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$ f:M_{2.2}(\mathbb{R}) \rightarrow \mathbb{R^3}, $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \begin{pmatrix} 2a - 4b \\ -6d \\ 8a-16b+2d \end{pmatrix}\beta =\big (\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} ,\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \big), \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \big), \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \big)$ $\beta'= ( e_{2}, 2e_{1}, -e_{3})$ of $\mathbb{R^{3}}$ ( these are the standard unit vectors of $\mathbb{R^{3}}$)

What is the transformation matrix $M_{\beta'}^{\beta}(f)= ?$

Well my thought is that

$f(\left( \begin{array}{c} 0\\ 1\\ 0 \end{array} \right)) =\left( \begin{array}{c} -4\\ 0\\ -16 \end{array} \right) = \square \left( \begin{array}{c} 1\\ 0\\ 8 \end{array} \right) +$$\square \left( \begin{array}{c} -4\\ 0\\ 2 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ 0\\ 0 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ -6\\ 2 \end{array} \right)$

$f(\left( \begin{array}{c} 2\\ 0\\ 0 \end{array} \right)) =\left( \begin{array}{c} 4\\ 0\\ 16 \end{array} \right) = \square \left( \begin{array}{c} 1\\ 0\\ 8 \end{array} \right) +$$\square \left( \begin{array}{c} -4\\ 0\\ 2 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ 0\\ 0 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ -6\\ 2 \end{array} \right) $

$f(\left( \begin{array}{c} 0\\ 0\\ -1 \end{array} \right)) =\left( \begin{array}{c} 0\\ 6\\ -2 \end{array} \right) = \square \left( \begin{array}{c} 1\\ 0\\ 8 \end{array} \right) +$$\square \left( \begin{array}{c} -4\\ 0\\ 2 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ 0\\ 0 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ -6\\ 2 \end{array} \right) $

And from the linear combinations of the $squares$ I can fill the transformation matrix in the columns and in this way I will get the transformation matrix. Is my idea correct? And how does the transformation matrix look like? Thank you in advance

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  • $\begingroup$ what is the sens of $f\left(\begin{pmatrix}0\\1\\0\end{pmatrix}\right)$ whereas $f:\mathcal M_{2\times 2}(\mathbb R)\to \mathbb R^3$ ? $\endgroup$
    – Surb
    Commented Dec 20, 2021 at 15:46
  • $\begingroup$ I'm just trying to find out what these notations exactly mean and I have no clue if that my way of solution correct is $\endgroup$
    – Herrpeter
    Commented Dec 20, 2021 at 15:48

1 Answer 1

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Your idea is not correct, since the domain of $f$ is $\Bbb R^{2\times2}$; so it makes no sense to talk about, say$$f\left(\begin{bmatrix}0\\1\\0\end{bmatrix}\right).$$

Note that\begin{align}f\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)&=\begin{bmatrix}2\\0\\8\end{bmatrix}\\&=0\times e_2+1\times(2e_1)-8(-e_3).\end{align}Therefore, the entries of the first column of the matrix that you're after are $0$, $1$, and $-8$.

Can you take it from here?

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  • $\begingroup$ $\begin{align}f\left(\begin{bmatrix}0&1\\0&0\end{bmatrix}\right)&=\begin{bmatrix}-4\\0\\0\end{bmatrix}\\&=0\times e_2+-4\times(2e_1) \times0(-e_3).\end{align}$$\begin{align}f\left(\begin{bmatrix}0&0\\1&0\end{bmatrix}\right)&=\begin{bmatrix}0\\0\\0\end{bmatrix}\\&=0\times e_2+0\times(2e_1) \times0(-e_3).\end{align}$$\begin{align}f\left(\begin{bmatrix}0&0\\0&1\end{bmatrix}\right)&=\begin{bmatrix}0\\-6\\2\end{bmatrix}\\&=-6\times e_2+0\times(2e_1) \times-2(-e_3).\end{align}$ so that will be the rest? $\endgroup$
    – Herrpeter
    Commented Dec 20, 2021 at 15:56
  • $\begingroup$ and the transformation matrix:$\begin{pmatrix} 0 & 0 & 0 & 0\\ 1 & -4 & 0 & -6 \\ -8 & 0 & 0 & 2 \\ \end{pmatrix}$ $\endgroup$
    – Herrpeter
    Commented Dec 20, 2021 at 15:58
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    $\begingroup$ Something like that. I noticed that$$\begin{bmatrix}-4\\0\\0\end{bmatrix}=0\times e_2+\color{red}{(-2)}\times(2e_1)+0\times(-e_3).$$ $\endgroup$ Commented Dec 20, 2021 at 15:59
  • $\begingroup$ $\begin{pmatrix} 0 & 0 & 0 & -6\\ 1 & -2 & 0 & 0 \\ -8 & 16 & 0 & -2 \\ \end{pmatrix}$ how can I compute kernel and image of $f$? $\endgroup$
    – Herrpeter
    Commented Dec 20, 2021 at 19:45
  • $\begingroup$ That's another question. I suggest that you post it as such. $\endgroup$ Commented Dec 20, 2021 at 19:58

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