# For A in RR_(p xx n) and B in RR_(m xx n), show that ccN(A) sube ccN(b) iff BA^+A=B.

Teagan Huffman 2022-09-20 Answered
For $A\in {\mathbb{R}}^{p×n}$ and $B\in {\mathbb{R}}^{m×n}$, show that $\mathcal{N}\left(A\right)\subseteq \mathcal{N}\left(B\right)$ iff $B{A}^{+}A=B$. The "+" note indicates Moore-Penrose pseudo-inversion.
My attempt:
Suppose $B{A}^{+}A=B$ and $x\in \mathcal{N}\left(A\right)$. Then Ax=0. We have
$Bx=\left(B{A}^{+}A\right)x=B{A}^{+}\left(Ax\right)=0.$
This shows that $x\in \mathcal{N}\left(B\right)$
Suppose $\mathcal{N}\left(A\right)\subseteq \mathcal{N}\left(B\right)$. Let $x\in {\mathbb{R}}^{n}$ where $x\in \mathcal{N}\left(A\right)\subseteq \mathcal{N}\left(B\right)$. Thus we have Ax=0. But note that $KAx=0$ for all K matrices with p columns. Thus assuming $B{A}^{+}$ has p columns we have
$Bx=Ax=B{A}^{+}\left(Ax\right)=B{A}^{+}Ax.$
In the second half, I'm not sure if I was able to do it or not. Tell me what I need to fix!
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Haylee Krause
If $B{A}^{+}A=B$, then $\mathcal{N}\left(A\right)\subseteq \mathcal{N}\left(B\right)$ thus:
Suppose $\mathcal{N}\left(A\right)\subseteq \mathcal{N}\left(B\right)$. We will show that $B{A}^{+}A=B$. Let $x\in {\mathbb{R}}^{n}$. Let us decompose $x={x}_{1}+{x}_{2}$ where ${x}_{1}\in \mathcal{N}\left(A{\right)}^{\perp }$ and ${x}_{2}\in \mathcal{N}\left(A\right)$. For ${x}_{2}$ we know that $B{A}^{+}\underset{0}{\underset{⏟}{A{x}_{2}}}=B{A}^{+}\mathbf{0}=\mathbf{0}$
Let us observe what $B{A}^{+}A$ does on ${x}_{1}$. Observe that,
${x}_{1}\in \mathcal{N}\left(A{\right)}^{\perp }\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}A{x}_{1}\in \mathcal{R}\left(A\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{A}^{+}A{x}_{1}\in \mathcal{N}\left(A{\right)}^{\perp }.$
But since $\mathcal{N}\left(A{\right)}^{\perp }$ is isomorphic to $\mathcal{R}\left(A\right)$, this means that our ${x}_{1}$ that was since to $\mathcal{R}\left(A\right)$ and back to $\mathcal{N}\left(A{\right)}^{\perp }$ is still our ${x}_{1}$ we had starting out. Thus,
${A}^{+}A{x}_{1}={x}_{1}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}B{A}^{+}A{x}_{1}=B{x}_{1}.$
Lastley, let us observe what B does to x,
$Bx=B\left({x}_{1}+{x}_{2}\right)=B{x}_{1}+\underset{0}{\underset{⏟}{B{x}_{2}}}=B{x}_{1}.$
We have,
$B{A}^{+}Ax=B{A}^{+}A{x}_{1}=B{x}_{1}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}B{A}^{+}A=B.$