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dim V vectors in V is a basis of V (see 2.17); thus (v 1 ,...,v dim V ) is a
basis of V. With respect to this basis consisting of eigenvectors, T has
a diagonal matrix. Diagonal Matrices 89
We close this section with a proposition giving several conditions
on an operator that are equivalent to its having a diagonal matrix with
respect to some basis.

5.21 Proposition: Suppose T ∈L(V). Let λ 1 ,...,λ m denote the For complex vector
distinct eigenvalues of T. Then the following are equivalent: spaces, we will extend
this list of equivalences
(a) T has a diagonal matrix with respect to some basis of V; later (see Exercises 16
(b) V has a basis consisting of eigenvectors of T; and 23 in Chapter 8).
(c) there exist one-dimensional subspaces U 1 ,...,U n of V, each in-
variant under T, such that

V = U 1 ⊕· · · U n ;

(d) V = null(T − λ 1 I) ⊕· · · null(T − λ m I);
(e) dim V = dim null(T − λ 1 I) + ··· + dim null(T − λ m I).

Proof: We have already shown that (a) and (b) are equivalent.
Suppose that (b) holds; thus V has a basis (v 1 ,...,v n ) consisting of

eigenvectors of T. For each j, let U j = span(v j ). Obviously each U j
is a one-dimensional subspace of V that is invariant under T (because
each v j is an eigenvector of T). Because (v 1 ,...,v n ) is a basis of V,
each vector in V can be written uniquely as a linear combination of
(v 1 ,...,v n ). In other words, each vector in V can be written uniquely
as a sum u 1 +· · ·+ u n , where each u j ∈ U j . Thus V = U 1 ⊕ ··· ⊕ U n .
Hence (b) implies (c).
Suppose now that (c) holds; thus there are one-dimensional sub-
spaces U 1 ,...,U n of V, each invariant under T, such that

V = U 1 ⊕ ··· ⊕ U n .

For each j, let v j be a nonzero vector in U j . Then each v j is an eigen-
vector of T. Because each vector in V can be written uniquely as a sum
u 1 +· · ·+u n , where each u j ∈ U j (so each u j is a scalar multiple of v j ),
we see that (v 1 ,...,v n ) is a basis of V. Thus (c) implies (b).