Linear Algebra
M1 - FIB
Contents:
5. Matrices, systems of linear equations and determinants
6. Vector space
7. Linear maps
8. Diagonalization
Anna de MierMontserrat Maureso
Dept. Matematica Aplicada II
Translation: Oriol Cano, Josep Elgueta
5. Matrices, systems of linear equations anddeterminants
5.1 Matrices: basic operations and row echelon form
1
Review of matrix algebra
2
Scalars
By a field of scalars K we will mean a set of numbers with twooperations (addition and multiplication) such that
- the usual properties are satisfied (commutative, associative ,distributive, identity elements)
- they are invertible (we can subtract and divide)
Examples: R,Q,Zp,C
3
Matrices
Let m, n ≥ 1 be integers. A m × n matrix with elements in thefield K consists of mn elements of K arranged in a table with mrows and n columnsWe will denote by aij the element in the ith row and jth columnA generic matrix is represented by:
A =
a11 a12 · · · a1na21 a22 · · · a2n
......
. . ....
am1 am2 · · · amn
We will also use the notation A = (aij)m×n
The set of all m × n matrices will be denoted by Mm×n(K)
4
Types of matrices
I A matrix of type 1× n is called a row matrix
I A matrix of type m × 1 is called a column matrix
I The null matrix Om,n (or simply O) is the matrix of typem × n with all elements equal to 0
I A matrix of type n× n is called a square matrix. The set of allsquare n × n matrices with elements in K is denoted byMn(K). A square matrix (aij)n×n is
I upper triangular if aij = 0 for all i > jI lower triangular if aij = 0 for all i < jI diagonal if it is both upper and lower triangular
I The matrix Diag(λ1, . . . , λn) is the diagonal matrix (dij)n×n
with dii = λi for all i
I The identity matrix In is the diagonal matrix Diag(1, 1, . . . , 1)
5
Matrix addition
Let A,B ∈Mm×n(K) with A = (aij) and B = (bij)
Their sum is the matrix A + B = (cij) ∈Mm×n(K) defined by
cij = aij + bij
PropertiesIf A,B,C ∈Mm×n(K), the following holds:
I (Associative) (A + B) + C = A + (B + C )
I (Commutative) A + B = B + A
I (Identity element) A + O = O + A = A
I (Additive inverse) There exists a matrix B such thatA + B = B + A = O
(This matrix B is denoted −A)
6
Scalar multiplication
Let A ∈Mm×n(K) with A = (aij) and let λ ∈ K be a scalar
The multiplication of A by the scalar λ is the matrixλA = (bij) ∈Mm×n(K) defined by
bij = λaij
PropertiesIf λ, µ ∈ K and A,B ∈Mm×n(K), the following holds:
I (Pseudo-associative) λ(µA) = (λµ)A
I (Distributive 1) λ(A + B) = λA + λB
I (Distributive 2) (λ+ µ)A = λA + µA
I (Identity) 1A = A
Note that (−1)A = −A
7
Transposition
Let A = (aij)m×n ∈Mm×n(K)
Its transpose is the matrix At = (bij)n×m ∈M(K)n×m defined bybij = aji
Clearly (At)t = A
A square matrix A issymmetric if At = Askew-symmetric if At = −A
8
Matrix multiplication
Let A = (aij)m×n ∈Mm×n(K) and B = (bij)n×p ∈Mn×p(K)
Their product is the matrix AB = (cij)m×p ∈Mm×p(K) with
cij =n∑
k=1
aikbkj = ai1b1j + ai2b2j + · · ·+ ainbnj
Observations
I The product of two matrices may not exist
I BA may not exist even when AB exists
I If AB and BA are defined, it may be AB 6= BA
I Multiplication is an internal map in Mn(K)
9
Properties of matrix multiplication
For matrices A,B,C the following holds (wherever the operationsare defined):
I (Associative) (AB)C = A(BC )
I (Distributive) A(B + C ) = AB + AC and(A + B)C = AC + BC
I (Identity element) IA = A = AI , where I is the identity matrixof the required type
I (Relation with the transpose) (AB)t = BtAt
If A ∈Mn(K), we will denote by Ak the product AA · · ·A(ie, A2 = AA,A3 = AAA, etc.)
10
Inverse matrix
Let A,B ∈Mn(K). We say that B is the inverse matrix of Awhen
AB = BA = In
If this holds we say that A is invertible and we denote by A−1 theinverse matrix
Observations
I If the inverse exists, it is unique
I Not all matrices have an inverse
I Invertible matrices have no zero rows or columns
11
Properties of the inverse matrix
If A and B are invertible matrices of the same type and λ is anonzero scalar, the following holds:
I the matrix A−1 is invertible and (A−1)−1 = A
I the matrix Ak is invertible and (Ak)−1 = (A−1)k
I the matrix λA is invertible and (λA)−1 = (λ)−1A−1
I the matrix At is invertible and (At)−1 = (A−1)t
I the product AB is invertible and (AB)−1 = B−1A−1
12
Elementary row operations and row echelon form
13
Elementary row operations
Let A ∈Mm×n(K)
An elementary row operation of A consists of one of thefollowing three operations:
(I) switching two rows of A
(II) multiplying a row of A by a nonzero scalar
(III) adding to one row of A the result of multiplying another rowby a nonzero scalar
A matrix is (row) elementary if it can be obtained from anidentity matrix by performing only one elementary row operation
14
Matrix equivalence
TheoremLet T be an elementary row operation and let M ∈Mm×n(K).The result of applying the operation T to the matrix M is EM,where E is the elementary matrix obtained by applying T to theidentity Im
A matrix B is (row) equivalent to a matrix A if B can beobtained from A by a finite sequence of elementary row operations
So, if B is equivalent to A we can write
B = ErEr−1 · · ·E2E1A,
where all Ei are elementary matrices
15
Row echelon form
A matrix is in (row) echelon form if it satisfies the following:
- if it has a zero row (all entries equal to zero), all the rowsbelow it are also zero
- in each nonzero row, the first nonzero element is equal to 1(it is called the leading coefficient or pivot of the row)
- the pivot of a row is always to the right of the pivot on theprevious row
TheoremEvery matrix is equivalent to a matrix in echelon form
The rank of a matrix A is the number of nonzero rows of anymatrix in echelon form equivalent to A
16
Application: computing the inverse (I)
LemmaIf E is an elementary matrix, then E is invertible and its inverseE−1 is also an elementary matrix
Proof:
(I) If B is an elementary matrix corresponding to a type (I) rowoperation (switching rows i and j), we have BB = I
(II) If Cλ is the elementary matrix corresponding to a type (II) rowoperation (multiplying a row by λ 6= 0), we haveCλCλ−1 = I = Cλ−1Cλ
(II) If Dk is the elementary matrix corresponding to a type (III)row operation (adding to row i row j multiplied by k), wehave DkD−k = I = D−kDk
17
Application: computing the inverse (II)
TheoremLet A ∈Mn(K) and let M be a matrix in row echelon formequivalent to A. Then A is invertible if and only if all the elementsin the diagonal of M are equal to 1
CorollaryA matrix A ∈Mn(K) is invertible if and only if the rank of A is n
18
Gauss-Jordan method for computing the inverse
Let A ∈Mn(K)
The proof of the previous theorem implies thatif In = Er · · ·E2E1A, then A−1 = Er · · ·E2E1
Given A, its inverse A−1, if it exists, can be found as follows:
I Start with the matrix (A|In)
I Apply row operations to (A|In), until we get a matrix of theform (In|B)
I If we succeed, A−1 = B
I Otherwise, A is not invertible
19
5. Matrices, systems of linear equations anddeterminants
5.2 Systems of linear equations
20
Systems of linear equations
A linear equation in the variables x1, . . . , xn is an expression ofthe form
a1x1 + a2x2 + · · ·+ anxn = b,
where a1, . . . , an, b belong to the field of scalars K
A solution is (s1, . . . , sn) ∈ Kn such that
a1s1 + a2s2 + · · ·+ ansn = b
(Obs. A linear equation can have no solution, one, or more thanone, even infinite, solutions)
21
Systems of linear equations
A system of linear equations is a set of linear equations (all ofthem with the same variables x1, . . . , xn)
The general form of a system of linear equations is:
a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2
......
am1x1 + am2x2 + · · ·+ amnxn = bm
A solution to the system is an n-tuple (s1, . . . , sn) ∈ Kn thatsatisfies all the equations
22
Solutions of a system
We will say that a system is
I inconsistent if it has no solution
I determined if it has only one solution
I underdetermined if it has more than one solution
The general solution of a system is the set of all its solutions
Two systems are equivalent if they have the same set of solutions
23
Equivalent systems
Two systems with the same equations but sorted differently areequivalent
And if in a system
I we multiply an equation by a (nonzero) scalar, or
I we add to an equation a multiple of another one
the resulting system is equivalent to the first one
24
Matrix form of a systemGiven the system
a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2
......
am1x1 + am2x2 + · · ·+ amnxn = bm
the system matrix and the matrices of variables and of constantterms are
A =
a11 a12 · · · a1na21 a22 · · · a2n
......
. . ....
am1 am2 · · · amn
x =
x1x2...xn
b =
b1b2...bm
We can write the system as a product of matrices:
Ax = b
25
Augmented matrix
The augmented matrix is the matrix (A|b), ie,
(A|b) =
a11 a12 · · · a1n b1a21 a22 · · · a2n b2
......
. . ....
...am1 am2 · · · amn bm
Obs. If row operations are applied to the augmented matrix of asystem, the resulting system is equivalent to the first one
So, every system of linear equations is equivalent to another onewhose augmented matrix is in echelon form
26
Solving a system in row echelon formThe row echelon form of a consistent system looks like
x1 + c12x2 + c13x3 + · · ·+ c1rxr + · · ·+ c1nxn = d1x2 + c23x3 + · · ·+ c2rxr + · · ·+ c2nxn = d2
......
xr + · · ·+ crnxn = dr
(rearranging the variables if needed)
Variables x1, . . . , xr are called basic and the others are called free
The system is solved “upwards” as follows
The basic variable xr is isolated in terms of the free variables:
xr = dr − cr ,r+1xr+1 − · · · − crnxn
Now we can isolate xr−1 in terms of xr and the rest of the freevariables, etc
27
General solution of a system in row echelon form
In a system in row echelon form we can express all the basicvariables in terms of the free ones (and the coefficients):
x1 = f1 + e1,r+1xr+1 + · · ·+ e1,nxn
x2 = f2 + e2,r+1xr+1 + · · ·+ e2,nxn...
...
xr = fr + er ,r+1xr+1 + · · ·+ er ,nxn
This is the general solution of the system
Obs. For each assignment of a scalar value to each of the freevariables xr+1, . . . , xn we get a particular solution of the system
We say that the system has n − r degrees of freedom
28
Parametric form of the general solutionIf the general solution of a system is
x1 = f1 + e1,r+1xr+1 + · · ·+ e1,nxn
x2 = f2 + e2,r+1xr+1 + · · ·+ e2,nxn...
...
xr = fr + er ,r+1xr+1 + · · ·+ er ,nxn
the parametric form of the solution is the expression
x1x2...xrxr+1
...xn
=
f1f2...fr0...0
+ xr+1
e1,r+1
e2,r+1...
er ,r+1
1...0
+ · · ·+ xn
e1,ne2,n
...er ,n
0...1
29
Discussing a system: the Rouche-Frobenius Theorem
TheoremLet be a system of linear equations with associated matrixA ∈Mm×n(K) and augmented matrix (A|b)
Let r , r ′ be the ranks of A, (A|b) respectively.
Then,
I if r < r ′, the system is inconsistent
I if r = r ′ = n, the system is determined
I if r = r ′ < n, the system is underdetermined with n − rdegrees of freedom
The rank of the associated matrix of a consistent linear system iscalled the rank of the system
30
Homogeneous systems
A system of linear equations is homogeneous if all the constantterms are equal to 0
Obs. A homogeneous system is always consistent (since we havethe trivial solution x1 = · · · = xn = 0)
CorollaryLet A be the associated matrix of a homogeneous system in nvariables; let r be the rank of A. Then
I if r = n, the system is determined and the only solution is thetrivial one
I if r < n, the system is underdetermined and it has non-trivialsolutions
31
Solving the system: Gaussian elimination
To find the general solution of any system of linear equations weproceed as follows:
1. Find the augmented matrix (A|b)
2. Find a matrix in row echelon form M equivalent to (A|b)
3. Apply the Rouche-Frobenius theorem to determine whetherthe system is consistent or not
4. In the case the system is consistent, find the general solutionfrom the equivalent system with augmented matrix M
32
5. Matrices, systems of linear equations anddeterminants5.3 Determinants
33
Definition of determinantLet A = (aij) ∈Mn(K). A minor of A is any matrix obtained bydeleting a certain number of rows and the same number ofcolumns of A
The minor associated to the element aij is the matrix Aij
obtained by deleting row i and column j from A.The minor Aij is an (n − 1)× (n − 1) square matrix.
The determinant of A is recursively defined by
- if n = 1, then det(A) = a11
- if n ≥ 2, then
det(A) = a11 det(A11)− a12 det(A12) + · · ·+(−1)1+ja1j det(A1j) + · · ·+ (−1)n+1a1n det(A1n)
The cofactor of the element aij is (−1)i+j det(Aij)
34
Computing the determinant
(Instead of det(A), sometimes we write |A|)
I Matrices 2× 2 and 3× 3:∣∣∣∣a bc d
∣∣∣∣ = a det((d))− b det((c)) = ad − bc
∣∣∣∣∣∣
a b cd e fg h i
∣∣∣∣∣∣= a
∣∣∣∣e fh i
∣∣∣∣− b
∣∣∣∣d fg i
∣∣∣∣+ c
∣∣∣∣d eg h
∣∣∣∣
= a(ei − fh)− b(di − fg) + c(dh − eg)
= aei + cdh + bfg − ceg − afh − bdi
I One can prove by induction:If A has a zero row or column then det(A) = 0If A = Diag(a1, a2, . . . , an), then det(A) = a1a2 . . . an
35
TheoremLet A ∈Mn(K) and i , j ∈ [n]. Then
det(A) =n∑
k=1
aik(−1)i+k det(Aik)
(Expansion of the determinant along row i)
det(A) =n∑
k=1
akj(−1)k+j det(Akj)
(Expansion of the determinant along column j)
36
Determinants and row operations
Let A,B ∈Mn(K). If B is the matrix obtained from A
I switching two rows, then det(B) = − det(A) (row operationof type (I))
I multiplying the i-th row of A by λ, then det(B) = λ det(A)(row operation of type (II))
I adding to a row a multiple of another one, thendet(B) = det(A) (row operation of type (III))
CorollaryIf M is obtained from A by performing row operations,
det(M) = K det(A), where K 6= 0
So, if A and M are equivalent matrices, then
det(A) 6= 0 ⇔ det(M) 6= 0
37
Characterization of invertible matrices
TheoremA matrix A ∈Mn(K) is invertible if and only if det(A) 6= 0
CorollaryA matrix A ∈Mn(K) has rank n if and only if det(A) 6= 0
TheoremLet A ∈Mm×n(K). The rank of A is r if and only if the largestminor of A with nonzero determinant has type r × r
38
Determinants and operations with matrices
If A,B ∈Mn(K), then
I det(AB) = det(A) det(B)
I det(At) = det(A)
I if A is invertible, det(A−1) = det(A)−1
However, in general det(A + B) 6= det(A) + det(B)
39
6. Vector spaces
40
Rn and its operations
Rn = {
x1x2...xn
: xi ∈ R, 1 ≤ i ≤ n}
Let x =
x1x2...xn
and y =
y1y2...yn
be elements of Rn and λ ∈ R
Addition in Rn:
x + y =
x1 + y1x2 + y2
...xn + yn
Scalar Multiplication in Rn:
λx =
λx1λx2
...λxn
(i.e., both operations are defined componentwise)41
PropertiesAddition in Rn satisfies the following properties, for allu, v ,w ∈ Rn:
s1) (associative) u + (v + w) = (u + v) + w
s2) (commutative) u + v = v + u
s3) (identity element) u + 0 = u where 0 = (0 0 . . . 0)t
s4) (additive inverse) for all u there exists u′ such that u + u′ = 0
The scalar multiplication in Rn satisfies the following properties,for all u, v ∈ Rn and λ, µ ∈ R:
p1) λ(µu) = (λµ)u
p2) λ(u + v) = λu + λv
p3) (λ+ µ)u = λu + µu
p4) 1u = u
(All properties hold because they hold in R and the operations aredefined componentwise)
42
6.2 Vector spacesA vector space over a field K consists of
1. a non empty set E
2. a map E × E → E (addition +) and
3. a map K× E → E (scalar multiplication ·)such that the following holds for all u, v ,w ∈ E and all λ, µ ∈ K:
e1) (associative) u + (v + w) = (u + v) + w
e2) (commutative) u + v = v + u
e3) (identity element) there exists a unique element 0E ∈ E suchthat u + 0E = u
e4) (additive inverse) for each u ∈ E there exists a unique u′ ∈ Esuch that u + u′ = 0E
e5) λ(µu) = (λµ)u
e6) λ(u + v) = λu + λv
e7) (λ+ µ)u = λu + µu
e8) 1u = u, where 1 is the identity element of the product in K43
Some examples of vector spaces
I Rn
I Zn2: strings of n bits
Addition is bit by bit: e.g.,
0110
+
1110
=
1000
Scalar multiplication: 0u = 0Zn2
and 1u = uI Mm×n(K) (the m × n matrices with entries in the field K)I Upper triangular matrices in Mn(R)I P(R): the set of polynomials with coefficients in RI Pd(R): the polynomials with degree at most d with
coefficients in RI The trivial vector space with only the zero vector: {0E}I The solutions of a homogeneous system of linear equations
44
PropertiesIf v ∈ E and λ is a scalar, the following holds:
I 0v = 0E
I λ 0E = 0E
I If λ v = 0E , then λ = 0 or v = 0E
I The additive inverse of v is (−1)v ; usually we will write −v
45
6.3 Vector subspaces and linear combinations
A subset S ⊆ E is a vector subspace (VS) if
(s1) S 6= ∅(s2) for all u, v ∈ S , u + v ∈ S
(s3) for all u ∈ S and all λ ∈ K, λu ∈ S
The vector 0E belongs to all vector subspaces
Some examples of vector subspaces
I Pd(R) is a vector subspace of the vector space P(R)
I Upper triangular matrices of Mn(R) are a VS of Mn(R)
I The solutions of a homogeneous system of linear equations inn variables and coefficients in K is a VS of Kn
46
Intersection of subspaces
Lemma If S and S ′ are vector subspaces of E , then S ∩ S ′ is alsoa subspace of E .
The union of vector subspaces is not a vector subspace in general,as in the case
S = {(xx
): x ∈ R}, and S ′ = {
(x−x
): x ∈ R};
We have
(11
)+
(2−2
)6∈ S ∪ S ′
Linear combination
Given vectors u1, . . . , uk in E , a linear combination of u1, . . . , ukis any vector of the form
λ1u1 + · · ·+ λkuk ,
where λ1, . . . , λk are scalars
47
Subspaces spanned by some vectors
Let u1, . . . , uk be vectors of E . The subspace spanned byu1, . . . , uk is the set
〈u1, . . . , uk〉 = {λ1u1 + λ2u2 + · · ·+ λkuk : λ1, . . . , λk ∈ K},
i.e., the set of all linear combinations of u1, . . . , uk
PropositionThe subspace 〈u1, . . . , uk〉 is a vector subspace. Moreover, it is thesmallest subspace that contains u1, . . . , uk
If a space S can be written as S = 〈u1, . . . , u`〉, we will say that{u1, . . . , u`} is a spanning set of S . The spanning set of a spaceis not unique
Observe that v is a linear combination of u1, . . . , uk if and only ifv ∈ 〈u1, . . . , uk〉
48
Examples of 〈u1, . . . , u`〉
I Rn = 〈
10...0
,
01...0
, . . . ,
00...1
〉
I The matrix space Mm×n(R) is spanned by the matrices Mij
that have all their entries equal to 0, except the (i , j) entry,which is equal to 1, 1 ≤ i ≤ n and 1 ≤ j ≤ mFor example, M2(R) = 〈M11,M12,M21,M22〉, where
M11 =
(1 00 0
),M12 =
(0 10 0
),M21 =
(0 01 0
),M22 =
(0 00 1
)
I Pd(R) = 〈1, x , . . . , xd〉(These examples can be generalized to any field K)
49
I To span the upper triangular matrices, it is enough to use theprevious matrices Mij with i ≤ j
I A subspace whose vectors are given in terms of parameters
{a + (b − a)x + (c − b)x2 + (a− c)x3 : a, b, c ∈ R}={a(1− x + x3) + b(x − x2) + c(x2 − x3) : a, b, c ∈ R}=〈1− x + x3, x − x2, x2 − x3〉
50
6.4 Linear independenceLet u1, . . . , uk ∈ E . The equation
λ1u1 + λ2u2 + · · ·+ λkuk = 0E
always has the solution λ1 = · · · = λk = 0.If this is the only solution we will say that the vectors u1, . . . , ukare linearly independent (LI)
If there is a solution with some λi 6= 0, we will say that the vectorsare linearly dependent (LD)
(We will also say that the set {u1, . . . , uk} is LI or LD, resp.)
Examples:
I The vector 0E is linearly dependent
I Given a vector u 6= 0E , the vector u is linearly independent
I If u is any given vector and λ is a scalar, {u, λu} is LD51
To determine if a set of vectors u1, u2, . . . , uk of Rn is linearlyindependent we proceed as follows:
(1) we form a matrix A with the given vectors, putting them incolumns
(2) we compute the rank r of A
(3) I if r = k, then the k vectors are LII if r < k , then they are LD; if we have computed the rank
putting the matrix A in row echelon form, then the vectorsthat correspond to the columns with a pivot are a maximal LIsubset; if we have computed the rank by minors, the vectorsthat correspond to the columns of the largest minor of A withnonzero determinant are a maximal LI subset.
52
In general, to determine if a set of vectors u1, u2, . . . , uk of avector K-space E is linearly independent we proceed as follows:
(1) from the vector equation
λ1u1 + λ2u2 + · · ·+ λkuk = 0E
we obtain a homogeneous system with unknownsλ1, λ2, . . . , λk
(2) we discuss the system, if it isI determined, the vectors u1, u2, . . . , uk are LII underdetermined, the vectors u1, u2, . . . , uk are LD
53
Properties
Let S = {u1, . . . , uk} be a set of vectors of a K-vector space E
I If 0E is in S , then u1, . . . , uk are LD
I If u1, . . . , uk are LI, then 0E is not in S
I If u1, . . . , uk are LI, every subset of S is LI
I If u1, . . . , uk are LD, every set that contains S is LD
TheoremIf u1, . . . , uk are LD and u1 is a linear combination of the othervectors in S , then
〈u1, u2, . . . , uk〉 = 〈u2, . . . , uk〉
54
Characterizations
TheoremA set of vectors S is LD if, and only if, there exists a vector v in Sthat is a linear combination of the other vectors in S
CorollaryLet v ∈ E . If u1, . . . , uk are LI, then v , u1, . . . , uk are LI if, andonly if, v 6∈ 〈u1, . . . , uk〉
55
6.5 Bases and dimensionLet E be a K-vector space. A set of vectors B = {b1, b2, . . . , bn}is a basis of E if
(b1) B is linearly independent
(b2) E = 〈b1, b2, . . . , bn〉, i.e., b1, b2, . . . , bn span E
The canonical basis
I of Kn is {
10...0
,
01...0
, . . . ,
00...1
}
I ofMm×n(K) is given by the mn matrices Mij that have all theentries equal to 0 except the (i , j) entry, which is equal to 1
I of Pd(K) is {1, x , x2, . . . , xd}({xd , xd−1, . . . , 1} is also a canonical basis; we will need tospecify which one we use)
56
Let B = {b1, . . . , bn} be a basis of E
PropositionEach vector of E can be written in a unique way as a linearcombination of the vectors in B
Let v ∈ E . If v = α1b1 + · · ·+ αnbn, we say that
vB =
α1...αn
is the vector of coordinates of v in the basis B
PropositionLet {u1, . . . , uk} be a set of vectors of E that are LI. Then k ≤ n
CorollaryEvery basis of E has n elements
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Dimension
The cardinal of any basis of a vector space E (or of a VS) will becalled the dimension of the space, denoted by dim(E)
The dimensions of the spaces that we will mostly work with are:dim(Kn) = n, dim(Mm×n(K)) = nm, and dim(Pd(K)) = d + 1
The dimension of the subspace {0E} is 0
Let E be of dimension n and let W = {w1, . . . ,wn} be a subset ofE
I if W is a LI set, then W is a basis of E
I if W spans E , then W is a basis of E
If S is a subspace of E then
I dim(S) ≤ dim(E )
I if dim(S) = dim(E ), S = E
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Change of basis
Let B = {b1, · · · , bn} and B ′ = {b′1, · · · , b
′n} be two bases of a
K-vector space E . Let u be a vector of EWe look for the relationship between the coordinate vectors uBand uB′ .
We call change-of-basis matrix from B to B′ the matrix whosecolumns are the vectors of coordinates (b1)B′ , . . . , (bn)B′ . It isdenoted by PB
B′
PBB′ =
......
...(b1)B′ (b2)B′ . . . (bn)B′
......
...
Then
I uB′ = PBB′uB
I PB′B =
(PBB′)−1
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7. Linear Maps
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7.1 Definitions, examples and propertiesLet E and F be two K-vector spaces. A map f : E → F is linearif:
(a1) for all u, v ∈ E , f (u + v) = f (u) + f (v)(a2) for all u ∈ E and all λ ∈ K, f (λu) = λf (u)
If E = F , we say that f is an endomorphism
ExamplesI Trivial map. f : E → F given by f (u) = 0F , u ∈ E , is linearI Identity map. IE : E → E given by IE (u) = u, u ∈ E , is linearI Let A ∈Mm×n(K).
The map f : Kn → Km given by f (v) = Av , v ∈ Kn, is linearI The following is a non-linear map
f :M2×2(R)→ P2(R), f
((a bc d
))= x2−(a+d)x+(2c−b)
I The map f : R2 → R2, f
(xy
)=
(x2y2
x + y
)is not linear
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PropertiesLet f : E → F be a linear map. Then
I f (0E ) = 0FI f (
∑λivi ) =
∑λi f (vi ), for all u ∈ E
I if S is a subspace of E , f (S) is a subspace of F
I if S ′ is a subspace of F , f −1(S ′) is a subspace of E
PropositionLet B = {b1, . . . , bn} be a basis of E . Then f is uniquelydetermined by f (b1), . . . , f (bn)
i.e., the image of any vector u ∈ E can be obtained from theimage of a basis:if u = α1b1 + · · ·+ αnbn, then f (u) = α1f (b1) + · · ·+ αnf (bn)
CorollaryIf S = 〈v1, . . . , vk〉 is a subspace of E , then
f (S) = 〈f (v1), . . . , f (vk)〉62
Let B = {b1, . . . , bn} be a basis of E , W a basis of F and m thedimension of F
The matrix associated to f in the bases B and W is the matrixin Mm×n(K) whose columns are the coordinates in the basis W ofthe images of the vectors in the basis B. It is denoted by MB
W(f)
MBW (f ) =
......
...f (b1)W f (b2)W . . . f (bn)W
......
...
∈Mm×n(K)
To find the vector of coordinates of the image of a vector u ∈ Ewe just need to compute the matrix product
f (u)W = MBW (f )uB .
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7.2 Characterization of linear maps
A linear bijective map is called an isomorphism
Characterization of the type of linear mapLet f : E → f be a linear map and let M be the matrix associatedto f in given bases of E and of F .
I f is injective ⇔ rank(M) = dim(E )
I f is surjective ⇔ rank(M) = dim(F )
I f is an isomorphism ⇔ rank(M) = dim(E ) = dim(F )
I If E and F have the same dimension, thenf is an isomorphism ⇔ f is injective ⇔ f is surjective
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7.3 Composition of linear maps
Let E , F and G be vector spaces with bases B,W and V ,respectively
Proposition
If f : E → F and g : F → G are linear maps, the compositiong ◦ f : E → G is also a linear map and
MBV (g ◦ f ) = MW
V (g)MBW (f )
Proposition
If f : E → F is an isomorphism, the inverse map f −1 : F → E isalso linear and
MWB (f −1) = (MB
W (f ))−1
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7.4 Change of basis
We look for the change to the matrix associated to a linear mapwhen we change the basis in the source and/or the target space.
Let f : E → F be a linear map, B and B ′ bases of E , and W andW ′ bases of F
EBf−−−−→
MBW (f )
FW
IE
xPB′B
PWW ′
yIF
EB′f−−−−−→
MB′W ′ (f )
FW ′
f = IF ◦ f ◦ IEMB′
W ′(f ) = PWW ′ MB
W (f ) PB′B
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8. Diagonalization
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The problem of diagonalization
Let f : E → E be an endomorphism. Is there any basis B of E inwhich the matrix MB(f ) is simple? More concretely, diagonal?
DefAn endomorphism f : E → E is diagonalizable if there exists abasis B of E such that MB(f ) is diagonal.
Obs. The matrix MB(f ) may be non-diagonal, but if we know thatthe endomorphism f diagonalizes in another basis B ′, then thematrix
(PB′B )−1MB(f )PB′
B
is diagonal.So, being diagonalizable is equivalent to the existence of aninvertible matrix P such that P−1MB(f )P is diagonal.
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Eigenvalues and eigenvectors
DefThe scalar λ is an eigenvalue of an endomorphism f if there existsa vector v 6= 0E such that f (v) = λv .
All vectors v 6= 0E such that f (v) = λv are called eigenvectors ofeigenvalue λ.
TheoremThe endomorphism f : E → E diagonalizes if and only if there is abasis of E completely made of eigenvectors of f .
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Computing eigenvaluesLet M be the matrix associated to f : E → E in a basis B
DefThe characteristic polynomial of the endomorphism f is
pf (x) = det(M − xIn)
TheoremThe eigenvalues of f are the roots of the characteristic polynomial
The algebraic multiplicity of an eigenvalue λ is the multiplicity ofλ as a root of pf (x) and it is denoted by mλ
The equation pf (x) = 0 is called the characteristic equation
TheoremThe characteristic polynomial does not depend on the basis inwhich we compute the matrix M
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Eigenspaces
Let λ be an eigenvalue of the endomorphism f : E → EThe eigenspace of the eigenvalue λ is the set
Eλ = {u ∈ E : f (u)− λu = 0E}
Properties
I Eλ is a vector subspace of E
I 1 ≤ dim(Eλ) ≤ mλ
The dimension of Eλ is called the geometric multiplicity of λ
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Characterization of diagonalizable endomorphisms
Let f : E → E be an endomorphism of a vector space E ofdimension n.
TheoremThe endomorphism f is diagonalizable if and only if it has neigenvalues (counting multiplicities) and for each eigenvalue thealgebraic and the geometric multiplicity coincide.
CorollaryIf f has n different eigenvalues, then it is diagonalizable.
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Diagonalization algorithmTo decide whether the endomorphism f : E → E is diagonalizableor not, we will proceed as follows:
(1) We find the matrix associated to f in any basis and wecalculate the characteristic polynomial pf (x).
(2) We find the eigenvalues and their multiplicities solvingpf (x) = 0.
(3) If the multiplicities of the eigenvalues add up to less thandim(E ), the endomorphism does not diagonalize. Otherwisego to (4).
(4) For each eigenvalue λ, we find the eigenspace Eλ and itsdimension dim(Eλ).
(5) If for all λ we have mλ = dim(Eλ), the endomorphismdiagonalizes. Otherwise it does not.
If the endomorphism diagonalizes, a basis in which it diagonalizesis given by the union of the bases of the spaces Eλ.
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