Show that $\min_X \|X\|_2 = \frac{1}{\sigma_1}$

Question

Given $ A\in \mathbb{C}^{m\times n}$ with $\sigma_1$ as biggest singular value, and $\det(I-AX) = 0$ where $ X\in \mathbb{C}^{n\times m} $, can you show that $$\min_X \|X\|_2 = \frac 1{\sigma_1}\:?$$ I believe that $\|X\|_2 \geq \frac1{\sigma_1}$ follows from $1$ being an eigenvalue of $AX$ and the definition of the matrix norm, however I can't show that this bound is attained in all cases. I thought that maybe the pseudo-inverse would give it, but I think that gives the reciprocal of the smallest singular value, not the largest.

Answer 1

If $\det(I-AX)=0$, there is a $y$ with $\|y\|_2=1$ such that $AXy=y$ so $$ 1=\|y\|_2=\|AXy\|_2\leq \|A\|_2\|X\|_2=\sigma_1\|X\|_2. $$

Therefore, for any such $X$ (assuming it exists), $$ \|X\|_2\geq \frac{1}{\sigma_1}. $$

It remains to show that there is a solution with $\|X\|_2=1/\sigma_1$ (thus solving also the existence problem). Let $u_1$ and $v_1$ be the singular vectors of unit norm corresponding to $\sigma_1$ (such that $Av_1=\sigma_1 u_1$). Let's try to see what happens if we take $$ X_\star=\frac{1}{\sigma_1}v_1u_1^*. $$ We certainly have $\|X_\star\|_2=\frac{1}{\sigma_1}$ and $$ AX_\star=\frac{1}{\sigma_1}Av_1u_1^*=u_1u_1^*. $$ Now $$ (I-AX_\star)u_1=(I-u_1u_1^*)u_1=0, $$ which shows that $I-AX_\star$ is singular and hence $\det(I-AX_\star)=0$.