I have this question:
Let $Z\in M_{2\times2}(\mathbb{R})$ be defined as
$$Z = \left( \begin{align} 1 &&1\\1 &&1 \end{align} \right)$$
and consider $T: M_{2\times2}(\mathbb{R}) \to M_{2\times2}(\mathbb{R})$ such that $$T(X) = ZX$$
Determine a basis for $\ker(T)$
Now I have transformation matrix with respect to the order basis as:
$$\left[ \begin{align} 1&&0&&1&&0\\0&&1&&0&&1\\1&&0&&1&&0\\0&&1&&0&&1 \end{align} \right]$$
Not sure if that is helpful for this, but I don't know how to work it out.
Is $\ker(T) = \emptyset$ or just the identity? I feel like there is no $\ker$.
The kernel of $T$ is $$ \ker T=\{A\in M_{2\times2}(\mathbb{R}):T(A)=0\} $$ that is $ZA=0$. You can find it using the representing matrix with respect to the basis $\{E_1,E_2,E_3,E_4\}$ where $$ E_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}\quad E_2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\quad E_3=\begin{bmatrix}0&0\\1&0\end{bmatrix}\quad E_4=\begin{bmatrix}0&0\\0&1\end{bmatrix} $$ which is the matrix you wrote. Now row reduce it: $$ \begin{bmatrix} 1&0&1&0\\0&1&0&1\\1&0&1&0\\0&1&0&1 \end{bmatrix} \to \begin{bmatrix} 1&0&1&0\\0&1&0&1\\0&0&0&0\\0&0&0&0 \end{bmatrix} $$ which means that the solutions of the homogeneous system are generated by the vectors $$ \begin{bmatrix}-1\\0\\1\\0\end{bmatrix} \quad\text{and}\quad \begin{bmatrix}0\\-1\\0\\1\end{bmatrix} $$ Taking into account the meaning of the representing matrix, we can say that $$ \{-E_1+E_3,-E_2+E_4\}= \left\{ \begin{bmatrix}-1&0\\1&0\end{bmatrix}, \begin{bmatrix}0&-1\\0&1\end{bmatrix} \right\} $$ is a basis of the kernel of $T$.