# Smallest singular value interpretation as inverse spectral norm?

The largest singular value of a finite dimensional matrix is its spectral norm (L2 operator norm). In other words, it is the maximum scaling that A does to any vector. Does the smallest singular value have any interesting properties? For instance, if the singular values are all lower bounded by 0, then could $$\sigma_{min}$$ it be interpreted as a kind of "inverse" spectral norm; i.e. the minimum the matrix must scale any vector?

Yes: $$\sigma_{min}^2$$ is the least eigenvalue of $$A^* A$$, which is the minimum of $$x^* A^* A x = \|A x\|^2$$ for $$x$$ with $$\|x\| = 1$$.
• Can you explain why this is the case and provide a source? Does this imply that if $A$ has n distinct eigenvalues, then the smallest one (by magnitude) is the minimum amount A must scale the vector? – information_interchange Apr 6 at 4:52
• 1) If $T$ is hermitian and $\lambda$ its least eigenvalue, $T - \lambda I$ is positive semidefinite, so $x^*(T-\lambda I) x \ge 0$. 2) No, it does not. This is about singular values, not eigenvalues. – Robert Israel Apr 6 at 13:24
• Interesting, I don't see why $x^*(T-\lambda I)x \geq 0$ shows that the $\lambda$ is the smallest scaling that can be done. How is the definition of PSD related to scaling? What if T is non-square, let alone Hermitian? – information_interchange Apr 6 at 15:31
• I'm using $T = A^* A$. $x^* (A^* A - \lambda I) x = \|A x\|^2 - \lambda \|x\|^2$, so if this $\ge 0$, it says $\|Ax\| \ge \sqrt{\lambda} \|x\|$. – Robert Israel Apr 6 at 16:07
• Okay, that makes sense. So the key is $x^TA^TAx = \|Ax\|^2$, and then we have an expression for the norm and can conclude immediately based on the fact that $A^TA-\lambda I$ is PSD. But you haven't shown that $\lambda$ is the smallest such scaling possible. For instance, your argument also holds if we plug in $\lambda_{max}$, but we know that clearly the minimum scaling is not the largest singular value. – information_interchange Apr 6 at 23:49