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Given the linear operator: $f\left(a+bx+cx^2\right)\rightarrow \begin{bmatrix}b+c&a\\ b&c\end{bmatrix}$, find the matrix representation of this linear operator in respect to the bases $\left(\begin{bmatrix}1&0\\ \:0&0\end{bmatrix}\begin{bmatrix}1&1\\ \:0&0\end{bmatrix}\begin{bmatrix}1&1\\ \:1&0\end{bmatrix}\begin{bmatrix}1&1\\ \:1&1\end{bmatrix}\right)$ and $\left\{1,x,x^2\right\}$.

I have a few questions/misunderstadings with this problem. First of the operator $f$ clearly can't map the whole $M_{2,2}$ space and by providing $f$ with the vectors from the standard basis we get:

$f\left(1\cdot a\right)=\begin{bmatrix}0&1\\ 0&0\end{bmatrix}$

$f\left(x\cdot b\right)=\begin{bmatrix}1&0\\ 1&0\end{bmatrix}$

$f\left(x\cdot b\right)=\begin{bmatrix}1&0\\ 0&1\end{bmatrix}$

So the matrix of this linear transformation is:

$F=\begin{bmatrix}0&1&1\\ 1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}$

Now what is the correlation of these two bases? I know that each column represents the coordinates of one the vectors from the base $\left\{1,x,x^2\right\}$ in the standard base of $M_{2,2}$, but how am I to rewrite F using the non-standard base given above?

EDIT: I have also noticed that the vectors from the non-standard base given above can be written as linear-combos of the vectors in the std base, but even so I can't apply the formula $F'=S^{-1} F S$, since $F$ isn't of the right format.

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    Sorry but I cannot make sense of what you write. Consider the matrices $$U=\begin{bmatrix}1&0\\ \:0&0\end{bmatrix}\quad V=\begin{bmatrix}1&1\\ \:0&0\end{bmatrix}\quad W=\begin{bmatrix}1&1\\ \:1&0\end{bmatrix}\quad Z=\begin{bmatrix}1&1\\ \:1&1\end{bmatrix}$$ then $\{U,V,W,Z\}$ is a free system hence $(U,V,W,Z)$ is indeed a basis of $M_{2,2}$, and $$f(1)=V-U\quad f(x)=U-V+W\quad f(x^2)=U-W+Z$$ hence the matrix of $f$ in the bases $(1,x,x^2)$ and $(U,V,W,Z)$ is $$\begin{bmatrix}-1&1&1\\1&-1&0\\0&1&-1\\0&0&1\end{bmatrix}$$ – 2017-01-08
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    Note that $a+bx+cx^2$ is the span of the vectors $1,x,x^2$ with the scalars $a,b,c$ For a linear transformation we will have $$f(a+bx+cx^2)=af(1)+bf(x)+cf(x^2)$$ In other words it suffices to figure out where the basis vectors are sent under the transformation $$f(1)=f(1+0x+0x^2)\mapsto\begin{bmatrix}0&1\\0&0\end{bmatrix} \\f(x)=f(0+1x+0x^2)\mapsto\begin{bmatrix}1&0\\1&0\end{bmatrix} \\f(x^2)=f(0+0x+1x^2)\mapsto\begin{bmatrix}1&0\\0&1\end{bmatrix}$$ – 2017-01-08
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    @Did I have mixed up, the change of basis matrix with this. Thank you, your comment will suffice as the correct answer and if you so wish you can post it as such so I can mark it. – 2017-01-09

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