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I need to find a 2x3 matrix that does transformations from $\mathbb{R}^3$ to $\mathbb{R}^2$ given that it performs the following 3 transformations: enter image description here

I know how to find 3x3 matrices by finding where $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$, and $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ go geometrically but for some reason I'm blanking on how to do this with something like a 2x3. What am I missing?

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  • $\begingroup$ The process is no different for a $2\times3$ matrix. $\endgroup$ – amd Oct 3 '18 at 18:39
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Let $$v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \quad v_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \quad v_3 = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}.$$

Our goal is it to find a matrix $ A = \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}$ such that $$Av_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad Av_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \quad Av_3 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ as described in your notes.

Let us consider the first equation:

It is $Av_1 = \begin{pmatrix} a \\ d \end{pmatrix} \stackrel{!}{=} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, so we obtain $a = 1, \, d = 0$.

We do the same thing with the second and third equation:

$$Av_2 = \begin{pmatrix} b+c \\ e+f \end{pmatrix} \stackrel{!}{=} \begin{pmatrix} 0 \\ 1 \end{pmatrix} , \quad Av_3 = \begin{pmatrix} -b+c \\ -e+f \end{pmatrix} \stackrel{!}{=} \begin{pmatrix} 0 \\ 0 \end{pmatrix}.$$

Now we want to find out what $b,c,e,f$ exactly are.

We start with $b$ and $c$: The second and third equation give us $$\begin{align} b+c &= 0 \\ -b+c &= 0 \end{align}$$

which yields (by computing the linear system of equations) $b=c=0$.

We do the same procedure with $e$ and $f$: The second and third equation give us $$\begin{align} e+f &= 1 \\ -e+f &= 0 \end{align}$$

which gives us $e = f = \frac{1}{2}$ this time.

So our final result is the matrix

$$ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} \end{pmatrix}.$$

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Relative to the bases $\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\1\end{pmatrix},\begin{pmatrix} 0\\-1\\1\end{pmatrix}\}$ for $\mathbb R^3$and the standard basis for $\mathbb R^2$, the matrix is: $\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}$.

To get it relative to both standard bases, use the change of basis matrix: $\begin{pmatrix} 1&0&0\\0&1&-1\\0&1&1\end{pmatrix}^{-1}=\frac12\cdot\begin{pmatrix}2&0&0\\0&1&1\\0&1&1\end{pmatrix}$.

So, rel the standard bases, we get: $\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}\cdot\frac12 \begin{pmatrix}2&0&0\\0&1&1\\0&1&1\end{pmatrix}=\begin{pmatrix}1&0&0\\0&\frac12&\frac12\end{pmatrix}$.

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