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 $\|x\| = 1$.

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – information_interchange Apr 6 at 4:52
  • $\begingroup$ 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. $\endgroup$ – Robert Israel Apr 6 at 13:24
  • $\begingroup$ 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? $\endgroup$ – information_interchange Apr 6 at 15:31
  • $\begingroup$ 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 $\|Ax\| \ge \sqrt{\lambda} \|x\|$. $\endgroup$ – Robert Israel Apr 6 at 16:07
  • $\begingroup$ 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. $\endgroup$ – information_interchange Apr 6 at 23:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.