one of then being constant, equal to the eigenvector.
Example. Solve the recurrence relation yn+3 = 9 y
n+2 � 24 yn+1 + 20y
n
.Looking for solutions of the type y
n
= �n we obtain �n+3 = 9�n+2�
24�n+1+20�n which implies (disregarding � = 0) that �3�9�2+24��20 =
0 which factors (�� 5) (�� 2)2 = 0 therefore �1 = 5 and �2 = �3 = 2. Thegeneral solution is z
n
= c15n + c22n + c3n2n.
4.3. Stability. Clearly the constant zero sequance xk
= 0 is a solution ofany linear homogeneous discrete equation (76): 0 is an equilibrium point(a steady state).
42 RODICA D. COSTIN
As in the case of di↵erential equations, an equilibrium point of a di↵erenceequation is called asymptotically stable, or an attractor, if solutions startingclose enough to the equilibrium point converge towards it.
For linear di↵erence equations this means that limk!1 x
k
= 0 for allsolutions x
k
. This clearly happens if and only if all the eigenvalues �j
of Msatisfy |�
j
| < 1.If all eigenvalues have either |�
j
| < 1 or |�j
| = 1 and for eigenvalues ofmodulus 1, the dimension of the eigenspace equals the multiplicity of theeigenvalue, 0 is a stable point (or neutral).
In all other cases the equilibrium point is called unstable.
this is one of the most famous sequence of numbers, studied for more that 2milennia (it fist appeared in ancient Indian mathematics), which describescountless phenomena in nature, in art and in sciences.
The Fibonacci numbers are defined by the recurrence relation
(80) Fk+2 = F
k+1 + Fk
with the initial condition F0 = 0, F1 = 1.Substituting F
k
= �k into the recurrence (80) it follows that �2 = � + 1with solutions
�1 =1 +
p
5
2= � = the golden ratio, �2 =
1�p
5
2
Fk
is a linear combination of �k
1 and �k
2: Fk
= c1�k
1 + c2�k
2. The values ofc1, c2 can be found from the initial conditions:
Note that the ratio of two consecutive Fibonacci numbers converges tothe golden ratio:
