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Let $T: M_{22} \rightarrow M_{22}$ be defined by:

$$T \begin{pmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix}\end{pmatrix} = \begin{bmatrix} 2c & a+c \\ b-2c & d \end{bmatrix}, $$ $B$ be the standard basis for $M_{22}$,

and

$$B' = \left \{ \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right \} $$

Find the matrix $[T]_B$ of the linear transformation $T$ with respect to the basis $B$.

I'm having a hard time understanding exactly how to approach these questions. Any sort of direction would be very helpful.

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    $\begingroup$ The vanilla way to do it is to basically plug in elements of the basis in order to T then express it as linear combo of the elements of the basis and take transpose. that is your $[T]_{\beta '}$ $\endgroup$ – Tiger Blood Apr 7 '16 at 1:22
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If we map the matrices to four vectors $$ \phi(M) = \phi \left( \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \right) = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} $$ we have $$ A \phi(M) = \phi(T(M)) \iff \\ A \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \begin{pmatrix} 2c \\ a+c \\ b-2c \\ d \end{pmatrix} $$ for some matrix $A$ which leads to $$ A = \begin{pmatrix} 0 & 0 & 2 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

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the basis for $B: e_1 = a, e_2 = b, e_3=c, e_4=d$

$[T]_B =\pmatrix{0&0&2&0\\1&0&1&0\\0&1&-2&0\\0&0&0&1\\}$

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