Sum of M matrices

For a symmetric $M$-matrix $M$ and a positive diagonal matrix $D$, I'm trying to evaluate whether $$x^TDMx = \frac{1}{2}x^T(DM+MD)x > 0$$ for all $x \in \mathbb{R}^n$. Since $M$-matrices are $D$-stable, I believe it is true that $DM$ and $MD$ are both $M$-matrices. So, is it true that $DM + MD$ is positive definite? If not, what would a counter-example be, and what would be a sufficient condition such that this inequality holds?

What if I strengthened $M$ to be a diagonally dominant matrix, with strict dominance in at least one row?

• If $M$ is positive definite and $D$ is positive diagonal, then both $MD$ and $DM$ are positive definite. Sum of $N$ positive definite matrices is also positive definite, so you are right. – Seyhmus Güngören Jan 15 '15 at 0:20
• $MD$ and $DM$ are not symmetric matrices, so I cannot see how they could be positive definite in the typical sense, unless you are using some other result of which I am not aware. – John Jan 15 '15 at 0:25
• Check this one. math.stackexchange.com/questions/4336/… what I am saying is true but not in general. – Seyhmus Güngören Jan 15 '15 at 0:47
• So if I'm understanding things properly, if $DM$ is diagonalizable, my desired property will hold? – John Jan 15 '15 at 0:57
• @john No, it won't. Product of two SPD matrices is always diagonalizable. – Algebraic Pavel Jan 15 '15 at 1:01

If $M$ is an M-matrix, then there is a positive definite diagonal matrix $D$ such that $DM+MD$ is positive definite (when $M$ is symmetric, that $D$ could trivially be the identity matrix). I'm afraid that this is not true for any p.d. $D$. Take, e.g., $$M=\pmatrix{2 & -1 \\ -1 & 2}, \quad D=\pmatrix{20 & 0 \\ 0 & 1}.$$ This covers also the case of a diagonally dominant $M$.

I believe that there's no sufficient condition for this to hold true unless $M$ is diagonal. Assume for now that $n=2$. Let $$M=\pmatrix{\alpha&-\gamma\\-\gamma&\beta}, \quad \alpha,\beta,\gamma>0.$$ Take $$D=\pmatrix{\delta&0\\0&1}, \quad \delta>0,$$ and assume $x:=[\xi,1]^T$. We have $$x^TDMx=\alpha\delta\xi^2 -\gamma(1+\delta)\xi + \beta.$$ This quadratic function of $\xi$ attains negative values if the discriminant $$\tag{1} d(\delta):=\gamma^2(1+\delta)^2-4\alpha\beta\delta$$ is positive for some $\delta$. Indeed, it is positive for large enough $\delta$. Hence if $\gamma>0$, there is always a diagonal matrix $D$ and a vector $x$ such that $x^TDMx<0$.

If $n>2$, you can reduce the problem to the $2\times 2$ case by looking for an $x$ having nonzeros in the positions $i$ and $j$ for which $m_{ij}<0$ ($i\neq j$) which will effectively reduce the proof the what we just did here. This gives:

For any $n\times n$ ($n\geq 2$) symmetric M-matrix $M=(m_{ij})$ such that $m_{ij}<0$ for some $i\neq j$, there is a diagonal matrix $D$ such that $DM+MD$ is not positive definite.

Also:

Let $M$ be a symmetric M-matrix such that $DM+MD$ is positive definite for any positive definite diagonal matrix $D$. Then $M$ is diagonal.

• Thanks for the interesting counter-example. I guess the intuition is that if $$\frac{d_{\rm max}}{d_{\rm min}} >\!\!> 1$$ the result can fail. I wonder if one can state a sufficient condition like $$\frac{d_{\rm max}}{d_{\rm min}} \leq \frac{\lambda_{\rm max}(M)}{\lambda_{\rm min}(M)}$$ – John Jan 17 '15 at 18:58