Let $\mathbf{A}$ be a $2×2$ matrix with complex entries. If $\det (\mathbf{A})= 0$ and $\operatorname{trace}(\mathbf{A})\ne 0$ , then show that $\ker (\mathbf{A})\cap\operatorname{range}(\mathbf{A})=\{0\}$ and $\mathbb{C}^2=\operatorname{span}\{\ker(\mathbf{A}), \operatorname{range}\mathbf{A}\}$.

I take a linear transformation $T:\mathbb{C}\rightarrow \mathbb{C}$ with $T (1)= a+bi $ and $T (i)=d+ci $ Where $$ \mathbf{A}= \begin {pmatrix} a&d\\ b&c\\ \end {pmatrix} $$

Then I'm stuck.

Please help.


The characteristic polynomial of $A$ is $x^2-\operatorname{tr}(A)x$, which has two roots: $0$ and $\operatorname{tr}(A)$. So, there are non-zero vectors $v$ and $w$ such that $A.v=0$ and that $A.w=\operatorname{tr}(A)w$. They are linearly independent, $\ker A=\mathbb{C}v$, and $\operatorname{range}A=\mathbb{C}w$. Therefore, $\ker A\cap\operatorname{range}A=\{0\}$.


You know that $\dim\ker\mathbf{A}+\dim\operatorname{range}\mathbf{A}=2$; you also know that $\dim\ker\mathbf{A}>0$, because $\det\mathbf{A}=0$.

If $\dim\ker\mathbf{A}=2$, then $\mathbf{A}$ is the zero matrix, but then its trace is $0$. Hence we know that $\dim\ker\mathbf{A}=1$ and therefore also $\dim\operatorname{range}\mathbf{A}=1$.

Suppose $\ker\mathbf{A}\cap\operatorname{range}\mathbf{A}\ne\{0\}$. The intersection can have dimension at most $1$, so the only possibility is that $\operatorname{range}\mathbf{A}=\ker\mathbf{A}$, which implies $\mathbf{A}^2=\mathbf{0}$ and again the trace of $\mathbf{A}$ is $0$.

Therefore $\ker\mathbf{A}\cap\operatorname{range}\mathbf{A}=\{0\}$ and consequently $\ker\mathbf{A}+\operatorname{range}\mathbf{A}=\mathbb{C}^2$.


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