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Linear Algebra

Exercise 2.2.2

Answers

1.

We have

T (1, 0) = (2, 3, 1) = 2 (1, 0, 0) + 3 (0, 1, 0) + 1 (0, 0, 1)

and

T (0, 1) = (- 1, 4, 0) = - 1 (1, 0, 0) + 4 (0, 1, 0) + 0 (0, 0, 1)

. Hence we get

{[T]}_{β}^{γ} = (\begin{matrix} 2 & - 1 \\ 3 & 4 \\ 1 & 0 \end{matrix}) .

2.

{[T]}_{β}^{γ} = (\begin{matrix} 2 & 3 & - 1 \\ 1 & 0 & 1 \end{matrix}) .

3.

{[T]}_{β}^{γ} = (\begin{matrix} 2 & 1 & - 3 \end{matrix}) .

4.

{[T]}_{β}^{γ} = (\begin{matrix} 0 & 2 & 1 \\ - 1 & 4 & 5 \\ 1 & 0 & 1 \end{matrix}) .

5.

{[T]}_{β}^{γ} = (\begin{matrix} 1 & 0 & \dots & 0 \\ 1 & 0 & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 1 & 0 & \dots & 0 \end{matrix}) .

6.

{[T]}_{β}^{γ} = (\begin{matrix} 0 & 1 \\ ⋮ \\ 1 & 0 \end{matrix}) .

7.

{[T]}_{β}^{γ} = (\begin{matrix} 1 & 0 & \dots & 0 & 1 \end{matrix}) .

jephianlin

2011-06-27 00:00

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