Find the basis and dimension of Im(T) and Ker(T) for the following Linear Transformation

Question

Let $T:V\to W$ be a Linear Transformation then find the basis and dimension of $Im(T)$ and $Ker(T)$ where$$ T(A)=MA ,M=\begin{bmatrix} 1 & -1 \\ -2 & 2 \end{bmatrix}$$ My Approach:$e_{1}=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},e_{2}=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},e_{3}=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},e_{4}=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$ are the standard basis of $M_{4\times4}$ $$T(e_{1})=\begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix},T(e_{2})=\begin{bmatrix} 0 & 1 \\ 0 & -2 \end{bmatrix},T(e_{3})=\begin{bmatrix} -1 & 0 \\ 2 & 0 \end{bmatrix},T(e_{4})=\begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}$$From here i couldn't think of how to go further.Any Hints or solution will be appreciated.
Thanks in advance.

Answer 1

You started in the right way. Now notice that $T(e_1)=-T(e_3)$ and $T(e_2)=-T(e_4)$, but $T(e_1)$ and $T(e_3)$ are linearly indipendent, then they generate the image $Im(T)$ and hence the image has dimension $2$. Since $V$ has dimension $2$, you can deduce that $$\dim Ker(T)=\dim V-\dim Im(T)=4-2=2.$$ Now it remains to find a basis of the $Ker(T)$. You can now notice that $T(e_1+e_3)=T(e_1)+T(e_3)=0$ and similarly $T(e_2+e_4)=T(e_2)+T(e_4)=0$. Since $e_1+e_3$ and $e_2+e_4$ are linearly indipendent they generate the kernel.

Answer 2

You have two options:

Option one:

You can calculate the matrix $M_T$ representing the transformation $T$. Note that, as $T$ maps from a $4$ dimensional space to a $4$ dimensional space, the matrix representing $T$ will be a $4\times 4$ matrix. The first column of the matrix $M_T$ will be the vector representing $T(e_1)$ in your chosen basis, so in the standard basis, the first column of $M_T$ is $$\begin{bmatrix}1\\0\\-2\\0\end{bmatrix}$$

Once you have $M_T$ all written down, you simply have to calculate the basis and image of the matrix using standard methods, then convert back to your basis "vectors" (the $2\times 2$ matrices).

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Option two:

Do it all in the matrix representation. Write $A=\begin{bmatrix}a& b\\c&d\end{bmatrix}$ and write down that $MA$ is equal to. Then, you can see that $A$ is in the kernel of $T$ if $MA=0$, and you have $4$ linear equations that need to be fulfilled. Write them all down, and look at what they mean.

A similar thing can be done to get the image of $T$.