Page 92 - Linear Algebra Done Right

Invariant Subspaces
Theorem: Let T ∈L(V). Suppose λ 1 ,...,λ m are distinct eigen-
5.6
values of T and v 1 ,...,v m are corresponding nonzero eigenvectors.
Then (v 1 ,...,v m ) is linearly independent. 79
Proof: Suppose (v 1 ,...,v m ) is linearly dependent. Let k be the
smallest positive integer such that
5.7 v k ∈ span(v 1 ,...,v k−1 );

the existence of k with this property follows from the linear dependence
lemma (2.4). Thus there exist a 1 ,...,a k−1 ∈ F such that

5.8 v k = a 1 v 1 + ··· + a k−1 v k−1 .

Apply T to both sides of this equation, getting

λ k v k = a 1 λ 1 v 1 + ··· + a k−1 λ k−1 v k−1 .
Multiply both sides of 5.8 by λ k and then subtract the equation above,
getting
0 = a 1 (λ k − λ 1 )v 1 + ··· + a k−1 (λ k − λ k−1 )v k−1 .

Because we chose k to be the smallest positive integer satisfying 5.7,
(v 1 ,...,v k−1 ) is linearly independent. Thus the equation above implies
that all the a’s are 0 (recall that λ k is not equal to any of λ 1 ,...,λ k−1 ).
However, this means that v k equals 0 (see 5.8), contradicting our hy-
pothesis that all the v’s are nonzero. Therefore our assumption that
(v 1 ,...,v m ) is linearly dependent must have been false.

The corollary below states that an operator cannot have more dis-
tinct eigenvalues than the dimension of the vector space on which it
acts.

5.9 Corollary: Each operator on V has at most dim V distinct eigen-
values.

Proof: Let T ∈L(V). Suppose that λ 1 ,...,λ m are distinct eigenval-
ues of T. Let v 1 ,...,v m be corresponding nonzero eigenvectors. The
last theorem implies that (v 1 ,...,v m ) is linearly independent. Thus
m ≤ dim V (see 2.6), as desired.