How to show $A=\begin{bmatrix}A_1\\A_2\end{bmatrix}$ is non-singular when $N(A_1)=R(A_1^T)$?

Question

Suppose $A=\begin{bmatrix}A_1\\A_2\end{bmatrix}$ is a block matrix such that $N(A_1)=R(A_2^T)$. How can we show it is nonsingular.

My try: We know that a matrix is nonsingular if $Ax=0$ has only the solution $x=0$. Therefore, assume that $Ax=0$ so we have to show $x=0$ as follows

$$Ax=\begin{bmatrix}A_1\\A_2\end{bmatrix}x=\begin{bmatrix}A_1x\\A_2x\end{bmatrix}=0$$

Since $A_1x=0$ we conclude that $x$ is in $N(A_1)$.

How can I use the fact that $N(A_1)=R(A_2^T)$ to come up with $A_2x=0$. I know that $R(A_2^T)=N(A_2)$ and I am done but we are not allowed to use this fact because we have not proved it yet.

The answer would be proof of $R(A_2^T)=N(A_2)$ or an intuitive way bypassing $R(A_2^T)=N(A_2)$.

Answer 1

Well, $A_2x=0$ is also given by $Ax=0$.
You should prove $x=0$.

Also, the correct form of the identity is $$R(B^T) \ =\ N(B)^\perp$$ and the exercise is basically equivalent to it.

But the adjoint property enables a quick proof for $R(B^T)^\perp =N(B)$:

Both conditions are equivalent to that $$0=\langle x, B^Ty\rangle =x^TB^Ty=\langle Bx, y\rangle$$ holds for all $y$.