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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 $\Re(\lambda_i) < 0$?

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  • $\begingroup$ @user1551 That's interesting, could I ask you to add more to your comment? How does the $i$-th diagonal entry of $V^T U$ relates to the trace of $A$? If all the eigenvalues have positive real part, will there be a negative diagonal entry of $V^T U$? $\endgroup$ – Astor Apr 26 '17 at 14:44
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Let $Q=V^TU$. It suffices to consider the matrix $Q\Sigma$, because it is similar to (and hence has identical eigenvalues as) $U\Sigma V^T$.

If some $i$-th diagonal entry of $Q$ is negative, you may just magnify the $i$-th diagonal entry of $\Sigma$ to force the trace of $Q\Sigma$ to become negative, so that $Q\Sigma$ possesses an eigenvalue with negative real part.

If $Q$ has a nonnegative diagonal, what you want to achieve is not always possible, and I don't know what conditions would make $Q\Sigma$ or $A$ possess an eigenvalue with negative real part. Anyway here is a counterexample. Consider any $Q\in SO(2,\mathbb R)$ that has a nonnegative diagonal. Since $Q\Sigma$ has positive determinant, if it has an eigenvalue $\lambda$ with negative real part, it must possess either two real negative eigenvalues or a conjugate pair of non-real eigenvalues with negative real parts. In either case the trace of $Q\Sigma$ would be negative, which is a contradiction to the assumption that $Q\Sigma$ has a nonnegative diagonal.

For a more concrete counterexample, consider $$ Q=\frac1{\sqrt{2}}\pmatrix{1&-1\\ 1&1},\ \Sigma=\pmatrix{a\\ &b}. $$ By positive scaling, it suffices to consider the product $\pmatrix{1&-1\\ 1&1}\pmatrix{a\\ &1}=\pmatrix{a&-1\\ a&1}$. The eigenvalues of this matrix are $\dfrac{a+1\pm\sqrt{a^2-6a+1}}2$, whose real parts are always nonnegative when $a>0$.

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  • $\begingroup$ Great contribution! The question then seems to reduce to determine if $Q$ can have a negative diagonal element! By the way, I believe that I will end up proving that what I want to do cannot be done and the "positive real part of the eigenvalues" property is independent of the singular values. $\endgroup$ – Astor Apr 26 '17 at 15:02
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    $\begingroup$ @Astor You seem to have over-interpreted my answer You don't need a negative diagonal. What you need is only one negative diagonal entry of $Q$. And if all diagonal entries of $Q$ happen to be non-negative, what you want to achieve is not always possible (as illustrated by my answer), but I ain't saying that it is always impossible --- just impossible in some cases. $\endgroup$ – user1551 Apr 26 '17 at 15:25
  • $\begingroup$ Got it, thanks. My hunch is that what I ask for cannot be done though and the "real part positive definiteness" is codified into $U$ and $V$. $\endgroup$ – Astor Apr 26 '17 at 15:33
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There are $3×3$ examples where the real part of a complex-conjugate pair of eigenvalues changes sign depending on the singular values.

The key is to find a matrix with purely imaginary eigenvalues and perturb the singular values.

The question has been answered in MathOverflow. See here.

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