MA 242 December 8, 2008

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MA 242 December 8,2008More on symmetric matrices
Theorem.Let v
1
and v
2
be eigenvectors associated to distinct eigenvalues of a symmetric
matrix A.Then v
1
v
2
= 0.
Theorem.(Spectral Theorem for symmetric matrices) If A is an n n symmetric matrix,
then
1.A has n real eigenvalues (counted with multiplicity),
2.the geometric multiplicity of each eigenvalue is the same as its algebraic multiplicity,
and
3.distinct eigenspaces are mutually orthogonal.
Consequently,any symmetric matrix is orthogonally diagonalizable.
Note:Any matrix A that is orthogonally diagonalizable is symmetric.
1
MA 242 December 8,2008Example.Consider
A=
2
6
4
2 4 2
4 2 2
2 2 1
3
7
5
:
Once again with the aid of Mathematica,we see that A has two distinct real eigenvalues,
 = 7 and  = 2.We also have three linearly independent eigenvectors
v
1
=
2
6
4
2
2
1
3
7
5
;v
2
=
2
6
4
1
0
2
3
7
5
;and v
3
=
2
6
4
1
1
0
3
7
5
:
2
MA 242 December 8,2008Suppose that we have an orthogonal diagonalization of the form D= P
T
AP.Then
A= PDP
T
yields the spectral decomposition of A.
Lemma.Let
A=
2
6
4a
1a
2:::a
n3
7
5 and B =
2
6
4b
1b
2:::b
n3
7
5
be two n n matrices.Then AB
T
= a
1
b
T
1
+a
2
b
T
2
+:::+a
n
b
T
n
.
3
MA 242 December 8,2008Now apply the lemma to the product PDP
T
where P is the orthogonal matrix
P =
2
6
4u
1u
2:::u
n3
7
5:
Example.Let's determine the spectral decomposition for the 2 2 example
A=

3 1
1 3

:
We use the orthonormal basis of eigenvectors
u
1
=
2
6
6
6
4
1 p2
1p2
3
7
7
7
5
and u
2
=
2
6
6
6
4

1p2
1p2
3
7
7
7
5
:
4
MA 242 December 8,2008Example.Let's determine the spectral decomposition for the rst example
A=
2
6
4
5 2 0
2 4 2
0 2 3
3
7
5
:
We use the orthonormal basis of eigenvectors
u
1
=
2
6
6
6
6
6
6
6
6
4

23

23
13
3
7
7
7
7
7
7
7
7
5
;u
2
=
2
6
6
6
6
6
6
6
6
4
23

13
23
3
7
7
7
7
7
7
7
7
5
;and u
3
=
2
6
6
6
6
6
6
6
6
4

13
23
23
3
7
7
7
7
7
7
7
7
5
:
5