# 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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