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A linear transformation T is defined by

T: $\Bbb R^3$ $\rightarrow$ $\Bbb R^2$ $\Rightarrow$ T$\begin{pmatrix}x\\y\\z\end{pmatrix}$ = $\begin{pmatrix}2x+y\\y+2z\end{pmatrix}$

Find bases $\mathscr B_3'$ and $\mathscr B_2'$ for $\Bbb R^3$ and $\Bbb R^2$ respectively such that $Mat_{\mathscr B_3' , \mathscr B_2'}$ (T) = $\begin{pmatrix} 1&0&-1\\0&1&2 \end{pmatrix}$

So far I have attempted to answer this by producing an arbitrary basis for $\Bbb R^3$, and then using the definition of the transformation matrix, express it as a linear combination of the basis vectors that will make up the basis for $\Bbb R^2$. If my arbitrary basis for $\Bbb R^3$ is {$\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}$,$\begin{pmatrix}a_2\\b_2\\c_2\end{pmatrix}$,$\begin{pmatrix}a_3\\b_3\\c_3\end{pmatrix}$}

then I can simplify this to $(\alpha_2+2\alpha_3) \begin{pmatrix}2a_2+b_2\\b_2+2c_2\end{pmatrix}$ + $(\alpha_1-\alpha_3) \begin{pmatrix}2a_1+b_1\\b_1+2c_1\end{pmatrix}$

I'm not sure what else to try or how to proceed if this is the correct approach

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  • $\begingroup$ I suggest working backwards from a basis for the image of $T$, as in the answer to your previous question. $\endgroup$ – amd Oct 29 '18 at 23:41
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I do not think your approach is correct.

We can simplify the answer by taking the ordered bases $$\begin{align} \mathscr B_3' & = \left\{ \varepsilon_1 = \begin{pmatrix}1\\0\\0\end{pmatrix}, \varepsilon_2 = \begin{pmatrix}0\\1\\0\end{pmatrix}, \varepsilon_3 = \begin{pmatrix}0\\0\\1\end{pmatrix}\right\},\\\\ \mathscr B_2' & = \left\{ \begin{pmatrix}\beta_{11}\\ \beta_{12} \end{pmatrix}, \begin{pmatrix}\beta_{21}\\ \beta_{22} \end{pmatrix}\right\}\end{align}$$ and solving for $\mathscr B_2'.$ The simplification is that with our choice for $\mathscr B_3',$ coordinate matrices in $\Bbb R^3$ have the same entries as vectors in $\Bbb R^3,$ so we do not have to worry about converting from one to the other.

Now the $i$th column of Mat$_{\mathscr B_3', \mathscr B_2'}(T)$ is $[T\varepsilon_i]_{\mathscr B_2'}$ where $[ \cdot ]_{\mathscr B_2'}$ denotes the coordinate matrix with respect to ordered basis $\mathscr B_2'.$ That gives us the following three systems of equations: $$\begin{align} \begin{pmatrix} 1\\ 0\end{pmatrix} & = \left[ T \begin{pmatrix} 1\\ 0\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\ & = \left[ \begin{pmatrix} 2\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\ 1 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} + 0 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & = \begin{pmatrix} 2 \\ 0 \end{pmatrix},\\\\ \end{align}$$


$$\begin{align} \begin{pmatrix} 0\\ 1\end{pmatrix} & = \left[ T \begin{pmatrix} 0\\ 1\\ 0\end{pmatrix} \right]_{\mathscr B_2'}\\\\ & = \left[ \begin{pmatrix} 1\\ 1\end{pmatrix} \right]_{\mathscr B_2'}\\\\ 0 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} + 1 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & = \begin{pmatrix} 1 \\ 1 \end{pmatrix},\\\\ \end{align}$$


$$\begin{align} \begin{pmatrix} -1\\ 2\end{pmatrix} & = \left[ T \begin{pmatrix} 0\\ 0\\ 1\end{pmatrix} \right]_{\mathscr B_2'}\\\\ & = \left[ \begin{pmatrix} 0\\ 2\end{pmatrix} \right]_{\mathscr B_2'}\\\\ -1 \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} + 2 \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix} & = \begin{pmatrix} 0 \\ 2 \end{pmatrix}.\end{align}$$ Solving for the four $\beta$s, we get $$\mathscr B_2' = \left\{ \begin{pmatrix} 2\\ 0\end{pmatrix}, \begin{pmatrix} 1\\ 1\end{pmatrix}\right\}.$$ As amd stated in his answer to your previous question, other bases will satisfy the hypotheses of your question.

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I do not think your approach is correct.

We can simplify the answer by taking the ordered bases $$\begin{align} \mathscr B_3' & = \left\{ \begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\end{pmatrix}\right\},\\\\ \mathscr B_2' & = \left\{ \begin{pmatrix}\beta_{11}\\ \beta_{12} \end{pmatrix}, \begin{pmatrix}\beta_{21}\\ \beta_{22} \end{pmatrix}\right\}\end{align}$$ and solving for $\mathscr B_2'.$ The simplification is that with our choice for $\mathscr B_3',$ coordinate matrices in $\Bbb R^3$ have the same entries as vectors in $\Bbb R^3,$ so we do not have to worry about converting from one to the other. Then with $[ \cdot ]_{\mathscr B_2'}$ denoting the coordinate matrix with respect to ordered basis $\mathscr B_2',$ we have $$\begin{align} \left[ T \begin{pmatrix} x\\ y\\ z\end{pmatrix} \right]_{\mathscr B_2'} & = \mbox{Mat}_{\mathscr B_3', \mathscr B_2'}(T) \begin{pmatrix} x\\ y\\ z\end{pmatrix}\\\\ \left[ \begin{pmatrix} 2x + y\\ y + 2z\end{pmatrix} \right]_{\mathscr B_2'} & = \begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & 2\end{pmatrix} \begin{pmatrix} x\\ y\\ z\end{pmatrix}\\\\ & = \begin{pmatrix} x - z\\ y + 2z\end{pmatrix}\\\\ \begin{pmatrix} 2x + y\\ y + 2z\end{pmatrix} & = (x - z) \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} + (y + 2z) \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix}\end{align}$$

To solve for the four $\beta$s, equate the coefficients of $x, y,$ and $z$ and solve to get $$\mathscr B_2' = \left\{ \begin{pmatrix} 2\\ 0\end{pmatrix}, \begin{pmatrix} 1\\ 1\end{pmatrix}\right\}.$$ As amd stated in his answer to your previous question, other bases will satisfy the hypotheses of your question.

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