$A$,$B$ orthogonal projection matrices following $\text{Col}(B) \subset \text{Col}(A)$. Prove $AB$ = $BA$ = $B$.

Question

We are given that both $A$ and $B$ are orthogonal projection matrices (i.e. for all $v \in \text{Col}(B)$ we have $Bv = v$, and for all $v \perp \text{Col}(B)$ we have $Bv = 0$) and that $\text{Col}(B) \subset \text{Col}(A)$. My task is to prove $AB = B$ and $BA = B$.

If I can prove that $AB = B$, then the second part is pretty straightforward:

Proof: Suppose $AB = B$ with $A,B$ defined above. Then $A^T = A$ and $B^T = B$, so we have \begin{align} (AB)^{T}&=B^{T} \\ B^TA^T &= B \\ BA &= B \end{align}

How can I prove $AB = B$? I was thinking something along the lines of the following:

Proof: Suppose $x \in \text{Col}(B)$. Then we have $Bx = x$. This expression implies that $ABx = Ax = x = Bx$. Thus, $AB = B$.

What are the flaws in this logic?

Answer 1

For the other part:

$\forall x\in V$, $x=x_a+x_a'$ where $x_a\in \text{Col}(A)$ and $x_a'\in \text{Col}(A)^\perp$.

Since $\text{Col}(B)\subset \text{Col}(A)\Rightarrow \text{Col}(A)^\perp \subset \text{Col}(B)^\perp$

So we have $x_a'\in \text{Col}(B)^\perp\Rightarrow Bx_a'=0$

$$BAx=BAx_a=Bx_a=B(x_a+x_a')=Bx\Rightarrow BA=B$$

Answer 2

I'm not sure if this is rigorous but intuitively. We consider two cases: $x\in Col(B)$ and $x\notin Col(B)$. If $x\in Col(B)$ then A also projects $x$ into $Col(B)$ (so $x=Ax\in Col(B)$ as well) and so $Bx=x$, hence $BAx=Bx=x$. If $x \notin Col(B)$ we immediately have $Bx=0$ and we have two more cases. If $x\in Col(A)$ then A projects it into a larger subspace, and if we let $y=Ax$, we have $y\notin Col(B)$ and so $BAx=By=0=Bx$ because B is an orthogonal projection matrix. If $x\notin Col(A)$ then it is even simpler, as $Ax=0$, and so $BAx=B0=0=Bx$.