A Linear transformation from the vector space of polynomials Of degree at most $2$ with real cofficients to the vector space of $ 2\times 2$ matrices Let $\mathbb{P}$$_{2}$$\left(\mathbb{R}\right)$ be the vector space
of polynomials in x with real cofficients.Let $\mathbb{M}$$_{2}\left(\mathbb{R}\right)$
be the vector space of $2\times2$ real matrix.If a Linear Transformation T$\colon$$\mathbb{P}$$_{2}$$\left(\mathbb{R}\right)$ $\rightarrow$$\mathbb{M}$$_{2}$$\left(\mathbb{R}\right)$is
defined as
T$\left(f\right)$=$\begin{bmatrix}
f\left(0\right)-f\left(2\right) & 0\\
0 & f\left(1\right)
\end{bmatrix}$.
Then Prove that
$\text{Range}(T)=Span\left\{ \begin{pmatrix}0 & 0\\
0 & 1
\end{pmatrix},\begin{pmatrix}-2 & 0\\
0 & 1
\end{pmatrix}\right\} $
My approach: Too much difficult for me. I couldn't even make a start.
 A: For any degree two polynomial $f(x)=ax^2+bx+c$, 
$$T(f(x))=\begin{bmatrix}
    -4a-2b       & 0  \\
    0       & a+b+c  \\
    \end{bmatrix}=(2a+b)\begin{bmatrix}
    -2       & 0  \\
    0       & 1  \\
    \end{bmatrix}+(c-a)\begin{bmatrix}
    0       & 0  \\
    0       & 1  \\
    \end{bmatrix}$$
Which give us our result.
A: Let
$$f(x)=ax^2+bx+c\in P_2(\Bbb R)\implies \begin{cases}f(2)-f(0)=4a+2b+c-c=4a+2b\\{}\\f(1)=a+b+c\end{cases}\implies$$
$$Tf=T(ax^2+bx+c):=\begin{pmatrix}4a+2b&0\\0&a+b+c\end{pmatrix}$$
Observe that
$$f\in\ker T\iff\begin{cases}4a=-2b\iff2a=-b\\\text{and also}\\a+b+c=0\iff3a=-c\end{cases}\;\implies\dim\ker T=1\implies\dim\,\text{Im}\,T=2$$
by the dimensions theorem
So we can simply choose
$$a=b=0,\,c=1\implies \begin{cases}T(1)=\begin{pmatrix}0&0\\0&1\end{pmatrix}\\{}\\a=-2,\, b=3,\;c=0\implies T(x^2+x)=\begin{pmatrix}\!-2&0\\0&1\end{pmatrix}\end{cases}\;\;\in\text{ Im}\,T$$
and since both matrices above are linearly independent and $\;\dim\text{ Im}\,T=2\;$ we're done.
