Page 143 - Linear Algebra Done Right

Self-Adjoint and Normal Operators
Proposition: An operator T ∈L(V) is normal if and only if
7.6
proposition implies
∗
Tv = T v Note that this ∗ 131
that null T = null T
for every normal
for all v ∈ V.
operator T.
Proof: Let T ∈L(V). We will prove both directions of this result
at the same time. Note that
∗
T is normal ⇐⇒ T T − TT ∗ = 0
⇐⇒ (T T − TT )v, v = 0 for all v ∈ V
∗
∗
⇐⇒ T Tv, v = TT v, v for all v ∈ V
∗
∗
2 ∗ 2
⇐⇒ Tv = T v for all v ∈ V,
where we used 7.4 to establish the second equivalence (note that the
operator T T − TT ∗ is self-adjoint). The equivalence of the first and
∗
last conditions above gives the desired result.
Compare the next corollary to Exercise 28 in the previous chapter.
That exercise implies that the eigenvalues of the adjoint of any operator
are equal (as a set) to the complex conjugates of the eigenvalues of the
operator. The exercise says nothing about eigenvectors because an
operator and its adjoint may have different eigenvectors. However, the
next corollary implies that a normal operator and its adjoint have the
same eigenvectors.

7.7 Corollary: Suppose T ∈L(V) is normal. If v ∈ V is an eigen-
vector of T with eigenvalue λ ∈ F, then v is also an eigenvector of T ∗
¯
with eigenvalue λ.
Proof: Suppose v ∈ V is an eigenvector of T with eigenvalue λ.
Thus (T − λI)v = 0. Because T is normal, so is T − λI, as you should
verify. Using 7.6, we have

¯
0 = (T − λI)v = (T − λI) v = (T − λI)v ,
∗
∗
¯
and hence v is an eigenvector of T ∗ with eigenvalue λ, as desired.
Because every self-adjoint operator is normal, the next result applies
in particular to self-adjoint operators.