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Let $A$ be a square matrix of size $n \times n$ ($n>2$) and let $B$ be $A$ if we delete the last row and column (size $(n-1) \times (n-1)$). Let $\sigma (A)$ be the least singular value of $A$ and $\sigma(B)$ the least singular value of $B$. Is it true that $\sigma(A) \leq \sigma(B)$?

I have seen the interlacing theorem but I am not sure if it can be used for non-symmetric matrices. My main concern is example 2 in 17-12 of http://math.ecnu.edu.cn/~jypan/Teaching/books/SVD.pdf

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    $\begingroup$ No. Take a $2\times 2$ matrix with $a_{11}=0$, so $\sigma(B)=0$, and make $A$ non-singular. (And interlacing won't help, this is false even for symmetric matrices, for the same reason.) $\endgroup$
    – Christian Remling
    Commented Apr 5, 2015 at 4:24
  • $\begingroup$ Thanks so much, but I forgot to put a constraint on the size of $A$ (at least size $3\times 3$), this was a trivial case. $\endgroup$
    – user31317
    Commented Apr 5, 2015 at 6:23

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No: $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\0 & 1 & 0 \end{pmatrix}$

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