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6.1 Eigenvalues and Diagonalization
DefinitionsA is n x n. is an eigenvalue of A if
AX = X has non zero solutions X (called eigenvectors)
If is an eigenvalue of A, the set E = E(A) = {X | X in n, AX = X}is a vector space called the eigenspace associated w/
(i.e. E is all eigenvectors corresponding to & 0 vector) is eigenvalue if E has at least one non-zero vector.Can also write AX = X as (In - A)X = 0
Example
Show that = -3 is an eigenvalue of A, and find the
eigenspace E-3. A
5 8 16
4 1 8
4 4 11
Write out (In - 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 (In - A)X = 0, and 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,
(In-A) cannot be invertible for to be an eigenvalue.
So det (In-A) =0.
Definition
The characteristic polynomial of the n x n matrix A is:cA(x) = det(xI - A)
Theorem 1
A (n x n). The eigenvalues of A are the real roots of the characteristic polynomial of A --the real numbers satisfying:
cA() = det(In - A) = 0The eigenspace E = {X | (I - A)X = 0} consists of all solutions to a system of n linear equations in n variables.
The eigenvectors corresponding to 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 to find eigenspace: X such that ( 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.
cAT (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.
TraceThe trace of a square matrix A (tr A) is the sum of the entries on the main diagonal of A.
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
a1jbj 1
j1
n
a2 jb j2j1
n
...
anjbjn
j1
n
tr(AB) aijb jij1
n
i1
n
b jiaiji1
n
j1
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.
�
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.
Use of thm 5
Proving similarity is not always easy. But if we can find a simple (often diagonal) matrix to which both A and B are both similar, then: A~D and B~D means D~B by (2)
and A~B then by (3)