Let $A$ be a $2×2$ matrix with complex entries and $\det(A) = 0$ and $\mathrm{trace}(A)\neq 0$ Let $A$ be a $2×2$ matrix with complex entries. Suppose that $\det(A) = 0$
and $\mathrm{tr}(A)\neq 0$. Show the following:
a) $\ker(A) \cap \mathrm{range}(A)= \{\mathbb{0}\}$
b)  $\mathbb{C}^2 =  \mathrm{span}(\ker(A) \cup \mathrm{range}(A))$
My thinking:
I take $
A=      \begin{pmatrix}
        0 & 0  \\
        1 & 1  \\
        \end{pmatrix}
$. Here I got that $\ker(A) = 1$ and $\mathrm{rank}(A) = 1$  as  $\ker(A) \cap \mathrm{range}(A)= 1$  as I am not getting $\{0\}$ for option a).
And for option b) I do not know  how  to approach…
 A: The kernel of a $2\times 2$ complex matrix is a subset of $\mathbb C^2$, so it cannot be $1$. The dimension of the kernel in your case is, like you wrote, $1$, but the kernel itself, for your case, is equal to $$\ker(A)=\left\{\begin{bmatrix}z\\-z\end{bmatrix}|z\in\mathbb Z\right\}$$
Similarly, the range of a matrix is another subspace, and in the case of your matrix, the range of the matrix is $$\mathrm{range}(A)=\left\{\begin{bmatrix}0\\-z\end{bmatrix}|z\in\mathbb Z\right\}$$
you can easily see that the intersection of those two sets is $$\left\{\begin{bmatrix}0\\0\end{bmatrix}\right\}$$
which is also written as $\{0\}$ (but where $0$ is the zero of the vector space, not the scalar field!

To actually prove option $a$, you should take an element $v\in\ker(A)\cap \mathrm{range}(A)$ and prove that $v=0$. Since $v\in\ker(A)$, you know what $Av$ is equal to, and since $v\in\mathrm{range}(A)$, you also know that $v=Aw$ for some vector $w$. Now some mathematical work should take place.
A: Start with
$$A=\begin{bmatrix}a & b\\c & d\end{bmatrix}$$
We know that $a+d\neq 0$ and $ad=bc$. Compute
$$A^2=\begin{bmatrix}a^2+bc & ab+bd\\ac+cd & bc+d^2\end{bmatrix}=(a+d)A$$
Now consider $v\in \ker{A}\cap\operatorname{range}{A}$; one has $Av=0$ and $v=Aw$ for some $w\in \Bbb{C}^2$, so $A^2w=0=(a+d)Aw$. But $a+d\neq 0$, therefore $v=Aw=0$
Now any $v\in\Bbb{C}^2$ can be written as
$$v=\underbrace{v-\left({1\over a+d}Av\right)}_{s}+{1\over a+d}Av$$
$1/(a+d)Av\in \operatorname{range}{A}$ and $As=0$
