Let $A$ be an $n \times n$, non-symmetric, real, weak diagonally dominant M-Matrix. Its diagonal is strictly positive, its off-diagonal is negative or zero and all its columns sum to zero. $A$ has rank less than $n$.

$A$ has a singular value decomposition $A = W \Sigma V^T$ where $W$ and $V^T$ are unitary matrices (i.e. $VV^T=WW^T=I$).

Because $A$ is low rank, $V^T$ and $W$ are not unique.

Is the following assessment true?:

For any $n \times n$, non-symmetric, real, weak diagonally dominant, low rank M-Matrix $A$ there exists a singular value decomposition $A = W \Sigma V^T$ such that $WV^T$ is positive semi-definite (i.e. its eigenvalues have positive real part).

Example:

$A = \left( \begin{array}{ccc} 1 & -1 & -2 \\ 0 & 2 & -1 \\ -1 & -1 & 3 \end{array} \right), W = \left(\begin{array}{ccc} -0.46 & 0.67 & 0.58 \\ -0.35 & -0.74 & 0.58 \\ 0.81 & 0.064 & 0.58 \end{array} \right),$ $ V^T=\left(\begin{array}{ccc} -0.31 & -0.26 & 0.91 \\ 0.26 & -0.94 & -0.18 \\ 0.91 & 0.18 & 0.37 \end{array} \right), \Sigma=\left(\begin{array}{ccc} 4.1 & 0 & 0 \\ 0 & 2.3 & 0 \\ 0 & 0 & 0 \end{array} \right)$

Eigenvalues of $WV^T$: $0.84+0.54i, 0.84-0.54j, 1$

Changing the sign of the third row of $V^T$ conserves the properties of all matrices but the last eigenvalue of $WV^T$ changes from $1$ to $-1$ making $WV^T$ lose positive definiteness.