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Exercise 5.4.31

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1.

Compute

T (e_{1}) = (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}), T^{2} (e_{1}) = (\begin{matrix} 0 \\ 12 \\ 6 \end{matrix})

and

T^{2} (e_{1}) = - 6 e_{1} + 6 T (e_{1}) .

This means that the characteristic polynomial of $T_{W}$ is $t^{2} - 6 t + 6$ .

2.

We know that

\dim (ℝ^{3} ∕ W) = 3 - \dim (W) = 1

by Exercise 1.6.35. So every nonzero element in $ℝ^{3} ∕ W$ is a basis. Since $e_{2}$ is not in $W$ , we have $e_{2} + W$ is a basis for $ℝ^{3} ∕ W$ . Now let $β = {e_{2} + W}$ . We may compute

\bar{T} (e_{2} + W) = (\begin{matrix} 1 \\ 3 \\ 2 \end{matrix}) + W = - e_{2} + W

and ${[\bar{T}]}_{β} = (\begin{matrix} 1 \end{matrix})$ . So the characteristic polynomial of $\bar{T}$ is $- t - 1$ .

3.

Use the result in Exercise 5.4.28, we know the characteristic polynomial of

T

- (t + 1) (t^{2} - 6 t + t) .

jephianlin

2011-06-27 00:00

Exercise 5.4.31

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