Weird condition for null space and range implying invertibility

by AstlyDichrar   Last Updated November 18, 2018 22:20 PM

The question is:

Let $A = \begin{bmatrix} A_1 \\ A_2\end{bmatrix}\in \mathbb{M}_{n\times n}(\mathbb{C})$ (an $n\times n$ matrix with entries on $\mathbb{C}$) and suppose that $\mathcal{N}(A_1)=\mathcal{R}(A_2^\top)$ ($\mathcal{N}$ being the null space, and $\mathcal{R}$ the range). Prove that $A$ is invertible.

My thought process was, to prove that $A$ is invertible, it seems reasonable that from what we're given I'm gonna try to prove that $n(A)=0$ (the dimension of the null space of $A$ is $0$).

Well, we have that $\dim(\mathcal{N}(A))=\dim(\mathcal{N}(A_1)\cap\mathcal{N}(A_2))$, which is equal to $\dim(\mathcal{N}(A_1))+\dim(\mathcal{N}(A_2))-\dim(\mathcal{N}(A_1)+\mathcal{N}(A_2))$. By the hypothesis, $\mathcal{N}(A_1)=\mathcal{R}(A_2^\top)$, and $\mathcal{R}(A_2^\top)=\mathcal{R}(A_2)$, so we're left with

$$\dim(\mathcal{N}(A))=n-\dim(\mathcal{N}(A_1)+\mathcal{N}(A_2))$$

but I can't find a way to justify why $\dim(\mathcal{N}(A_1)+\mathcal{N}(A_2))=n$, which is what it has to be if $A$ is invertible...



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