Exercise 5.4.32

Answers

As Hint, we induct on the dimension n of the space. For n = 1, every matrix representation is an upper triangular matrix. Suppose the hypothesis is true for n = k 1. Now we consider the case n = k. Since the characteristic polynomial splits, T must has an eigenvector v corresponding to eigenvalue λ. Let W = span({v}). By Exercise 1.6.35 we know that the dimension of VW is equal to k 1. So we apply the induction hypothesis to T¯ defined in Exercise 5.4.27. This means we may find a basis

α = {u2 + W,u3 + W,,uk + W}

for VW such that [T¯]α is an upper triangular matrix. Following the argument in Exercise 5.4.30, we know that

β = {v,u2,u3,,uk}

is a basis for V . And we also know the matrix representation is

[T]β = (λ O [T ¯ ]α ) ,

which is an upper triangular matrix since [T¯]α is upper triangular.

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2011-06-27 00:00
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