# Singular values and trace norm of a submatrix

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: $$$$\sigma_i \leq \lambda_{i-2}$$$$ while other sources seem to make the stronger claim: $$$$\sigma_m \leq \lambda_{m-1} \leq \sigma_{m-1} \leq \ldots \leq \sigma_2 \leq \lambda_1 \leq \sigma_1.$$$$ So which one is it? And if the latter does not hold, can we at least prove the following? $$$$\sum_{i=2}^m \sigma_i \leq \sum_{i=1}^{m-1} \lambda_i$$$$

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.).