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 $\Re(\lambda\_i) < 0$?***



Can I modify the singular values of a matrix in order to get a negative eigenvalue?
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. 


