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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 $a\in[1,2]$? .........................................

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 $a=1$)

.....................................................

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?

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    @DavideGiraudo: Could you provide a proof for the eigenvalues of matrix $a diag(V)-V$ in case of $w_{ij}=1$?2012-02-15

2 Answers 2

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In the case $w_{ij}=1$ we have $V=\pmatrix{n&-1&-1&\cdots &-1 \\\ -1&n&-1&\ldots &-1 \\\ \vdots&\vdots&\ddots& &-1\\\ -1&-1&-1&\ldots&n}$ and $M_a:=a\operatorname{diag}(V)-V=\pmatrix{n(a-1)&1&1&\cdots &1 \\\ 1&n(a-1)&1&\ldots &1 \\\ \vdots&\vdots&\ddots& &1\\\ 1&1&1&\ldots&n(a-1)}.$ We can compute the determinant $\det(M_a-XI_n)$ adding to the first line all the other one. We get $\det(M_A-XI_N)=(n(a-1)-(n-1)X)(n(a-1)-1-X)^{n-1},$ and if we want $M_a$ positive semi-definite we should have $n(a-1)-1\geq 0$ so $a-1\geq\frac 1n$.

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    The diagonal entries of $V$ are $n-1$.2012-02-17
2

Claim: For a symmetric real matrix $A$, then $tr(X^TAX)\ge 0$ for all $X$ if and only if $A$ is positive semidefinite.

  • 0
    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 $a\cdot diag(V)-V$ still hold? Note the comment from DavideGiraudo from above, and his claim for case $w_{ij}=1$. Could something similar be derived for general $w_{ij}\geq 0$?2012-02-15