Given an $n \times k$ matrix $A$ and an $n \times (n-k)$ matrix $B$ such that $\text{span}(A) \oplus \text{span}(B) = \mathbb{R}^n$, how to get the projection matrix on $\text{span}(A)$ parallel to $\text{span}(B)$, in terms of $A$ and $B$?
Edit:
Below it is worked out that $P=\Big(\big(^−^ (^ )^{−1}^ \big)^{−1} ^ − (^ )^{−1}^ \big(^−^(^)^{−1}^ \big)^{-1}^ \Big)$.
- Does it look right?
- Can we possibly simplify it?
- Is the horrendous look the reason that we rarely see it fleshed out in textbooks, etc.?
Edit 2:
I have verified numerically that the formula is correct. Also, with SVD $A = U_A \Sigma_A V_A^T$, $B = U_B \Sigma_B V_B^T$ and $U_{AB} = U_A^T U_B$, it can be simplified that $P = U_A \Big(I - U_{AB}^T U_{AB} \Big)^{-1} U_A^T - U_A U_{AB} \Big(I - U_{AB}^T U_{AB} \Big)^{-1}U_B^T$.
But I have little intuition on how to understand the expression. If anyone can provide some, that'd be great.
Let $U = [A|B]^{-1}$ and denote by $\bar U$ the $k\times n$ upper part of $U$. Then $P = A\bar U$. Indeed, since $\bar U[A|B] = \left[\begin{matrix}I_k&0\end{matrix}\right]$, we have $A\bar Ua_j = Ae_j = a_j$ and $A\bar Ub_i = A(0) = 0$.