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Please let me know if my process or thinking is incorrect at any point.

Let $T:P_3 \rightarrow P_3$ be the linear transformation such that $$T(-2 x^2)= 3 x^2 + 3 x,\\T(0.5 x + 4)= -2 x^2 - 2 x - 3, \\ T(2 x^2 - 1) = -3 x + 2$$

The first question based on this information is :

Find $T(1)$.

I know that a transformation is defined by its behaviour on a basis. I took the basis $$\mathcal{B} = \{\begin{bmatrix}0\\0\\-2\end{bmatrix},\begin{bmatrix}4\\1/2\\0\end{bmatrix},\begin{bmatrix}-1\\0\\2\end{bmatrix}\}$$

Applying the transformation to the first basis vector gives $3 x^2 + 3 x$ which can be be made with the following linear combination:

$$\frac{45}{2}\begin{bmatrix}0\\0\\-2\end{bmatrix} + 6\begin{bmatrix}4\\1/2\\0\end{bmatrix} + 24\begin{bmatrix}-1\\0\\2\end{bmatrix}$$

Is this correct? Does this mean the first column in my matrix representation of the transformation is

$$\begin{bmatrix}\frac{45}2&.&.\\6&.&.\\24&.&.\end{bmatrix}?$$

If it isn't, please let me know why. If it is, am I not correct in saying that

$$T(1) = T(\begin{bmatrix}1\\0\\0\end{bmatrix}) = \begin{bmatrix}\frac{45}2&.&.\\6&.&.\\24&.&.\end{bmatrix}\begin{bmatrix}1\\0\\0\end{bmatrix} = \frac{45}2 + 6x + 24x^2?$$

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    $\begingroup$ The problem is that $1$ isn't represented as $\pmatrix{1\\0\\0}$ in your basis. $\endgroup$ – AlexR Apr 27 '15 at 13:07
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The first part is okay, but the second one (finding $T(1)$) isn't. Note that linear transformations give you $$T(1) = -T(-2x^2 + (2x^2 - 1)) = -T(-2x^2) - T(2x^2 - 1) \\ = -(3x^2 - 3x) - (-3x +2) = -3x^2 - 2$$

With your approach, you'd have to find that $$1 = -1\cdot (-2x^2) + 0\cdot (\frac12 x + 4) - 1\cdot(2x^2 - 1) = \pmatrix{-1\\0\\-1}_{\mathcal B}$$ So you'd need to find the third column of $[T]_{\mathcal B}$ as well.

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  • $\begingroup$ I was JUST about to ask you to alter yours to match my approach. I understand now. Thank you for giving me both versions. $\endgroup$ – aidandeno Apr 27 '15 at 13:18
  • $\begingroup$ There is a small mistake in your final line of calculations though. The last vector is $-1 + 2x^2$. $\endgroup$ – aidandeno Apr 27 '15 at 13:25
  • $\begingroup$ @aidandeno Yes, that was a typo. Thanks! $\endgroup$ – AlexR Apr 27 '15 at 13:29
  • $\begingroup$ Does it not change your final vector to $\pmatrix{-1\\0\\-1}?$ $\endgroup$ – aidandeno Apr 27 '15 at 13:33
  • $\begingroup$ @aidandeno It does, thanks again for double checking. $\endgroup$ – AlexR Apr 27 '15 at 13:34

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