Suppose a square symmetric matrix $V$ is given
$V=\left(\begin{array}{ccccc}
\sum w_{1s} & & & & \\
& \ddots & & -w_{ij} \\
& & \ddots & & \\
& -w_{ij} & & \ddots & \\
& & & & \sum w_{ns}
\end{array}\right) \in\mathbb{R}^{n\times n},$
with values $w_{ij}> 0$, hence with only positive diagonal entries.
Since the above matrix is diagonally dominant, it is positive semi-definite. However, I wonder if it can be proved that
$a\cdot diag(V)-V~~~~~a\in[1, 2]$
is also positive semi-definite. ($diag(V)$ denotes a diagonal matrix whose entries are those of $V$, hence all positive) In case of $a=2$, the resulting
$2\cdot diag(V)-V$
is also diagonally dominant (positive semi-definite), but is it possible to prove for $a\in[1,2]$?
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Note that the above proof would facilitate my actual problem; is it possible to prove
$tr[(X-Y)^T[a\cdot diag(V)-V](X-Y)]\geq 0$,
where $tr(\cdot)$ denotes matrix trace, for $X, Y\in\mathbb{R}^{n\times 2}$ and $a\in[1,2]$ ?
Also note that
$tr(Y^TVY)\geq tr(X^TVX)$ and $tr(Y^Tdiag(V)Y)\geq tr(X^Tdiag(V)X)$.
(if that facilitates the quest., assume $a=1$)
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Since the positive semi-definiteness could not generally be guaranteed for $a<2$, the problem casts to: for which restrictions on a does the positive semi-definiteness of a⋅diag(V)−V still hold?
Note the comment from DavideGiraudo, and his claim for case $w_{ij}=1$, for all $i,j$. Could something similar be derived for general $w_{ij}$≥0?
Best Answer
Claim: For a symmetric real matrix $A$, then $tr(X^TAX)\ge 0$ for all $X$ if and only if $A$ is positive semidefinite.