<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] Your question is fuzzy: $f$ has one matrix with respect to the two given bases. So there is only one case.
Let's call the matrix we want to find $M_f$ and write the coefficients of the two given vectors down:
The vector $a + bx + cx^2$ with respect to the basis $(1,x,x^2)$ is $(a,b,c)$.
The vector $\begin{bmatrix}
        b+c & a  \\
        b & c \\
        \end{bmatrix}$ with respect to the basis $({\begin{bmatrix}
        1 & 0  \\
        0 & 0 \\
        \end{bmatrix},\begin{bmatrix}
        1 & 1  \\
        0 & 0 \\
        \end{bmatrix},\begin{bmatrix}
        1 & 1  \\
        1 & 0 \\
        \end{bmatrix},\begin{bmatrix}
        1 & 1  \\
        1 & 1 \\
        \end{bmatrix}})$ is $(b+c-a,a-b,b-c,c)$.
So, $M_f (a,b,c)^T = (b+c-a,a-b,b-c,c)^T$. 
If you denote the $3$ columns of $M_f$ by $c_1, c_2, c_3$ then you want $c_i$ to be such that 
$$ a c_1 + b c_2 + c c_3 = (b+c-a,a-b,b-c,c)$$
How to find the $c_i$? 
Note that a linear map is uniquely determined by what it maps basis vectors to. In this case, we have the basis vectors
$$ 1 = (1,0,0)^T = b_1, x = (0,1,0)^T=b_2, x^2 = (0,0,1)^T = b_3$$
and 
$$ M_f(b_1) = (-1,1,0,0), M_f(b_2) = (1,-1,1,0), M_f(b_3)=(1,0,-1,1)$$
Can you read the columns of $M_f$ off of this? 
</s>