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I realize the following problem can be summarized as to show

Adding rows to an $n \times n$ non-singular matrix will keep or increase its minimum singular values.

Let $\bf A$ be an $m \times n$ matrix in $\Bbb C^{m\times n}$ with $m>n$ and the first $n$ rows of it being linear independent. Thus ${\mathbf{A}} = \left( {\begin{array}{*{20}{c}} {{{\mathbf{A}}_1}} \\ {{{\mathbf{A}}_2}} \end{array}} \right)$ where ${\bf A}_1$ is a $n \times n$ non-singular matrix, and ${\bf A}_2$ is the remaining $m-n$ rows. Let $\sigma({\bf A}_1)$ be the set of all singular values of ${\bf A}_1$ and $\sigma({\bf A})$ be the set of all singular values of $\bf A$. The problem is to show

$\min \sigma({\bf A}_1) \le \min \sigma({\bf A})$.

Tony
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1 Answers1

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EDIT. Since $m>n$, the singular values of $A_1$ and $A$ are the square roots of the eigenvalues of $A_1^*A_1$ and $A^*A=A_1^*A_1+A_2^*A_2$

As quadratic forms, $A_1^*A_1+A_2^*A_2\geq A_1^*A_1$, that is, for every vector $x$: $x^T(A_1^*A_1+A_2^*A_2)x\geq x^TA_1^*A_1x$. That implies that $\min spectrum(A_1^*A_1)\leq \min spectrum(A^*A)$ and we are done.

  • Thanks. If a matrix $A$ with its every element no less than the corresponding elements of the other matrix $B$, then the singular values of $A$ no less corresponding singular values of $B$? – Tony Sep 27 '17 at 13:48
  • No. See my edit. –  Sep 27 '17 at 14:00
  • @loupblanc How do you know that $x^T(A^_1A_1+A^_2A_2)x\geq x^TA^_1A_1x$ implies that the $spectrum(A^_1A_1)\leq spectrum(A^*A)$? As far as I know, you can have a look at the inequality while using an eigenvector of matrix $A_1$ for example. But that doesn't mean that that same vector is also an eigenvector for $A$. So how do you know which one of the singular values is then smaller? I also tried to describe the eigenvector of the smallest singular value of $A_1$ through a linear combination of the ones eigenvectors of $A$, but that didn't get me anywhere either. – Thomas Feb 19 '20 at 16:03
  • @k1next . Your implication is false. Here, take $x$ as an eigenvector associated to the smallest eigenvalue of $A^*A$. –  Feb 27 '20 at 09:33