<s>[INST] Find bases for kernel and image of T where $T: P_2 \to M_2$ T is defined as $$T:P_2(\mathbb R) \to M_2 (\mathbb R) \ \text{where} \  T(ax^2 +bx+c)=\begin{pmatrix}-2a +c & b+c\\-3b-3c&6a-3c\\ \end{pmatrix}$$ and I need t find bases for $ Im(T) $ and $ker(T)$
I started with $im(T)$ and got
$$Im(T) = \begin{pmatrix}-2a +c & b+c\\-3b-3c&6a-3c\\ \end{pmatrix} \\
= a\begin{pmatrix}-2 & 0\\0&6\\ \end{pmatrix} +b \begin{pmatrix}0 &1\\-3 &0 \\ \end{pmatrix} +c \begin{pmatrix}1 & 1\\-3 & -3\\ \end{pmatrix} $$
using $$\begin{pmatrix}-2 & 0& 0&6\\ 
0 & 1& -3&0\\ 
1 &1& -3&-3\\
\end{pmatrix} $$ reduced to 
$$\begin{pmatrix}1 & 0& 0&-3\\ 
0 & 1& -3&0\\ 
0 &0& 0&1\\
\end{pmatrix} $$
Is the basis for $Im(T)=\{(1, 0 ,0 ,3),(0,1,-3,0),(0,0,0,1)\}$  ??
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To find $Ker(T) = \{(a,b,c) where \begin{pmatrix}-2a+c & b+c\\ 
-3b-3c& 6a-3c\\ 
\end{pmatrix} =0 \}$
by solving
$$(-2a+c)(6a-3c)-(b+c)(-3b-3c)=0$$
$$(2a-c)^2 -(b+c)^2=0$$
$$2a+b=0 \ \text{or} \ 2a-b-2c=0$$
but now what??
 [/INST] Let the basis of $P_2$ is $(1,x,x^2)$ and the basis of $M_2$ is $(a_{11}, a_{12}, a_{21}, a_{22})$ 
Where 
$a_{11}= \begin{pmatrix}1&0 \\ 0&0 \end{pmatrix}$ 
$a_{12}= \begin{pmatrix}0&1 \\ 0&0 \end{pmatrix}$ 
$a_{21}= \begin{pmatrix}0&0 \\ 1&0 \end{pmatrix}$ 
$a_{22}= \begin{pmatrix}0&0 \\ 0&1 \end{pmatrix}$ 
Then
 $T(x^2)=-2\times a_{11}+6\times a_{22}$
$T(x^3)=-2\times a_{11}+6\times a_{22}$
$T(x^2)=1\times a_{12}+-3\times a_{21}$
$T(1)=1\times a_{11}+1\times a_{12}+-3\times a_{21}+(-3)\times a_{22}$
Therefore the matrix of Linear transformation is:
$$\begin{pmatrix}-2 & 0& 1\\0&1&1\\ 0&-3&-3\\6&0&-3\end{pmatrix} \\ $$
Now find basis of column space of this matrix.( i.e, basis of image of T) 
Using elementry row transformation this matrix will reduce to :
$$\begin{pmatrix}-2 & 0& 1\\0&1&1\\ 0&0&0\\0&0&0\end{pmatrix} \\ $$
So basis of image of $T$ will be $((-2 ,0, 0, 6), (0 ,1, 0 ,0))$
Ans the basis of null space will be $ ((1/2,-1,1)$
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