Page 142 - Linear Algebra Done Right

Chapter 7. Operators on Inner-Product Spaces
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On a real inner-product space V, a nonzero operator T may satisfy
Tv, v = 0 for all v ∈ V. However, the next proposition shows that
this cannot happen for a self-adjoint operator.
7.4 Proposition: If T is a self-adjoint operator on V such that
Tv, v = 0
for all v ∈ V, then T = 0.
Proof: We have already proved this (without the hypothesis that
T is self-adjoint) when V is a complex inner-product space (see 7.2).
Thus we can assume that V is a real inner-product space and that T is
a self-adjoint operator on V. For u, w ∈ V, we have

T(u + w), u + w −PT(u − w), u − w
7.5 Tu, w = ;
4
this is proved by computing the right side, using

Tw, u = w, Tu
= Tu, w ,
where the first equality holds because T is self-adjoint and the second
equality holds because we are working on a real inner-product space.
If Tv, v = 0 for all v ∈ V, then 7.5 implies that Tu, w = 0 for all
u, w ∈ V. This implies that T = 0 (take w = Tu).

An operator on an inner-product space is called normal if it com-
mutes with its adjoint; in other words, T ∈L(V) is normal if
TT ∗ = T T.
∗
Obviously every self-adjoint operator is normal. For an example of a
normal operator that is not self-adjoint, consider the operator on F 2
whose matrix (with respect to the standard basis) is

2 −3
.
3 2
Clearly this operator is not self-adjoint, but an easy calculation (which
you should do) shows that it is normal.
We will soon see why normal operators are worthy of special at-
tention. The next proposition provides a simple characterization of
normal operators.