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Exercise 5.4.32
Answers
As Hint, we induct on the dimension of the space. For , every matrix representation is an upper triangular matrix. Suppose the hypothesis is true for . Now we consider the case . Since the characteristic polynomial splits, must has an eigenvector corresponding to eigenvalue . Let . By Exercise 1.6.35 we know that the dimension of is equal to . So we apply the induction hypothesis to defined in Exercise 5.4.27. This means we may find a basis
for such that is an upper triangular matrix. Following the argument in Exercise 5.4.30, we know that
is a basis for . And we also know the matrix representation is
which is an upper triangular matrix since is upper triangular.