Let $P\in\mathbb{R}^{n\times m}$ with $n>m$ and $Q\in\mathbb{R}^{n\times k}$ with $n>k$ such that $P^T P = I_m$ and $Q^T Q = I_k$. Also, assume $\text{ker}P^T \cap \text{ker}Q^T = \emptyset$. Then, prove the following claim:
There exists $c>1$ such that the matrix $$M = (I_n - cQQ^T) PP^T (I_n - cQQ^T) - (I_n - cQQ^T)$$ is positive definite. (That is, $v^T M v > 0$ for all $v\in\mathbb{R}^n$ such that $v\neq 0$ or, equivalently, all eigenvalues of $M$ are in open right half complex plane.)
Is the above claim true or false? If true, how to prove it?
Remark 1. The matrix $(I_n - cQQ^T) PP^T (I_n - cQQ^T)$ is positive semidefinite for all $c$ because it is in the form of $H^T H$.
Remark 2. The matrix $(I_n - cQQ^T)$ is positive semi-definite for $c=1$ and positive definite for $0\leq c <1$. But since we consider $c>1$, it comes out to be a non-definite matrix, which means it has both positive and negative eigenvalues.
Let $P=w=Q$ with $\|w\|=1$, $c>1$, and let $v\cdot w=0$, $v\ne0$. Then $$Mv=(I-cww^T)ww^T(I-cww^T)v-(I-cww^T)v=-v$$ $$\therefore v^TMv<0$$
More generally, if $v\in\ker P^T\cap\ker Q^T$, then $v^TMv\le0$.
Answer to modified question with $\ker P^T\cap\ker Q^T=\{0\}$.
Let $m=1$, $n>2$, let $P=w$ with $\|w\|=1$; let $Q$ be such that $Q^Tw=0$. Then, as before $Mw=0$.