lower bound on the norm of (correlated) matrix multiplication

Question

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.

Answer 1

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.