258 CHAPTER 12. SPECTRAL THEORY

such thatr

∑j=1

c jw j = 0 (12.7)

where each c j ̸= 0. Say Mw j = µ jw j where

{µ1, · · · ,µr} ⊆ {λ 1, · · · ,λ k} ,

the µ j being distinct eigenvalues of M. Out of all such subsets, let this one be such that ris as small as possible. Then necessarily, r > 1 because otherwise, c1w1 = 0 which wouldimply w1 = 0, which is not allowed for eigenvectors.

Now apply M to both sides of 12.7.

r

∑j=1

c jµ jw j = 0. (12.8)

Next pick µk ̸= 0 and multiply both sides of 12.7 by µk. Such a µk exists because r > 1.Thus

r

∑j=1

c jµkw j = 0 (12.9)

Subtract the sum in 12.9 from the sum in 12.8 to obtainr

∑j=1

c j

(µk−µ j

)w j = 0

Now one of the constants c j

(µk−µ j

)equals 0, when j = k. Therefore, r was not as small

as possible after all. ■Here is another proof in case you did not follow the above.

Theorem 12.1.16 Suppose Mvi = λ ivi, i = 1, · · · ,r , vi ̸= 0, and that if i ̸= j, then λ i ̸= λ j.Then the set of eigenvectors, {v1, · · · ,vr} is linearly independent.

Proof: Suppose the conclusion is not true. Then in the matrix(v1 v2 · · · vr

)not every column is a pivot column. Let the pivot columns be {w1, · · · ,wk}, k < r. Thenthere exists v ∈ {v1, · · · ,vr} , Mv =λ vv, v /∈{w1, · · · ,wk} , such that

v =k

∑i=1

ciwi. (12.10)

Then doing M to both sides yields

λ vv =k

∑i=1

ciλ wiwi (12.11)

But also you could multiply both sides of 12.10 by λ v to get

λ vv =k

∑i=1

ciλ vwi.