12.3. SPECTRAL THEORY OF SELF ADJOINT OPERATORS 303

Proof : If vk is one of the orthonormal basis vectors, Avk = λkvk. Also,∑i

λivi ⊗ vi (vk) =∑i

λivi (vk, vi) =∑i

λiδikvi = λkvk.

Since the two linear transformations agree on a basis, it follows they must coincide. ■By Theorem 11.4.5 this says the matrix of A with respect to this basis {vi}ni=1 is the

diagonal matrix having the eigenvalues λ1, · · · , λn down the main diagonal.The result of Courant and Fischer which follows resembles Corollary 12.3.3 but is more

useful because it does not depend on a knowledge of the eigenvectors.

Theorem 12.3.6 Let A ∈ L (X,X) be self adjoint where X is a finite dimensional innerproduct space. Then for λ1 ≤ λ2 ≤ · · · ≤ λn the eigenvalues of A, there exist orthonormalvectors {u1, · · · , un} for which

Auk = λkuk.

Furthermore,

λk ≡ maxw1,··· ,wk−1

{min

{(Ax, x) : |x| = 1, x ∈ {w1, · · · , wk−1}⊥

}}(12.7)

where if k = 1, {w1, · · · , wk−1}⊥ ≡ X.

Proof: From Theorem 12.3.2, there exist eigenvalues and eigenvectors with {u1, · · · , un}orthonormal and λi ≤ λi+1.

(Ax, x) =

n∑j=1

(Ax, uj) (x, uj) =

n∑j=1

λj (x, uj) (uj , x) =

n∑j=1

λj |(x, uj)|2

Recall that (z, w) =∑

j (z, uj) (w, ui). Then let Y = {w1, · · · , wk−1}⊥

inf {(Ax, x) : |x| = 1, x ∈ Y } = inf

n∑

j=1

λj |(x, uj)|2 : |x| = 1, x ∈ Y

≤ inf

k∑

j=1

λj |(x, uj)|2 : |x| = 1, (x, uj) = 0 for j > k, and x ∈ Y

 . (12.8)

The reason this is so is that the infimum is taken over a smaller set. Therefore, the infimumgets larger. Now 12.8 is no larger than

inf

λkn∑

j=1

|(x, uj)|2 : |x| = 1, (x, uj) = 0 for j > k, and x ∈ Y

 ≤ λk

because since {u1, · · · , un} is an orthonormal basis, |x|2 =∑n

j=1 |(x, uj)|2. It follows, since

{w1, · · · , wk−1} is arbitrary,

supw1,··· ,wk−1

{inf{(Ax, x) : |x| = 1, x ∈ {w1, · · · , wk−1}⊥

}}≤ λk. (12.9)

Then from Corollary 12.3.3,

λk = inf{(Ax, x) : |x| = 1, x ∈ {u1, · · · , uk−1}⊥

}≤