Let $A \in \mathbb{R}^{n \times n}$ be a real nonsymmetric matrix with eigenvalues $\left\{\lambda_i : i=1..n\right\}$ with positive real part $\Re(\lambda_i) > 0$ $\forall i=1..n$

Let $A=U\Sigma V^T$ be the singular value decomposition of $A$ with singular values $\left\{\sigma_i : i=1..n\right\}$, $\sigma_i>0$.

Let $\Sigma'$ be a diagonal matrix with a different set of singular values $\left\{\sigma_i' : i=1..n\right\}$, $\sigma_i'>0$ and let $\left\{\lambda_i' : i=1..n\right\}$ be the eigenvalues of $A'= U\Sigma' V^T$.

Is there a set $\left\{\sigma_i' : i=1..n\right\}$ such that there is at least one $\lambda_i'$ with negative real part $\Re(\lambda_i) < 0$?


Not necessarily. For example, consider $$ A = \pmatrix{\cos(\theta) & -\sin(\theta)\cr \sin(\theta) & \cos(\theta)}$$ with eigenvalues $e^{\pm i \theta}$ having positive real part if $-\pi/2 < \theta < \pi/2$.
Since $A$ is orthogonal, its SVD has $U = A$, $\Sigma = V^T = I$. Then $$A' = A \pmatrix{\sigma_1' & 0 \cr 0 & \sigma_2'}$$ has trace $(\sigma_1' + \sigma_2') \cos(\theta) > 0$ and determinant $\sigma_1' \sigma_2' > 0$, and therefore its eigenvalues have positive real part.

EDIT: On the other hand, it is not always impossible. There are $3 \times 3$ examples where the real part of a complex-conjugate pair of eigenvalues changes sign depending on the $\sigma'_j$. For example, take $$ \eqalign{U &= \pmatrix{\sqrt{18-6\sqrt{3}}/6 & 0 & -\sqrt{18+6\sqrt{3}}/6 \cr -\sqrt{9+3\sqrt{3}}/6 & \sqrt{2}/2 & -\sqrt{9-3\sqrt{3}}/6\cr \sqrt{9+3\sqrt{3}}/6 & \sqrt{2}/2 & \sqrt{9-3\sqrt{3}}/6\cr}\cr V^T &= \pmatrix{-\sqrt{9-3\sqrt{3}}/6 & \sqrt{9-3\sqrt{3}}/6 & \sqrt{18+6\sqrt{3}}/6\cr \sqrt{2}/2 & \sqrt{2}/2 & 0\cr -\sqrt{9+3\sqrt{3}}/6 & \sqrt{9+3\sqrt{3}}/6 & -\sqrt{18-6\sqrt{3}}/6\cr}}$$ which come from the SVD of $$\pmatrix{0 & 0 & 1\cr 1 & 0 & -1\cr 0 & 1 & 1\cr}$$ For some $(\sigma'_1, \sigma'_2, \sigma'_3)$, such as $(1,1,0.3)$, the eigenvalues all have positive real part; for others, such as $(1,1,0.4)$, the real eigenvalue is positive but the complex conjugate pair have negative real part.

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  • $\begingroup$ Excellent counterexample! Many thanks. I need to answer then: is the property "having eigenvalues with positive real part" fully encoded in the $U$ and $V$ matrices? In other words, is that property independent of the singular values? $\endgroup$ – Astor Apr 26 '17 at 15:58
  • $\begingroup$ Actually a counterexample is already $A=\Sigma$ and $U=V=I$, so that $A'=\Sigma'$ $\endgroup$ – Pietro Majer Apr 26 '17 at 18:43
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    $\begingroup$ Yes, that is right. The real question is then: is it always impossible? $\endgroup$ – Astor Apr 26 '17 at 21:23
  • $\begingroup$ @RobertIsrael Thanks a million for your interest, I hope we can keep discussing for a bit? The matrix you propose in your edit: $A = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 1 \end{array} \right)$ does not have eigenvalues with positive real part which was the hypothesis. Do you think you could still provide a counterexample? $\endgroup$ – Astor May 2 '17 at 13:19
  • $\begingroup$ This matrix $A = U \Sigma V^T$ was chosen specifically so it had two complex eigenvalues with real part $0$ and one real and positive. If you change the $\Sigma$ a little bit, you can get the complex eigenvalues to have postive real part or negative real part. Thus the actual example with eigenvalues having positive real part would be not this $A$ but another with a different $\Sigma$, e.g. (as I indicated) $\text{diag}(1,1,0.3)$. $\endgroup$ – Robert Israel May 2 '17 at 15:17

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