This document discusses eigenvalues and diagonalization of matrices. Some key points: - Eigenvalues are values that satisfy the equation AX = λX, where X is a non-zero eigenvector. The eigenspace associated with λ contains all eigenvectors for that eigenvalue. - The characteristic polynomial of a matrix A is defined as det(xI - A). The eigenvalues of A are the roots of its characteristic polynomial. - Similar matrices have the same eigenvalues, determinant, rank, trace, and characteristic polynomial. Two matrices are similar if one can be obtained from the other by conjugation via an invertible matrix. - The trace of a matrix is the sum of its diagonal entries

1. 6.1 Eigenvalues and Diagonalization

2. Definitions
A 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

3. 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





















4. 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.

5. Definition
The characteristic polynomial of the n x n matrix A is:
cA(x) = det(xI - A)

6. 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) = 0
The 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.

7. 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.

8. 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

9. Example
If 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 − a11
x − a22
...
x − ann












det = (x-a11)(x-a22)…(x-ann)
So eigenvalues are{a11,a22,…,ann}

10. Example
Show 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 − AT
) = det((xI)T
− A)T
= det(xI − A)T
= det(xI − A) = cA
(x)

11. Theorem 2
If 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) = det
x − a −b
−b x − c





 = (x − a)(x − c) − b
2
= x2
− 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.

12. Similar Matrices
A, B (n x n) are similar (we say A~B) if B = P-1
AP
holds for some invertible matrix.
P is not unique.

13. Example
Find P-1
AP 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-1
AP=B.
So A = PBP-1
So A2
=(PBP-1
)(PBP-1
)=PB2
P-1
Generally An
=PBn
P-1
Life is made easy is B is diagonal since we just raise entries to n.

14. Interesting Fact
Similar Matrices will have the same determinant.
Proof:
P-1
AP = D
det(D) = det (P-1
AP) = (detP-1
)(detA)(detP) = (1/detP)(detA)
(det P) = det A.

15. Example
Show 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.

16. Trace
The trace of a square matrix A (tr A) is the sum of the entries
on the main diagonal of A.

17. 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
bj1
j=1
n
∑
a2 jbj2
j=1
n
∑
...
anj
bjn
j=1
n
∑


















tr(AB) = aijbji
j=1
n
∑






i=1
n
∑ = bjiaij
i=1
n
∑


j=1
n
∑ = tr(BA)

18. 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-1
AP
rank (B) = rank (P-1
AP) = rank(AP)=rankA since P is invertible
(and using cor 4 of thm 4 in 5.5)
tr B = tr (P-1
AP) = tr[(AP)P-1
] = tr (A) (uses thm 3)

19. Theorem 4 - cont
Characteristic polynomial
cB(x) = det (xI - B) = det(xI - P-1
AP)=det(P-1
xIP - P-1
AP)
(since xI = P-1
xIP -- 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.


20. 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-1
IP = I A≠

21. 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-1
AP
Let Q = P-1
, then B = QAQ-1
, so A= Q-1
BQ
Which means that B ~ A.

22. 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)