Matrix of linear operator. If $f:P_2[x] \rightarrow \mathbb{M}_{2x2}(\mathbb{R})$ is mapping defined as $f(a + bx + cx^2)=\begin{bmatrix} b+c & a \\ b & c \end{bmatrix}$. Determine is this linear operator, and if it is, determine it's matrix matrix if basis is $(1,x,x^2)$, and then, determine it's matrix if basis is $(\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix})$, and then find dimension and basis of $Imf$ and $Kerf$.
It is very easy to check if this is linear operator $ f(x+y)=f(x) + f(y)$ and $f(\alpha x)=\alpha f(x)$ since this is true for our mapping, it means that it is linear operator. When it comes to finding matrix representation, it is relatively easy in the first case where basis is $(1,x,x^2)$, then  $f(1)=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = (0,1,0,0)$, $f(x)=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} = (1,0,1,0)$,
$f(x^2)=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = (1,0,0,1)$, which means that matrix representation of given linear operator when the basis is $(1,x,x^2)$ is $F=\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
In the second case basis is given with vectors of codomain, i don't know how to form a matrix when i have basis with codomain vectors. How that can be done?
 A: Given a linear map $f \colon V \rightarrow W$ between two finite dimensional vector spaces, in order to represent $f$ by a matrix you need to provide an ordered basis $\mathcal{B} = (v_1, \dots, v_n)$ for $V$ and an ordered basis $\mathcal{C} = (w_1, \dots, w_m)$ for $W$. The matrix $[f]_{\mathcal{C}}^{\mathcal{B}}$ is then the matrix whose $i$-th column is the coordinate vector $[fv_i]_{\mathcal{C}}$ representing the vector $fv_i$ with respect to the basis $\mathcal{C}$. If $V = W$, then often one provides only one basis $\mathcal{B}$ with the implicit understanding that $\mathcal{B}$ should be used both for the domain and the co-domain.
In your case, $V \neq W$ and so there is no meaning in asking for the matrix representation of $f$ with respect to $\mathcal{B}$ without specifying what basis $\mathcal{C}$ is used for $W$. Similarly, it doesn't make sense to ask for the matrix representation of $f$ with respect to $\mathcal{C}$ without specifying $\mathcal{B}$.
Using both bases you provided, we have
$$ f(1) = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \implies [f(1)]_{\mathcal{C}} = \begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \\
f(x) = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} \implies [f(x)]_{\mathcal{C}} = \begin{pmatrix} 1 \\ -1 \\ 1 \\ 0 \end{pmatrix}, \\
f(x^2) = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \implies [f(x^2)]_{\mathcal{C}} =  \begin{pmatrix} 0 \\ 0 \\ -1 \\ 1 \end{pmatrix}$$
and thus
$$ [f]_{\mathcal{B}}^{\mathcal{C}} = \begin{pmatrix} -1 & 1 & 0 \\
1 & -1 & 0 \\
0 & 1 & -1 \\
0 & 0 & 1 \end{pmatrix}. $$
Clearly the three columns of $[f]_{\mathcal{B}}^{\mathcal{C}}$ are linearly independent and thus $3 = \operatorname{rank} [f]_{\mathcal{B}}^{\mathcal{C}} = \dim \operatorname{im}(f)$ and $\dim \ker(f) = 0$. Since $([f(1)]_{\mathcal{C}}, [f(x)]_{\mathcal{C}}, [f(x^2)]_{\mathcal{C}})$ are linearly independent in $\mathbb{R}^3$, this means that $(f(1), f(x), f(x^2))$ are linearly independent in $W$ and so they form a basis for $\operatorname{im}(T)$.
