<p>Let $M$ be symmetric and positive definite matrix (SPD). It is known [1] that <br/></p>

<p><strong>if</strong> $M$ is SPD and 
in addition satisfies $M_{ij}\leq 0$, for $i\neq j$ (called M-matrix) <br/> 
<strong>then</strong> there is a <strong>positive diagonal</strong> matrix $D$ such that <br/>
$DMD$ is symmetric, <strong>diagonally dominant</strong> and $M_{ij}\leq 0$ (called SDDM matrix).</p>

<p><strong>Q1</strong>: Let $M$ be SPD. Is there a <strong>diagonal</strong> matrix $D^{\prime}$ with $D^{\prime}_{ii} \neq 0~~ \forall i$ such that <br/>
$D^{\prime}MD^{\prime}$ is symmetric and <strong>diagonally dominant</strong> matrix?</p>

<p><strong>Q2</strong>: Let $M$ be SPD and $[M\mathbf{1}]_i\leq[\mathrm{diag}(M)\mathbf{1}]_i~~\forall i$ (a weaker condition than $M_{ij}\leq 0$). <br/>
Is there a <strong>diagonal</strong> matrix $D^{\prime}$ with $D^{\prime}_{ii} \neq 0~~ \forall i$ such that <br/>
$D^{\prime}MD^{\prime}$ is symmetric and <strong>diagonally dominant</strong> matrix?</p>

<p><strong>Q3</strong>: Is there a matrix $D^\prime$ from above such that all entries $D^{\prime}_{ii} \neq 0$ belong to <br/> a <strong>bounded</strong> range, say $[-1,1]\backslash{\{0\}}$.</p>

<p>[1]: Topics in Matrix Analysis
<a href="http://ebooks.cambridge.org/ebook.jsf?bid=CBO9780511840371" rel="nofollow">http://ebooks.cambridge.org/ebook.jsf?bid=CBO9780511840371</a></p>

CC BY-SA 3.0
2015-02-06T19:46:28.050Z
This question probably fits better at math.se or mathoverflow.
R B
CC BY-SA 3.0
2015-02-06T20:31:19.783Z
Cross-posted to http://mathoverflow.net/questions/195875 .
Emil Jeřábek
<linear-algebra><matrices><spectral-graph-theory><matrix-product>
Is there a diagonal matrix D such that DMD is SDD, where M is SPD matrix
Pavel
# Is there a diagonal matrix D such that DMD is SDD, where M is SPD matrix

(Asked by: username_0 on 2015-02-06)

Let $M$ be symmetric and positive definite matrix (SPD). It is known [1] that

**if** $M$ is SPD and
in addition satisfies $M\_{ij}\leq 0$, for $i\neq j$ (called M-matrix)   
**then** there is a **positive diagonal** matrix $D$ such that   
$DMD$ is symmetric, **diagonally dominant** and $M\_{ij}\leq 0$ (called SDDM matrix).

**Q1**: Let $M$ be SPD. Is there a **diagonal** matrix $D^{\prime}$ with $D^{\prime}\_{ii} \neq 0~~ \forall i$ such that   
$D^{\prime}MD^{\prime}$ is symmetric and **diagonally dominant** matrix?

**Q2**: Let $M$ be SPD and $[M\mathbf{1}]\_i\leq[\mathrm{diag}(M)\mathbf{1}]\_i~~\forall i$ (a weaker condition than $M\_{ij}\leq 0$).   
Is there a **diagonal** matrix $D^{\prime}$ with $D^{\prime}\_{ii} \neq 0~~ \forall i$ such that   
$D^{\prime}MD^{\prime}$ is symmetric and **diagonally dominant** matrix?

**Q3**: Is there a matrix $D^\prime$ from above such that all entries $D^{\prime}\_{ii} \neq 0$ belong to   
 a **bounded** range, say $[-1,1]\backslash{\{0\}}$.

[1]: Topics in Matrix Analysis
<http://ebooks.cambridge.org/ebook.jsf?bid=CBO9780511840371>

**Tags:** linear-algebra, matrices, spectral-graph-theory, matrix-product

> **Comments:**
> **username_1**: This question probably fits better at math.se or mathoverflow.  
> **username_2**: Cross-posted to http://mathoverflow.net/questions/195875 .  

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