Page 278 - 35Linear Algebra

278 Diagonalizing Symmetric Matrices

Example 140 For a general symmetric 2 × 2 matrix, we have:

a b λ − a −b
P λ = det
b d −b λ − d
= (λ − a)(λ − d) − b 2
2
2
= λ − (a + d)λ − b + ad
s
a + d a − d 2
2
⇒ λ = ± b + .
2 2
2
2
Notice that the discriminant 4b + (a − d) is always positive, so that the eigenvalues
must be real.

Now, suppose a symmetric matrix M has two distinct eigenvalues λ 6= µ
and eigenvectors x and y;

Mx = λx, My = µy.

T
T
Consider the dot product x y = x y = y x and calculate
T
T
x My = x µy = µx y, and
T
T
T
x My = (y Mx) (by transposing a 1 × 1 matrix)
T
= (y λx) T
= (λx y) T
= λx y.

Subtracting these two results tells us that:

T
T
0 = x My − x My = (µ − λ) x y.
Since µ and λ were assumed to be distinct eigenvalues, λ − µ is non-zero,
and so x y = 0. We have proved the following theorem.

Theorem 15.0.1. Eigenvectors of a symmetric matrix with distinct eigen-
values are orthogonal.

Reading homework: problem 1

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