Page 144 - Linear Algebra Done Right

Chapter 7. Operators on Inner-Product Spaces
132
If T ∈L(V) is normal, then eigenvectors of T
Corollary:
7.8
corresponding to distinct eigenvalues are orthogonal.
Proof: Suppose T ∈L(V) is normal and α, β are distinct eigen-
values of T, with corresponding eigenvectors u, v. Thus Tu = αu and
¯
Tv = βv. From 7.7 we have T v = βv. Thus
∗
¯
(α − β) u, v = αu, v −Pu, βv
∗
= Tu, v −Pu, T v
= 0.
Because α = β, the equation above implies that u, v = 0. Thus u and
v are orthogonal, as desired.
The Spectral Theorem

Recall that a diagonal matrix is a square matrix that is 0 everywhere
except possibly along the diagonal. Recall also that an operator on V
has a diagonal matrix with respect to some basis if and only if there is
a basis of V consisting of eigenvectors of the operator (see 5.21).
The nicest operators on V are those for which there is an ortho-
normal basis of V with respect to which the operator has a diagonal
matrix. These are precisely the operators T ∈L(V) such that there is
an orthonormal basis of V consisting of eigenvectors of T. Our goal
in this section is to prove the spectral theorem, which characterizes
these operators as the normal operators when F = C and as the self-
adjoint operators when F = R. The spectral theorem is probably the
most useful tool in the study of operators on inner-product spaces.
Because the conclusion of the spectral theorem depends on F,we
will break the spectral theorem into two pieces, called the complex
spectral theorem and the real spectral theorem. As is often the case in
linear algebra, complex vector spaces are easier to deal with than real
vector spaces, so we present the complex spectral theorem first.
As an illustration of the complex spectral theorem, consider the
2
normal operator T ∈L(C ) whose matrix (with respect to the standard
basis) is

2 −3
3 2 .
You should verify that