If $A=[u_1\cdots u_r]$ and $B=[v_1\cdots v_r]$ have independent columns, prove that $B^TA$ is invertible, where $u_i,v_i\in\mathbb{R}^n, n\geq r$.
If $A,B$ have independent columns then $Ax=0\implies x=0$ and $By=0\implies y=0$.
I can also prove that $rank(B^TA)\leq \min(rank(B^T),rank(A))=r$
We have to prove, $B^TAx=0\implies x=0$, it would be helpful to get a hint on how to approach this.
Note: This is needed to derive the expression for an oblique projection matrix in the form $P=A(B^TA)^{-1}B^T$, where $\{u_1,\cdots,u_r\}$ is a basis for the range of P, i.e., $Range(P)$ and $\{u_1,\cdots,u_r\}$ is a basis for the rowspace of P, i.e, $N^\perp$, and the vectorspace $V=Range(P)\oplus Kernel(P)$
Please check Page 13 for the proof.
Best Answer
Note that $x\in \ker B^TA$ is equivalent to $Ax \in \ker B^T$. Since $A$ has independent columns, if $x$ is non-zero, then $Ax$ is a non-zero vector in the column space of $A$. Being in the null space of $B^T$ is the same as being in the orthogonal complement of the column space of $B$ (since $B^Tx=0 \Leftrightarrow y^TB^Tx=0 \text{ for all } y \Leftrightarrow \langle x, By\rangle=0 \text{ for all } y$ ).
Then $Ax$ is in the intersection of column space of $A$ and the orthogonal complement of the column space of $B$. But if these two have non-trivial intersection, their sum has dimension which is insufficient to span the ambient space ($\dim V+W=\dim V+ \dim W- \dim V\cap W $, which in our case gives at most $r+(n-r)-1=n-1$). Thus the only $x$ that can be in $\ker B^TA$ is the zero $x$.