Suppose that $\{a_1, a_2, \cdots, a_n; b_1, b_2, \cdots, b_r\}$ is a basis for vector space $V\in\mathbb{R}^{m\times m}$, where $n+r=m$. There is no inner product defined in $V$.
Let $U=\text{span}(a_1, a_2, \cdots, a_n), W=\text{span}(b_1, b_2, \cdots, b_r)$, thus $V=U\oplus W$.
How to determine the projection $P$ along $W$ onto $U$?
I stuck halfway below:
Let $A_{m\times n}=(a_1, a_2,\cdots, a_n), B_{m\times r}=(b_1, b_2, \cdots, b_r)$. Since $\text{range}(P)=U$, I can assume that $P=AX$ for some $X_{n\times m}$. As $X$ has degree of freedom $nm$, I need $nm$ equations to determine $X$.
On the other hand, the complement of $P$ is $I-P$, which is a projector along $U$ onto $W$, this implies that $(I-P)=BY$ for some $Y_{r\times m}$. To determine $Y$, I need $rm$ equations.
So it seems that I have a choice here: determine $X$ or $Y$. Suppose $n<r$, then I choose to determine $X$:
$P^2=P\implies AXAX=AX\implies XA=I_n$. This gives $n^2$ equations. I still need $rn$ equations for determine $X$, those should come from the condition that $\text{range}(I-P)=W$, but I don't how to proceed from here.