If $ImA = (ImB)^{\bot}$ then $B^TA = 0$

Question

I have two square matrices $A$ and $B$ $n\times n$ which can be treated as linear maps in $\mathbb{R}^n$. And I need to prove that if $ImA = (ImB)^{\bot}$ then $B^TA = 0$.

My attempt: if $ImA = (ImB)^{\bot} \ $ then $\forall \ x \in ImA$ and $\forall \ z \in ImB \rightarrow z^Tx = 0 \ $ hence $\exists \ u, v \in \mathbb{R}^n$ such that $(Bu)^T(Av) = 0 \rightarrow u^TB^TAv = 0$ and as I understand the last expression doesn't imply that $B^TA = 0$.

However, If $B^TA = 0 $ then $\forall \ u, v \in \mathbb{R}^n \rightarrow u^TB^TAv = 0 \rightarrow (Bu)^TAv = 0 \rightarrow ImB \ \bot \ ImA \rightarrow ImA = (ImB)^{\bot}$. Thus I've "proved" the inverse of what I need. $\mathbf{But}$ in my textbook inverse is said to be $\mathbf{false}$.

Could you please tell me how to move on in proof and why my "proof" of inverse statement isn't correct? Thanks in advance!

Answer 1

$\exists \ u, v \in \mathbb{R}^n$ such that $(Bu)^T(Av) = 0 \rightarrow u^TB^TAv = 0$ and as I understand the last expression doesn't imply that $B^TA = 0$.

It does, if $u^T C v = 0$ for all $u,v$ , the $C$ must be zero. To see this look at what happens if you plug in $u=e_i$ and $v=e_j$

However, If $B^TA = 0 $ then $\forall \ u, v \in \mathbb{R}^n \rightarrow u^TB^TAv = 0 \rightarrow (Bu)^TAv = 0 \rightarrow ImB \ \bot \ ImA \rightarrow ImA = (ImB)^{\bot}$.

The last implication is false. $\text{Im} A \bot \text{Im} B$ does imply that $\text{Im} A$ ought to be the whole orthogonal complement of $\text{Im}B$. It only needs to be a subset of it.

As a hint try to prove the simpler fact $\ker A = (\text{Im}A^T)^\bot$ instead. The excersice you try to solve is then a direct consequence because then you can argue $\text{Im} A = (\text{Im} B)^\bot = \ker B^T$.

Answer 2

From $\text{Im}B\bot\text{Im}A$ it can be concluded that $\text{Im}B\subseteq(\text{Im}A)^{\bot}$ but not necessarily that $\text{Im}B=(\text{Im}A)^{\bot}$.

For instance let $A$ the the matrix with $0$ in all its entries. Then for every matrix $B$ we will have $\text{Im}B\bot\text{Im}A$ but not for every matrix $B$ the set $\text{Im}B$ will equal $\text{Im}(A)^{\bot}=\mathbb R^n$.

Answer 3

The key point for the proof is that it's not just "$\exists \ u, v \in \mathbb{R}^n$ such that $(Bu)^T(Av) = 0 \rightarrow u^TB^TAv = 0$". Instead, your condition implies that the statement $u^TB^TAv = 0$ is true for all $u$ and $v$. To complete the proof, test the action of $B^TA$ on all basis vectors.

In your false proof, you have only proved that $ImA \subseteq (ImB)^{\bot}$. There may be other vectors perpendicular to $Im(B)$ which are not in the image of $A$.