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Find the matrix representation of the linear transformation $\alpha:\mathbb{R}^3 \rightarrow \mathbb{R}^2$ defined by

$$\alpha : \begin{pmatrix} a\\b\\c \end{pmatrix}\rightarrow \begin{pmatrix} a+b+c\\b+c \end{pmatrix}$$ With respect to bases $\left\{ \begin{pmatrix} -1\\0\\2 \end{pmatrix},\begin{pmatrix} 0\\1\\1 \end{pmatrix},\begin{pmatrix} 3\\-1\\0 \end{pmatrix} \right\}$ of $\mathbb{R}^3$ and $\left\{ \begin{pmatrix} -1\\1 \end{pmatrix},\begin{pmatrix} 1\\0\end{pmatrix} \right\}$ of $\mathbb{R}^2$

My attemp: The first part is $$A_{B_1}=[\alpha(b_1) ~~\alpha(b_2)~~ \alpha(b_3)]=\left[\alpha\begin{pmatrix} -1\\0\\2 \end{pmatrix}~~\alpha\begin{pmatrix} 0\\1\\1 \end{pmatrix} ~~\alpha\begin{pmatrix} 3\\-1\\0 \end{pmatrix} \right]=\begin{pmatrix} 1 &2&2\\2 &2&-1 \end{pmatrix}$$ I'm confused about the second part , here $\alpha$ takes values in $\mathbb{R}^3$ so how can i proceed with finding matrix for vectors in $\mathbb{R}^2$ ?

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  • $\begingroup$ perhaps I'm misreading, but what do you mean by second part? I only see one question lol, which is to find the matrix representation of $\alpha$ relative to the given bases. Could you clarify EXACTLY what it is you're confused about (try to be specific so we can give better answers), because it seems like your confusion could be answered by re-examining the definition of matrix representation of a linear map relative to a choice of basis. $\endgroup$ – peek-a-boo Aug 22 '19 at 5:36
  • $\begingroup$ By second part I mean the representation with respect to the basis given in R2 $\endgroup$ – Zeno San Aug 22 '19 at 5:38
  • $\begingroup$ I suggest you take a look at this answer of mine math.stackexchange.com/questions/3257500/… Here, I gave a detailed answer about computing the matrix representation of a linear map, which I believe might be helpful for you. If you're still confused, let me know; I can try to answer specifically in your case (or someone else might do it in the meantime) $\endgroup$ – peek-a-boo Aug 22 '19 at 5:41
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What you have computed is the matrix for $\alpha$, with respect to the given basis on $\Bbb{R}^3$, but with respect to the standard basis on $\Bbb{R}^2$ (as opposed to the one given). If $B_2$ is the basis on $\Bbb{R}^2$, then the matrix you want is: $$\left[\begin{array}{c|c|c}\left[\alpha \begin{pmatrix}-1 \\ 0 \\ 2\end{pmatrix}\right]_{B_2} & \left[\alpha \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}\right]_{B_2} & \left[\alpha \begin{pmatrix}3 \\ -1 \\ 0\end{pmatrix}\right]_{B_2}\end{array}\right],$$ where $[v]_{B_2}$ refers to the coordinate column vector of $v$ with respect to basis $B_2$. That is, it will be $\begin{bmatrix} a \\ b \end{bmatrix}$, where $a$ and $b$ are the unique scalars such that $$v = a\begin{pmatrix} -1 \\ 1 \end{pmatrix} + b\begin{pmatrix} 1 \\ 0 \end{pmatrix}.$$ Finding such a vector requires solving a system of linear equations.

As an example, take the first column. We have, as you computed, $$\alpha \begin{pmatrix}3 \\ -1 \\ 0\end{pmatrix} = \begin{pmatrix}1 \\ 2 \end{pmatrix}.$$ Thus, we need to solve $$\begin{pmatrix}1 \\ 2 \end{pmatrix} = a\begin{pmatrix} -1 \\ 1 \end{pmatrix} + b\begin{pmatrix} 1 \\ 0 \end{pmatrix}.$$ If we turn this into a system of equations, the corresponding augmented matrix is $$\left[\begin{array}{cc|c}-1 & 1 & 1 \\ 1 & 0 & 2\end{array}\right] \sim \left[\begin{array}{cc|c}0 & 1 & 3 \\ 1 & 0 & 2\end{array}\right] \sim \left[\begin{array}{cc|c}1 & 0 & 2 \\ 0 & 1 & 3\end{array}\right],$$ hence $a = 2$ and $b = 3$, which is to say, $$\left[\alpha \begin{pmatrix}3 \\ -1 \\ 0\end{pmatrix}\right]_{B_2} = \left[\begin{pmatrix}1 \\ 2 \end{pmatrix}\right]_{B_2} = \begin{bmatrix} 2 \\ 3\end{bmatrix},$$ which is the first column of the matrix you want. The same procedure should find the other two columns.

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