lower bound on the norm of (correlated) matrix multiplication Suppose $X, U \in \mathbb{R}^{n \times r}$, $n>r$, where $U$ is a fixed matrix and $X$ is a variable, and both are of full column ranks. Let $\mathfrak{R} = \{ \Psi \in \mathbb{R}^{r \times r}: \Psi  \Psi^\top = \Psi^\top \Psi = I_r \}$ be the set of rotation matrices in dimension $r$. Also suppose there exists constants $c_1,c_2 >0$ such that $X$ satisifes:
(a) $\|X\|_2 > c_1$: this implies that $X$ is away from the zero matrix.
(b) $\min_{\Psi \in \mathfrak{R}} \|X-U \Psi\|_F > c_2$: this implies that $X$ is away from $U$ (after the best possible rotation).
$\textbf{Question}$:
Given condition (a) and (b), can we find a constant $c_3 > 0$ (probably in terms of $c_1$ and $c_2$) such that the following holds?
\begin{align}
&\|(XX^\top - UU^\top)X\|_F > c_3 
\end{align}
Remark: We have from (b) that $\|XX^\top - UU^\top \|_F > c_4$. You may use this instead of (b). You may also use other norms/singular value conditions in (a) and (b) if that helps.
 A: No, there is no lower bound. Take for $\epsilon\neq 0$ arbitrarily small:
$$ X=\left( \begin{matrix} 1 & 0  \\ 0 & \epsilon \\ 0 & 0\end{matrix} \right)  \ \ \mbox{and} \ \ U=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{matrix} \right) .$$
However, if you add a condition on $X^T X$ having a uniformly bounded inverse then your claim is correct.
Let $f(X)=\|(X X^T-U U^T)X\|_F$. 
If we add the constraint 
$\|(X^T X)^{-1}\|\leq M<+\infty$ to the others then we claim that $f$
will have a non-zero lower bound. Suppose the contrary. 
Then we may find a sequence $X_n$ satisfying the
criteria and such that $f(X_n) \rightarrow 0$.
It is easy to see that $X_n$ must be bounded. Using compactness
we may assume that $X_n$ converges to $X$ which therefore verifies:
$$ (X X^T-U U^T)X = 0 .$$ 
Because of our auxilary condition
 $X^T X$ is invertible so
$$ X = U \left(U^T X 
(X^T X)^{-1} \right) \equiv U A$$
with $A$ an $r\times r$ matrix.
Both $X$ and $U$ have  full rank
so $A$ is invertible and then 
$$ U (A A^T-I) U^T X= 0 $$
from which (using again full rank) $I=A A^T$ which contradicts the
condition regarding rotations.
