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let $T: M_{2,2} \rightarrow M_{2,2}$ be a linear transformation defined by:

$$T \left(\begin{bmatrix} a & b\\ c & d\\ \end{bmatrix}\right) = \begin{bmatrix}a + b& b + a \\ c - d&d+b\end{bmatrix} $$

Find the standard matrix for $S$ by using the standard basis of $M_{2,2}$

This is what i've done so far:

$$T \left(\begin{bmatrix} 1 & 0\\ 0 & 0\\ \end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&0\end{bmatrix} = x_1 $$

$$T \left(\begin{bmatrix} 0 & 1\\ 0 & 0\\ \end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&1\end{bmatrix} = x_2 $$

$$T \left(\begin{bmatrix} 0 & 0\\ 1 & 0\\ \end{bmatrix}\right) = \begin{bmatrix}0&0\\ 1&0\end{bmatrix} = x_3 $$

$$T \left(\begin{bmatrix} 0 & 0\\ 0 & 1\\ \end{bmatrix}\right) = \begin{bmatrix}0& 0\\ -1&1\end{bmatrix} = x_4 $$

But I'm not too sure where to go from here. All I know that these need to somehow become part of a larger matrix

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  • $\begingroup$ Would you know how to find the matrix representation of a transformation on $\mathbb R^4$? If so, think of the standard basis you've used as an ordered set of four column vectors in $\mathbb R^4$? and do the same thing. $\endgroup$ – Dan May 5 '15 at 13:49
  • $\begingroup$ @Dan so are you suggesting that the standard matrix for the transformation would simply be $[x_1 x_2 x_3 x_4]$? $\endgroup$ – PythonNewb May 5 '15 at 13:58
  • $\begingroup$ That's correct :) $\endgroup$ – Dan May 5 '15 at 14:02
  • $\begingroup$ @Dan so would the standard matrix be a 2 x 8 matrix or a 4 x 4 matrix? Still a little confused $\endgroup$ – PythonNewb May 5 '15 at 15:19
  • $\begingroup$ Yes, it would be a $4\times 4$ matrix. I'll provide an answer below if you'd like, but Omnomnomnom explains it correctly below. $\endgroup$ – Dan May 5 '15 at 15:27
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Let $v_1,v_2,v_3,v_4$ denote the standard basis. Note that $$ T(v_1) = T \left(\begin{bmatrix} 1 & 0\\ 0 & 0\\ \end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&0\end{bmatrix} = 1v_1 + 1v_2 +0v_3 + 0v_4\\ T(v_2) = T \left(\begin{bmatrix} 0 & 1\\ 0 & 0\\ \end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&1\end{bmatrix} = 1v_1 + 1v_2 + 0v_3 + 1v_4\\ T(v_3) = T \left(\begin{bmatrix} 0 & 0\\ 1 & 0\\ \end{bmatrix}\right) = \begin{bmatrix}0&0\\ 1&0\end{bmatrix} = 0v_1 + 0v_2 + 1v_3 + 0v_4\\ T(v_4) = T \left(\begin{bmatrix} 0 & 0\\ 0 & 1\\ \end{bmatrix}\right) = \begin{bmatrix}0& 0\\ -1&1\end{bmatrix} = 0v_1 + 0v_2 + (-1)v_3 + 1v_4 $$ It follows that the matrix of the transformation is given by $$ S = \pmatrix{ 1&1&0&0\\ 1&1&0&0\\ 0&0&1&-1\\ 0&1&0&1 } $$

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