Linear Algebra II (pdf, 500 kB)

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10.1. Lemma. Let V be an inner product space and f : V → V an endomorphism.
If f is orthodiagonalizable, then f is normal.
Proof. If f is orthodiagonalizable, then there is an orthonormal basis e1, . . . , en
of V such that f is represented by a diagonal matrix D with respect to this basis.
Now D is normal, hence so is f, by Cor. 9.18.
The proof of the other direction is a little bit more involved. We begin with the
following partial result.
10.2. Lemma. Let V be an inner product space, and let f : V → V be normal.
(1) f ∗ (v) = f(v).
(2) If f(v) = λv for some v ∈ V , then f ∗ (v) = ¯ λv.
(3) If f(v) = λv and f(w) = µw with λ = µ, then v ⊥ w (i.e., 〈v, w〉 = 0).
Proof. For the first statement, note that
f ∗ (v) 2 = 〈f ∗ (v), f ∗ (v)〉 = 〈f f ∗ (v) , v〉
For the second statement, note that
and so
= 〈f ∗ f(v) , v〉 = 〈f(v), f(v)〉 = f(v) 2 .
〈f ∗ (v), f ∗ (v)〉 = 〈f(v), f(v)〉 = |λ| 2 〈v, v〉
〈 ¯ λv, f ∗ (v)〉 = ¯ λ〈f(v), v〉 = ¯ λ〈λv, v〉 = |λ| 2 〈v, v〉
〈f ∗ (v), ¯ λv〉 = λ〈v, f(v)〉 = λ〈v, λv〉 = |λ| 2 〈v, v〉
〈 ¯ λv, ¯ λv〉 = |λ| 2 〈v, v〉
〈f ∗ (v)− ¯ λv, f ∗ (v)− ¯ λv〉 = 〈f ∗ (v), f ∗ (v)〉−〈 ¯ λv, f ∗ (v)〉−〈f ∗ (v), ¯ λv〉+〈 ¯ λv, ¯ λv〉 = 0 .
For the last statement, we compute
λ〈v, w〉 = 〈f(v), w〉 = 〈v, f ∗ (w)〉 = 〈v, ¯µw〉 = µ〈v, w〉 .
Since λ = µ by assumption, we must have 〈v, w〉 = 0.
This result shows that the various eigenspaces are orthogonal in pairs, and we
conclude that when f is a normal endomorphism of a complex inner product
space, it is orthodiagonalizable if it is just diagonalizable. It remains to prove
that this is the case.
10.3. Lemma. Let V be an inner product space over the field F = R or C, let
f : V → V be normal, and let p ∈ F [X] be a polynomial. Then p(f) is also
normal.
Proof. Let p(x) = amx m + · · · + a0. Then by Prop. 9.14,
p(f) ∗ = (amf ◦m + · · · + a1f + a0 idV ) ∗ = ām(f ∗ ) ◦m + · · · + ā1f ∗ + ā0 idV = ¯p(f ∗ ) ,
where ¯p is the polynomial whose coefficients are the complex conjugates of those
of p. (If F = R, then p(f) ∗ = p(f ∗ ).) Now p(f) and p(f) ∗ = ¯p(f ∗ ) commute since
f and f ∗ do, hence p(f) is normal.