Post on 16-Jul-2015
transcript
DefinitionsA is n x n. λ is an eigenvalue of A if
AX = λX has non zero solutions X (called eigenvectors)
If λ is an eigenvalue of A, the set Eλ = Eλ(A) = {X | X in ℜn, AX = λX}is a vector space called the eigenspace associated w/ λ
(i.e. Eλ is all eigenvectors corresponding to λ & 0 vector)λ is eigenvalue if Eλ has at least one non-zero vector.Can also write AX = λX as (λIn - A)X = 0
Example
Show that λ = -3 is an eigenvalue of A, and find the
eigenspace E-3. A =
5 8 16
4 1 8
−4 −4 −11
Write out (λIn - A)X = 0 and solve.
Get: X = s
−1
1
0
+ t
−2
0
1
So it is an eigenvalue since there is a non-zero solution. Eigenspace is:
E−3 = span
−1
1
0
,
−2
0
1
Discussion
Now we have (λIn - A)X = 0, and λ is an eigenvalue iff there exists a nonzero solution X.
Recall that a matrix U is invertible iff UX = 0 implies X = 0.
So, since we are looking for a nonzero solution above,
(λIn-A) cannot be invertible for λ to be an eigenvalue.
So det (λIn-A) =0.
Theorem 1
A (n x n). The eigenvalues of A are the real roots of the characteristic polynomial of A --the real numbers λ satisfying:
cA(λ) = det(λIn - A) = 0The eigenspace Eλ = {X | (λI - A)X = 0} consists of all solutions to a system of n linear equations in n variables.
The eigenvectors corresponding to λ are the nonzero vectors in the eigenspace.
Summary
So there are two issues: finding eigenvalues, and finding eigenspaces (and eigenvectors).
Finding the eigenvalues can be difficult - won’t do much here.
Spend more time dealing with eigenspaces.
Example
Find the characteristic polynomial, eigenvalues, and eigenspaces of A:
A =1 −2 3
2 6 −6
1 2 −1
Set up cA(x) = det (xI - A)
Eigenvalues will be the roots of the polynomial as those will give us where det is 0.
Then use those λ to find eigenspace: X such that (λ I-A)X=0
ExampleIf A is a triangular matrix, show that the eigenvalues of A are the entries on the main diagonal.
Proof: cA(x) = det (xI - A) = det ( a triangular matrix) = product of entries on main diagonal of (xI - A).
The matrix showing entries on main diagonal is:
x − a
11
x − a22
...
x − ann
det = (x-a11)(x-a22)…(x-ann)
So eigenvalues are{a11,a22,…,ann}
ExampleShow that A and AT have the same characteristic polynomial and thus the same eigenvalues.
Proof: From chapter 3, we know that a matrix and its transpose will have the same determinant.
cA
T (x) = det(xI − A T ) = det((xI) T − A)T
= det(xI − A)T = det(xI − A) = cA(x)
Theorem 2If A is a real symmetric matrix, each root of the characteristic polynomial cA(x) is real. (to be proven later)
Show this is true for a (2 x 2):
A =a b
b c
cA (x) = detx − a −b
−b x − c
= (x − a)(x − c) − b2
= x 2 − x(a + c) + (ac − b2 )
Recall that we can determine the nature of the roots from the discriminant: (b2-4ac) = (a+c)2-4(ac+b2) = a2+c2+2ac-4ac+4b2
=a2-2ac+c2+4b2 = (a-c)2 + 4b2 which is always pos so real roots.
Similar MatricesA, B (n x n) are similar (we say A~B) if B = P-1AP
holds for some invertible matrix.
P is not unique.
ExampleFind P-1AP in the following case, then compute An.
P =1 5
1 2
, A =
6 −5
2 −1
We are able to find a similar matrix B.
Then P-1AP=B.
So A = PBP-1
So A2=(PBP-1)(PBP-1)=PB2P-1
Generally An=PBnP-1
Life is made easy is B is diagonal since we just raise entries to n.
Interesting FactSimilar Matrices will have the same determinant.
Proof:
P-1AP = D
det(D) = det (P-1AP) = (detP-1)(detA)(detP) = (1/detP)(detA)(det P) = det A.�
ExampleShow that A and B are not similar.
A =1 2
2 1
,B =
1 1
−1 1
Just need to show that they do not have the same determinant.
Theorem 3
A,B (n x n), k is a scalar:
1. tr(A + B) = tr A + tr B and tr(kA) = k tr A
2. tr (AB) = tr (BA)
Proof:
1. (homework)
2. AB =
a1j
bj 1
j =1
n
∑
a2 jb j2j=1
n
∑...
anjb
jnj =1
n
∑
tr(AB) = aijb jij =1
n
∑
i =1
n
∑ = b jiaiji =1
n
∑
j =1
n
∑ = tr(BA)
Theorem 4
If A~B, they have the same determinant, the same rank, the same trace, the same characteristic polynomial, and the same eigenvalues. (similarity invariants)
Proof: Already shown that they have the same determinant.
Rank: Have B = P-1AP
rank (B) = rank (P-1AP) = rank(AP)=rankA since P is invertible (and using cor 4 of thm 4 in 5.5)
tr B = tr (P-1AP) = tr[(AP)P-1] = tr (A) (uses thm 3)
Theorem 4 - cont
Characteristic polynomial
cB(x) = det (xI - B) = det(xI - P-1AP)=det(P-1xIP - P-1AP)
(since xI = P-1xIP -- since xI is diagonal)
= det [P-1(xI - A)P]=(1/detP)(det(xI-A))(det P) = det(xI-A) = cA(x)
Eigenvalues: all matrices with the same characteristic poly will have the same eigenvalues since the eigenvalues are the roots of the characteristic polynomial.
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Fact
The invariants do not imply similarity.
Ex. I =1 0
0 1
, A =
1 2
0 1
Have same det,tr,rank,characteristic poly, eigenvalues, but are not similar since P-1IP = I A≠
Theorem 5
A,B,C (n x n). Then:
1. A~A for all A.
2. If A ~ B, then B~A
3. If A ~ B and B ~ C, then A~C.
Proof of 2 (others follow):
A~B ⇒ B = P-1AP
Let Q = P-1, then B = QAQ-1, so A= Q-1BQ
Which means that B ~ A.