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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.

EDIT 1

The following papers seem to address part of the problem:

A. Horn, R. Steinberg. Eigenvalues of the unitary part of a matrix. Pacific Journal of Mathematics Vol. 9, No. 2, June, 1959, pp. 541-550

A. Horn, On the eigenvalues of a matrix with prescribed singular values, Proc. Amer. Math. Soc, 5 (1954), 4-7.

R. Horn, G. Piazza, T. Politi. Explicit polar decompositions of complex matrices. Electronic Journal of Linear Algebra. Vol. 18, pp. 693-699, November 2009

The matrix $W V^T$ is also called phase gain in the context of control systems.

The answer to the problem seems to be there but eludes me still...

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    Is anything known about the case in which $A$ has full rank? – 2017-01-26
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    Having zero column sum implies that $A$ has at most rank $n-1$ but for a general invertible matrix the SVD decomposition is unique up to reflection of the singular vectors. – 2017-01-26
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    Sure, but my question is this: if $A$ is non-symmetric positive definite (and perhaps strictly diagonally dominant), is the (uniquely determined) $WV^T$ also positive definite? I figure that it might be well known. If not, what motivated this question? It's unusual to ask whether an orthogonal matrix is positive semidefinite. – 2017-01-26
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    Also: it may be useful to know that the possible matrices $WV^T$ are all neareast orthogonal neighbors to $A$. Also, $(W\Sigma W^T)(WV^T)$ and $(WV^T)(V\Sigma V^T)$ are both polar decompositions. – 2017-01-26
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    Thanks for your interest. It may be well known but, to be honest, I am struggling to find the answer for a general case. Seems to me that the answer should be no, since I think I can change the sign of the eigenvalue just because the row of $V^T$ I am changing sign of is the singular vector in the null space of $A$, I guess I couldn't modify it out of the null space without also changing the sign of the corresponding column of $W^T$ as to maintain the positivity of the singular value associated to the column/row. Unfortunately I don't have a formal proof. – 2017-01-26
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    In fact, if $A$ is invertible, then $WV^T$ is unique. My earlier comments give classifications of what $WV^T$ can be in general. – 2017-01-26
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    And again, it would be nice to know how you came up with this question – 2017-01-26
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    The question comes from a numerical method represented here as a matrix $I + A$, where $A$ is low rank, diagonally dominant, and all the other properties. I am looking for continuity and weak coercivity estimates, for which I intend to diagonalize $A$ by changing the basis of my spaces, then I have to argue positive definiteness of $W I V^T = W V^T$. – 2017-01-26
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    If you want to try numerical experiments, I'd check matrices of the form $$ A = \pmatrix{a & t\\0&1} $$ where $|t| < a$. Other than that, I've got nothing. Good luck. – 2017-01-26
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    @Omnomnomnom Thanks! I'll take a look into that. – 2017-01-26
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    @user1551 Exactly, it should say _positive or zero_, just look at the precise definition in the section "Characterizations" or to reference 4 in the Wikipedia article. – 2017-01-26
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    The definition in the "Characterizations" part still requires the matrix to be nonsingular, as $s>\rho(B)$. Plemmons does allow singular matrices, though. Anyway, I see that you do mean a singular matrix now. Thanks for the response. – 2017-01-26

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