Page 103 - Linear Algebra Done Right

Chapter 5. Eigenvalues and Eigenvectors
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At this stage of the proof we know that (a), (b), and (c) are all equiv-
alent. We will finish the proof by showing that (b) implies (d), that (d)
implies (e), and that (e) implies (b).
Suppose that (b) holds; thus V has a basis consisting of eigenvectors
of T. Thus every vector in V is a linear combination of eigenvectors
of T. Hence
5.22 V = null(T − λ 1 I) +· · ·+ null(T − λ m I).
To show that the sum above is a direct sum, suppose that
0 = u 1 + ··· + u m ,

where each u j ∈ null(T − λ j I). Because nonzero eigenvectors corre-
sponding to distinct eigenvalues are linearly independent, this implies
(apply 5.6 to the sum of the nonzero vectors on the right side of the
equation above) that each u j equals 0. This implies (using 1.8) that the
sum in 5.22 is a direct sum, completing the proof that (b) implies (d).
That (d) implies (e) follows immediately from Exercise 17 in Chap-
ter 2.
Finally, suppose that (e) holds; thus
5.23 dim V = dim null(T − λ 1 I) +· · ·+ dim null(T − λ m I).

Choose a basis of each null(T − λ j I); put all these bases together to
form a list (v 1 ,...,v n ) of eigenvectors of T, where n = dim V (by 5.23).
To show that this list is linearly independent, suppose

a 1 v 1 +· · ·+ a n v n = 0,
where a 1 ,...,a n ∈ F. For each j = 1,...,m, let u j denote the sum of
all the terms a k v k such that v k ∈ null(T − λ j I). Thus each u j is an
eigenvector of T with eigenvalue λ j , and
u 1 +· · ·+ u m = 0.

Because nonzero eigenvectors corresponding to distinct eigenvalues
are linearly independent, this implies (apply 5.6 to the sum of the
nonzero vectors on the left side of the equation above) that each u j
equals 0. Because each u j is a sum of terms a k v k , where the v k ’s
were chosen to be a basis of null(T − λ j I), this implies that all the a k ’s
equal 0. Thus (v 1 ,...,v n ) is linearly independent and hence is a basis
of V (by 2.17). Thus (e) implies (b), completing the proof.