show that $\mathcal{N}(A) \subseteq \mathcal{N}(B)$ iff $BA^+A = B$. For $A\in\mathbb{R}^{p\times n}$ and $B\in\mathbb{R}^{m\times n}$, show that $\mathcal{N}(A) \subseteq \mathcal{N}(B)$ iff $BA^+A = B$. Note "$^+$" indicates Moore-Penrose pseudo-inverse.
My attempt:
Suppose $BA^+A = B$ and $x \in\mathcal{N}(A)$. Then $Ax = 0$. We have
$$Bx = (BA^+A)x = BA^+(Ax) = 0.$$
This shows that $x\in \mathcal{N}(B)$.
Suppose $\mathcal{N}(A)\subseteq\mathcal{N}(B)$. Let $x\in\mathbb{R}^n$ where $x\in\mathcal{N}(A)\subseteq\mathcal{N}(B)$. Thus we have $Ax = 0$. But note that $KAx = 0$ for all $K$ matrices with $p$ columns. Thus assuming $BA^+$ has $p$ columns we have
$$Bx = Ax = BA^+(Ax) = BA^+Ax.$$
The second half I am honestly not sure if I was able to do that or not. Let me know what I need to fix! Feedback is great, thanks!
 A: I have been able to come up with a plausible answer. Please let me know what you think.
I believe my proof for "If $BA^+A = B$, then $\mathcal{N}(A) \subseteq \mathcal{N}(B)$" is sufficient. Thus I will not repeat it (Please see above).
Suppose $\mathcal{N}(A) \subseteq \mathcal{N}(B)$. We will show that $BA^+A = B$. Let $x\in\mathbb{R}^n$. Let us decompose $x = x_1 + x_2$ where $x_1\in\mathcal{N}(A)^\perp$ and $x_2\in\mathcal{N}(A)$. For $x_2$ we know that $BA^+\underbrace{Ax_2}_{0} = BA^+\mathbf{0} = \mathbf{0}$. Let us observe what $BA^+A$ does on $x_1$. Observe that,
$$x_1\in\mathcal{N}(A)^\perp \implies Ax_1\in\mathcal{R}(A) \implies A^+Ax_1 \in\mathcal{N}(A)^\perp.$$
But since $\mathcal{N}(A)^\perp$ is isomorphic to $\mathcal{R}(A)$, this means that our $x_1$ that was since to $\mathcal{R}(A)$ and back to $\mathcal{N}(A)^\perp$ is still our $x_1$ we had starting out. Thus,
$$A^+Ax_1 = x_1 \implies BA^+Ax_1 = Bx_1.$$
Lastley, let us observe what $B$ does to $x$,
$$Bx = B(x_1 + x_2) = Bx_1 + \underbrace{Bx_2}_{0} = Bx_1.$$
We have,
$$BA^+Ax = BA^+Ax_1 = Bx_1 \implies BA^+A = B.$$
