Let $O\in\mathbb{R}^{n\times m}$, $m>n$, be a matrix with orthonormal rows, that is $O O^\top =I_n$, where $\bullet^\top$ denotes transposition and $I_n$ the $n\times n$ identity matrix. Partition $O$ as follows $$ O=\left[O_1 | O_2\right], $$ where $O_1\in\mathbb{R}^{n\times n}$ and $O_2\in\mathbb{R}^{n\times (n-m)}$.

Let $\{\lambda_i(X)\}_{i=1}^n$ denote the set of eigenvalues of a matrix $X\in\mathbb{R}^{n\times n}$ and consider the following subset of the space of row-orthogonal matrices: $$ \mathscr{O}:=\left\{\,O=\left[O_1 | O_2\right]\in \mathbb{R}^{n\times m}, OO^\top =I_n\,:\, |\lambda_i(O_1)|<1,\, i=1,\dots,n\,\,\right\}. $$

My question. Are there other equivalent ways to characterize the subset $\mathscr{O}$ that do not directly involve the eigenvalues of $O_1$?

Any comment, help, suggestion or pointer to the literature is really appreciated. Thanks a lot!

  • $\begingroup$ What is $|\lambda(O_1)| < 1$ supposed to mean? $\endgroup$ – Omnomnomnom Jul 6 '18 at 16:16
  • $\begingroup$ @Omnomnomnom: It means that all the eigenvalues of $O_1$ have modulus strictly smaller than one. But let me edit the question to clarify this notation. $\endgroup$ – Ludwig Jul 6 '18 at 16:18
  • $\begingroup$ $O_2$ has full row-rank if and only if the largest singular value of $O_1$ is strictly less than $1$, which of course implies that $|\lambda(O_1)| < 1$. Would this be sufficient for your purposes? $\endgroup$ – Omnomnomnom Jul 6 '18 at 16:23
  • $\begingroup$ @Omnomnomnom: Actually, I would like to find conditions that hold for every column dimension $m>n$, and not only for $m\ge 2n$ (which is indeed the case when you require $O_2$ to be of full row-rank). $\endgroup$ – Ludwig Jul 6 '18 at 16:35
  • 2
    $\begingroup$ $O_1$ can be any matrix satisfying $\|O_1\| \leq 1$, where $\|\cdot\|$ denotes the spectral norm. If $O_2$ fails to have full row-rank, then we have $\|O_1\| = 1$. So it would seem that your question is equivalent to characterizing all matrices $A$ for which $|\lambda(A)| < \|A\|$. I'm not sure what results exist for this problem. $\endgroup$ – Omnomnomnom Jul 6 '18 at 16:41

A matrix $O_1$ will work as a block entry if and only if it satisfies $\|O_1\| \leq 1$. For a fixed $O_1$ satisfying $\|O_1\| \leq 1$, it suffices to take any $O_2$ satisfying $O_2 O_2^T = I - O_1O_1^T$.

As I state in the comment: in the "interesting case", we're looking for all matrices $O_1$ satisfying $|\lambda_{max}(O_1)| < \|O_1\| = 1$. To that end, it suffices to apply the criteria discussed on this post.


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.