Can every square matrix be written as product of an invertible matrix and a projection matrix?

Question

Let A be a square matrix. Will there always be an invertible matrix B and a projection matrix P such that A = BP?

Thanks

Answer 1

Sure: $A = (A+E)\Pi$ where $\Pi$ is the (orthogonal) projection onto $(\ker A)^\perp$ and E is any matrix with $\ker E = (\ker A)^{\perp}$ and $\operatorname{Im} E = \operatorname{Im}(A)^\perp$

Answer 2

The most direct answer: this result is implied by the Rank Normal Form of $A$
(which slightly overloads OP's notation)

$A= P\begin{bmatrix} I_r &\mathbf 0\\ \mathbf 0 & \mathbf 0\end{bmatrix} Q^{-1}= \Big(P Q^{-1}\Big) \Big( Q\begin{bmatrix} I_r &\mathbf 0\\ \mathbf 0 & \mathbf 0\end{bmatrix} Q^{-1}\Big)= BP' $

where $B$ is invertible and $P'$ is idempotent