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 $Im(T)=\{(1, 0 ,0 ,3),(0,1,-3,0),(0,0,0,1)\}$ ?? $$..................................................................... $$
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??
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)$