13.13. SPECTRAL THEORY OF SELF ADJOINT OPERATORS 369

Proof : If vk is one of the orthonormal basis vectors, Avk = λ kvk. Also,

∑i

λ ivi⊗ vi (vk) = ∑i

λ ivi (vk,vi) = ∑i

λ iδ ikvi = λ kvk.

Since the two linear transformations agree on a basis, it follows they must coincide. ■By Proposition 13.13.2 this says the matrix of A with respect to this basis {vi}n

i=1 is thediagonal matrix having the eigenvalues λ 1, · · · ,λ n down the main diagonal.

The result of Courant and Fischer which follows resembles Corollary 13.13.5 but ismore useful because it does not depend on a knowledge of the eigenvectors.

Theorem 13.13.8 Let A ∈L (X ,X) be self adjoint where X is a finite dimensional innerproduct space. Then for λ 1 ≤ λ 2 ≤ ·· · ≤ λ n the eigenvalues of A, there exist orthonormalvectors {u1, · · · ,un} for which

Auk = λ kuk.

Furthermore,

λ k ≡ maxw1,··· ,wk−1

{min

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

}}(13.26)

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

Proof: From Theorem 13.13.4, there exist eigenvalues and eigenvectors {u1, · · · ,un}which are orthonormal and λ i ≤ λ i+1.

(Ax,x) =n

∑j=1

(Ax,u j)(x,u j) =n

∑j=1

λ j (x,u j)(u j,x) =n

∑j=1

λ j∣∣(x,u j)

∣∣2Recall that (z,w) = ∑ j (z,u j)(w,ui). Then let Y = {w1, · · · ,wk−1}⊥

inf{(Ax,x) : |x|= 1,x ∈ Y}= inf

{n

∑j=1

λ j∣∣(x,u j)

∣∣2 : |x|= 1,x ∈ Y

}

≤ inf

{k

∑j=1

λ j∣∣(x,u j)

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

}. (13.27)

The reason this is so is that the infimum is taken over a smaller set. Therefore, the infimumgets larger. Now 13.27 is no larger than

inf

{λ k

n

∑j=1

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

}≤ λ k

because since {u1, · · · ,un} is an orthonormal basis, |x|2 = ∑nj=1

∣∣(x,u j)∣∣2 . It follows, since

{w1, · · · ,wk−1}

is arbitrary,

supw1,··· ,wk−1

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

}}≤ λ k. (13.28)