1
$\begingroup$

Let $A$ be an $ m \times n$ matrix where $ m \leq n $, and let $ B$ the matrix obtained from $A$ by removing both its first row and its first column. Let us denote the singular values of $A$ by: \begin{equation*} \sigma_1 \geq \ldots \geq \sigma_m \end{equation*} and the singular values of $B$ by: \begin{equation*} \lambda_1 \geq \ldots \geq \lambda_{m-1} . \end{equation*} My question is: which interlacing inequalities apply here? According to some books, we can only say: \begin{equation} \sigma_i \leq \lambda_{i-2} \end{equation} while other sources seem to make the stronger claim: \begin{equation} \sigma_m \leq \lambda_{m-1} \leq \sigma_{m-1} \leq \ldots \leq \sigma_2 \leq \lambda_1 \leq \sigma_1. \end{equation} So which one is it? And if the latter does not hold, can we at least prove the following? \begin{equation} \sum_{i=2}^m \sigma_i \leq \sum_{i=1}^{m-1} \lambda_i \end{equation}

$\endgroup$
1
$\begingroup$

The inequality $\sigma_1(A)\ge\sigma_1(B)\ge\sigma_2(A)\ge\sigma_2(B)\ge\cdots\ge\sigma_{m-1}(A)\ge\sigma_{m-1}(B)\ge\sigma_m(A)$ holds if $A$ is a (square) positive semidefinite matrix. It doesn't hold in general, not even if $A$ is Hermitian. E.g. when $$ A=\pmatrix{0&0&1\\ 0&0&0\\ 1&0&0}, $$ the three singular values of $A$ are $1,1,0$ but the two singular values of $B$ are $0,0$. In this counterexample, we also have $\sum_{i=2}^m\sigma_i(A)=1>0=\sum_{i=1}^{m-1}\sigma_i(B)$.

Another counterexample: let $$ A=\pmatrix{0&3&0\\ 2&0&-2\\ 1&0&1}. $$ The three singular values of $A$ are $3,2\sqrt{2},\sqrt{2}$ and the two singular values of $B$ are $\sqrt{5}$ and $0$. Here we have $\sigma_2(A)=2\sqrt{2}>\sqrt{5}=\sigma_1(B)$ and $\sum_{i=2}^m\sigma_i(A)=3\sqrt{2}>\sqrt{5}=\sum_{i=1}^{m-1}\sigma_i(B)$.

It is true that $\sigma_i(A)\le\sigma_{i-2}(B)$ for $3\le i\le\min\{m,n\}$. Actually, if we delete a row (resp. a column) of $A$ to obtain a matrix $C$, we get $\sigma_j(A)\le\sigma_{j-1}(C)$. Similarly, if we delete a column (resp. a row) of $C$ to obtain a matrix $B$, we get $\sigma_k(C)\le\sigma_{k-1}(B)$. Combine the two inequalities, we get $\sigma_i(A)\le\sigma_{i-2}(B)$.

Interestingly, the inequality $\sigma_j(A)\le\sigma_{j-1}(C)$ can be obtained from the interlacing inequality $\lambda_1(A)\ge\lambda_1(B)\ge\cdots\ge\lambda_{m-1}(A)\ge\lambda_{m-1}(B)\ge\lambda_m(A)$ for eigenvalues of Hermitian matrices. For a proof, see corollary 7.3.6 of Horn and Johnson's Matrix Analysis (2nd ed.).

$\endgroup$

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.