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Exercise 6.6.2

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We could calculate the projection of $(1, 0)$ and $(0, 1)$ are

\frac{⟨ (1, 0), (1, 2) ⟩}{∥ (1, 2) ∥^{2}} (1, 2) = \frac{1}{5} (1, 2)

and

\frac{⟨ (0, 1), (1, 2) ⟩}{∥ (1, 2) ∥^{2}} (1, 2) = \frac{2}{5} (1, 2)

by Theorem 6.6. So we have

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

On the other hand, we may do the same on $(1, 0, 0)$ , $(0, 1, 0)$ , and $(1, 0, 0)$ with respect to the new subspace $W = span ({(1, 0, 1)})$ . First compute

\frac{⟨ (1, 0, 0), (1, 0, 1) ⟩}{∥ (1, 0, 1) ∥^{2}} (1, 0, 1) = \frac{1}{2} (1, 0, 1),

\frac{⟨ (0, 1, 0), (1, 0, 1) ⟩}{∥ (1, 0, 1) ∥^{2}} (1, 0, 1) = 0 (1, 0, 1),

and

\frac{⟨ (0, 0, 1), (1, 0, 1) ⟩}{∥ (1, 0, 1) ∥^{2}} (1, 0, 1) = \frac{1}{2} (1, 0, 1) .

Hence the matrix would be

{[T]}_{β} = \frac{1}{2} (\begin{matrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{matrix}) .

jephianlin

2011-06-27 00:00

Exercise 6.6.2

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