I have a matrix equation in the form of
$$u^{T}A^{-1}=v^{T}B$$
where $u$ and $v$ are vectors, $A$ is a square matrix, and $B$ is a non-square matrix.
I'm told that while there is a unique solution for $u$ for a given $v$, $v$ has an infinite number of solutions for a given $u$.
Also, the general solution for $v$ is given as
$$v=B^{T+}A^{-T}u+(I-B^{T+}B^{T})w$$
Why does $v$ have an infinite number of solutions for a given $u$? Does it have to do with $B$ being non-square and having only a pseudoinverse?
Where is the term $(I-B^{T+}B^{T})w$ in the general solution for $v$ coming from?
Given the matrix $X\in{\mathbb C}^{m\times n},$ use its pseudoinverse to construct ortho-projectors into the nullspace $$P=I-X^+X\\Q=I-XX^+$$ The defining (and easily verified) properties of these projectors are $$\eqalign{ XP &= 0,\quad P^2=P=P^* \\ QX &= 0,\quad Q^2=Q=Q^* \\ }$$ Set $\,X=B^T$ and check the proposed solution $$\eqalign{ v &= \quad X^+A^{-T}u &+\; Pw \\ Xv &= XX^+A^{-T}u &+\; 0 \\ }$$ This satisfies the original matrix equation only if $\;XX^+=I_m$
In other words if $\;\operatorname{rank}(X) = m$.