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Exercise 5.4.28

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We use the notation given in Hint. Since $W$ is $T$ -invariant, we know the matrix representation of $T$ is

{[T]}_{β} = (\begin{matrix} B_{1} & B_{2} \\ O & B_{3} \end{matrix}) .

As the proof of Theorem 5.21, we know that $f (t) = \det ({[T]}_{β} - tI)$ and $g (t) = \det (B_{1} - tI)$ . It’s enough to show $h (t) = \det (B_{3} - tI)$ by showing $B_{3}$ is a matrix representation of $\bar{T}$ . Let

α = {v_{k} + W, v_{k + 1} + W, \dots, v_{n} + W}

be a basis for $V ∕ W$ by Exercise 1.6.35. Then for each $j = k, k + 1, \dots, n$ , we have

\bar{T} (v_{j}) = T (v_{j}) + W

= [\sum_{l = 1}^{k} {(B_{2})}_{lj} v_{l} + \sum_{i = k + 1}^{n} {(B_{3})}_{ij} v_{i}] + W

= \sum_{i = k + 1}^{n} {(B_{3})}_{ij} v_{i} + W = \sum_{i = k + 1}^{n} {(B_{3})}_{ij} (v_{i} + W) .

So we have $B_{3} = {[\bar{T}]}_{α}$ and

f (t) = \det ({[T]}_{β} - tI) = \det (B_{1} - tI) \det (B_{3} - tI) = g (t) h (t) .

jephianlin

2011-06-27 00:00

Exercise 5.4.28

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