Page 280 - 35Linear Algebra

280 Diagonalizing Symmetric Matrices

T
But P is an orthogonal matrix, so P −1 = P . Then:
 
x T
1
.
P −1 = P T =  . 
 . 
x T
n
 T 
x λ 1 x 1 ∗ · · · ∗
1
 T 
x λ 1 x 1 ∗ · · · ∗
T  2 
.
⇒ P MP =  . . .
 . . 
T
x λ 1 x 1 ∗ · · · ∗
n
 
λ 1 ∗ · · · ∗
 0 ∗ · · · ∗ 

=  . . . .

 . ∗ . 
0 ∗ · · · ∗
 
λ 1 0 · · · 0
 0 
 
=  .  .
ˆ
 . . M 
0
T
The last equality follows since P MP is symmetric. The asterisks in the
ˆ
matrix are where “stuff” happens; this extra information is denoted by M
ˆ
in the final expression. We know nothing about M except that it is an
(n − 1) × (n − 1) matrix and that it is symmetric. But then, by finding an
ˆ
(unit) eigenvector for M, we could repeat this procedure successively. The
end result would be a diagonal matrix with eigenvalues of M on the diagonal.
Again, we have proved a theorem:

Theorem 15.0.2. Every symmetric matrix is similar to a diagonal matrix
of its eigenvalues. In other words,

T
M = M ⇔ M = PDP T
where P is an orthogonal matrix and D is a diagonal matrix whose entries
are the eigenvalues of M.

Reading homework: problem 2

280