This is a linear transformation. I want to see what the kernel and the image are.
If we look at the basis vectors of $R^{2\times2}$ and we look where the transformation maps them, we see that only the matrix $\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$ is mapped to $\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$. I thought that only this would thus be kernel. It turns out that the image doesn't even exist and that the kernel is what I first thought the image is.
So the computation of the $A$ and $X$ expression is : $\begin{bmatrix}c & d-a \\ 0 & c \end{bmatrix}$
So the above is what the transformation does to a matrix $\begin{bmatrix}a & b \\ c & d \end{bmatrix}$ where the parameters are real numbers.
However if I look at the expression:
$$\alpha_1\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} +\alpha_2\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} + \alpha_3\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} + \alpha_4\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$$
I get the correct answer for Kernel basis which is :
$$\{ \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix},\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} \}$$
Can somebody please explain what is going on in this problem, why is the kernel what it is and why the image doesn't exist even if the transformation only maps the element $a_{i,j}$ to zero and not the whole matrix.
Best Answer
Actually,$$\mathcal T\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=\begin{bmatrix}c&d-a\\0&-c\end{bmatrix}.$$Therefore$$\ker\mathcal T=\left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}\,\middle|\,c=d-a=0\right\}=\left\{\begin{bmatrix}a&b\\0&a\end{bmatrix}\,\middle|\,a,b\in\Bbb R\right\}.$$And, yes, $\mathcal T$ has an image, which is $\left\{T(X)\,\middle|\,X\in\Bbb R^{2\times2}\right\}$. Of course, it is possible to describe this set explicitly. For instance:$$\operatorname{Im}\mathcal T=\left\{\begin{bmatrix}a&b\\0&-a\end{bmatrix}\,\middle|\,a,b\in\Bbb R\right\}.$$