Page 147 - Linear Algebra Done Right

The Spectral Theorem
where the first inequality holds by the Cauchy-Schwarz inequality (6.6).
2
The last inequality implies that (T +αT +βI)v = 0. Thus T +αT +βI
is injective, which implies that it is invertible (see 3.21). 2 135
We have proved that every operator, self-adjoint or not, on a finite-
dimensional complex vector space has an eigenvalue (see 5.10), so the
next lemma tells us something new only for real inner-product spaces.
7.12 Lemma: Suppose T ∈L(V) is self-adjoint. Then T has an
eigenvalue.

Proof: As noted above, we can assume that V is a real inner-
product space. Let n = dim V and choose v ∈ V with v = 0. Then Here we are imitating
the proof that T has an
n
2
(v, Tv, T v,...,T v)
invariant subspace of
dimension 1 or 2
cannot be linearly independent because V has dimension n and we have
(see 5.24).
n + 1 vectors. Thus there exist real numbers a 0 ,...,a n , not all 0, such
that
n
0 = a 0 v + a 1 Tv +· · ·+ a n T v.
Make the a’s the coefficients of a polynomial, which can be written in
factored form (see 4.14) as

a 0 + a 1 x + ··· + a n x n
2
2
= c(x + α 1 x + β 1 )...(x + α M x + β M )(x − λ 1 )...(x − λ m ),
where c is a nonzero real number, each α j , β j , and λ j is real, each
2
α j < 4β j , m + M ≥ 1, and the equation holds for all real x. We then
have
n
0 = a 0 v + a 1 Tv +· · ·+ a n T v
n
= (a 0 I + a 1 T + ··· + a n T )v
2
2
= c(T + α 1 T + β 1 I)...(T + α M T + β M I)(T − λ 1 I)...(T − λ m I)v.
2
Each T + α j T + β j I is invertible because T is self-adjoint and each
2
α j < 4β j (see 7.11). Recall also that c = 0. Thus the equation above
implies that
0 = (T − λ 1 I)...(T − λ m I)v.

Hence T − λ j I is not injective for at least one j. In other words, T has
an eigenvalue.