Let $A$ be a $2×2$ matrix with complex entries and $\det(A) = 0$ and $\mathrm{trace}(A)\neq 0$

Question

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…

Answer 1

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 $As=0$

Answer 2

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!

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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.