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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$\begingroup$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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