Assume we have a $m \times n$ matrix $A$ with real entries representing an operator $T$ on $n$ dimensional real vector space $V$. Then we select a $n-1$ dimensional subspace of $E$ of $V$ and restrict the $T$ to $E$. Say the matrix representing $T|_{E}$ is $A_{E}$. What is the range of singular values of $A_{E}$ in terms of singular values of $A$?


I assume that by singular values, you mean eigenvalues of $A^*A$ (more precisely, their positive square roots). There are variational characterisations \begin{align*} \lambda_i&=\min_{\begin{smallmatrix}U\subset V\\\dim U=i\end{smallmatrix}}\max_{\begin{smallmatrix}v\in U\\|v|=1\end{smallmatrix}}|Av|\;,\\ &=\max_{\begin{smallmatrix}U\subset V\\\dim U=n+1-i\end{smallmatrix}}\min_{\begin{smallmatrix}v\in U\\|v|=1\end{smallmatrix}}|Av|\;. \end{align*} If you restrict to $E\subset V$ and denote the singular values of $A_E$ by $\mu_j$, you get $\lambda_i\le\mu_i$ from the first and $\mu_i\le\lambda_{i+1}$ from the second equation. So $$\lambda_1\le\mu_1\le\lambda_2\le\cdots\le\mu_{n-1}\le\lambda_n\;.$$

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    $\begingroup$ By playing around with the case when $A$ is diagonal and $E$ is a coordinate subspace, one can also see that these interlacing inequalities are sharp. $\endgroup$
    – Terry Tao
    Apr 27 '16 at 19:30
  • $\begingroup$ thanks Sebastioan and Terry. I was hoping someone will give a one line cool proof by real stable polynomials. Sebastian's answer is more than enough though. $\endgroup$
    – alpx
    Apr 28 '16 at 10:33
  • $\begingroup$ @alpx Just note that the inequalities in the answer are only sharp for each individual singular value. For the whole tuple $(\mu_1,\dots,\mu_{n-1})$, there are more restrictions. For example, if $\lambda_1<\lambda_2<\lambda_3$, then you cannot get $\mu_1=\lambda_2=\mu_2$. If I am not mistaken, one needs some advanced geometry to find a full set of inequalities describing all possible tuples. $\endgroup$ Apr 28 '16 at 13:22
  • $\begingroup$ Something that tripped me up for a moment: note that these characterisations of singular values are with them ordered $\lambda_1\leq \lambda_2\leq...$ rather than the other way round (implied by the previous comment also). $\endgroup$
    – Kweku A
    Sep 30 '20 at 9:08

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