I am attempting to understand the solution to the following problem:
Let $A \in M_n(\mathbb{R})$ be a square matrix of rank $r < n$. Show that there exists a matrix $B \in M_n(\mathbb{R})$ such that for all $c \in \mathbb{R}$ $det(B + cA)$ is non zero.
I have inserted the approach used by the markscheme below, as well as my commentary on each step, where I highlight the things that I dont understand too well and would like some help with. (Please restrict your answers to this markscheme).
Firstly mark scheme uses matrix similarity:
1. $A= P^{-1} \begin{bmatrix} I_r & 0 \\ 0 & 0\end{bmatrix} Q$ where $P$ and $Q$ are invertible square matrices.
(I think that this part just means that any square matrix can can be written in reduced row echelon form.)
2. And then goes on to define a matrix $R$ (product of elementary matrices), such that:
$\begin{bmatrix} I_r & 0 \\ 0 &0\end{bmatrix} R = \begin{bmatrix} 0 & I_r \\ 0 & 0\end{bmatrix}$
(I think I know whats going on here: the matrix is flipped so that upon the addition of the identity matrix to this new matrix, the rank equals n).
Thus: $A= P^{-1} \begin{bmatrix} 0 & I_r \\ 0 &0\end{bmatrix}R^{-1}Q$
- Here is the crucial part which I do not understand:
"We can therefore define $B = P^{-1}R^{-1}Q$ such that $B + \lambda A$ has the same rank as $I_n + \lambda \begin{bmatrix} 0 & I_r \\ 0 & 0\end{bmatrix} $.
This just simply makes little sense to me. I think that I know what the mark scheme is getting at: to show that the matrix $B+ \lambda A$ is similar to $P^{-1}R^{-1}Q + \lambda P^{-1} \begin{bmatrix} 0 & I_r \\ 0 & 0\end{bmatrix} R^{-1}Q$
But, if its similar, upon multiplication by $P^{-1}$ and by $Q$ by both sides, I get:
$P^{-1}P^{-1}R^{-1}QQ + P^{-1}\lambda P^{-1} \begin{bmatrix} 0 & I_r \\ 0 & 0\end{bmatrix} R^{-1}QQ$ which is clearly not $B+ \lambda A$ (or is it?). If someone could please clarify whats going on here, I would be grateful.
The point is to make $$ \det \left(P^{-1}R^{-1}Q + \lambda P^{-1} \begin{bmatrix} 0 & I_r \\ 0 & 0\end{bmatrix} R^{-1}Q \right) \neq 0. $$ In fact, by multiplying $\det P$ on the left, and $\det (R^{-1}Q )^{-1}$ on the right, we have $$ \begin{align} \det P \det \left(P^{-1}R^{-1}Q + \lambda P^{-1} \begin{bmatrix} 0 & I_r \\ 0 & 0\end{bmatrix} R^{-1}Q \right) \det(R^{-1}Q)^{-1} =\det(I+\lambda \begin{bmatrix} 0 & I_r \\ 0 & 0\end{bmatrix}) = 1. \end{align} $$