Image of matrices to prove a matrix equality

Question

In a paper, they say that: Since $\textrm{Im}(M^\top) \subseteq \textrm{Im}[A^\top \quad B^\top]$, then

$$ M \bigg(-I + \begin{bmatrix} A\\ B \end{bmatrix}^\top \bigg( \begin{bmatrix} A\\ B \end{bmatrix} \begin{bmatrix} A\\ B \end{bmatrix}^\top \bigg)^\dagger \begin{bmatrix} A\\ B \end{bmatrix} \bigg) M^\top=0$$

I would like to kindly ask the following.

1. How can I understand the image of a matrix is a subset of the image of another matrix as the assumption here.

2. How can I prove the equality above.

Answer 1

Let $U = \begin{bmatrix} A^\top & B^\top \end{bmatrix}$. Then the claim becomes "$\text{Im}(M^\top) \subseteq \text{Im}(U)$ implies $M(-I + U(U^\top U)^\dagger U^\top)M^\top = 0$."

"$\text{Im}(M^\top) \subseteq \text{Im}(U)$" means that each column of $M^\top$ can be written as a linear combination of columns of $U$. In particular, this implies the existence of a matrix $V$ such that $M^\top = UV$. So, the equation can be rewritten as $$V^\top U^\top (-I + U(U^\top U)^\dagger U^\top) UV = 0$$ $$V^\top (-U^\top U + U^\top U(U^\top U)^\dagger U^\top U) V = 0$$

The middle term above is zero, since a property of the Moore-Penrose inverse is that $(U^\top U)(U^\top U)^\dagger U^\top U = U^\top U$.