<s>[INST] Find the matrix of linear transformation with respect to different basis Let $f:P_2[x]\rightarrow M_{2\times 2}(\mathbb{R})$ is a linear transformation defined as $f(a+bx+cx^2)=        \begin{bmatrix}
        b+c & a  \\
        b & c \\
        \end{bmatrix}$
Find the matrix of $f$ with respect to basis $        ({\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}})$ and $(1,x,x^2)$. Find a basis and dimension of $Im (f)$ and $Ker(f)$. 
In a standard basis for $P_2[x]$, the mapping is:
$$f        \begin{bmatrix}
        a  \\
        b  \\
        c  \\
        \end{bmatrix}=        \begin{bmatrix}
        0 + b + c \\
        a + 0 + 0 \\
        0 + b + 0 \\
        0 + 0 + c
        \end{bmatrix}\Rightarrow f=\begin{bmatrix}
        0 & b & c \\
        a & 0 & 0 \\
        0 & b & 0 \\
        0 & 0 & c
        \end{bmatrix}$$ 
Question: Is the matrix of $f$ the same for both basis  $ ({\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}})$ and $(1,x,x^2)$? If not, what is the method for finding the matrix of $f$ in both cases?
 [/INST] For the standard basis of $P_2[x]$, your answer is wrong: there can't be $a,b,c $ in the matrix, which should be:
$$\begin{pmatrix}
0&1&1\\1&0&0\\0&1&0\\0&0&1
\end{pmatrix}.$$
Now with the given basis of $P_2[x]$, first set 
$$E_1=\begin{pmatrix}
1&0\\0&0
\end{pmatrix},\enspace
E_2=\begin{pmatrix}
1&1\\0&0
\end{pmatrix},\enspace
E_3=\begin{pmatrix}
1&1\\1&0
\end{pmatrix},\enspace
E_4=\begin{pmatrix}
1&1\\1&1
\end{pmatrix}.
$$
Then note that
\begin{align*}
f(1)=\begin{pmatrix}
0&1\\0&0
\end{pmatrix}&=E_2-E_1,
&f(x)=\begin{pmatrix}
1&0\\1&0
\end{pmatrix}&=E_3-E_2+E_1,\\
f(x^2)=\begin{pmatrix}
1&0\\0&1
\end{pmatrix}&=E_4-E_3+E_1.\\
\end{align*}
whence the matrix of $f$:
$$\begin{pmatrix}
-1&1&1\\1&-1&0\\0&1&-1\\0&0&1
\end{pmatrix}.$$
</s>