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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!

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  • $\begingroup$ 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
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    $\begingroup$ $O_1$ 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
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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.

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