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I am reading a paper about colorization via optimization. Let us say that I can construct a symmetric matrix A with the characteristic that the sum of all elements in any row is 0 and all elements on its diagonal are positive. Well, as the sum of all rows of the A matrix is 0, the rows are linearly dependent and then it is not full rank. The A matrix is then singular. For my problem it's pretty obvious that this matrix is semidefinite positive, but I can't proof.

I've tried using the definition xTAx >= 0 without success. Checking at MATLAB, all the eigenvalues are >= 0, so the matrix is semidefinite positive, but I can't proof by this way too. I think that the right way involves the property that the sum of the rows is zero. Some suggestion?

Another question would involve the same semidefinite A matrix. Let C be a diagonal matrix with only a few elements equal to 1. I'm pretty sure that

A + C

is positive definite, and to answer that, I think the first question would be important.

PS: A + C is not diagonally dominant.

So, some suggestion to proof that A is semidefinite positive and A + C is definite positive with those informations?

Thanks.

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No.

$$ \left( \begin{array}{rrr} 1 & 2 & -3 \\ 2 & 1 & -3 \\ -3 & -3 & 6 \end{array} \right) , $$

Eigenvalues are $9,0,-1.$ Eigenvector $(1,-1,0)$ has eigenvalue $-1.$

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  • $\begingroup$ Hmmm, thanks. That's a counter example to show that with only this informations it's not possible to proof A is semidefinite positive... $\endgroup$ – Joao Feb 21 '14 at 13:32
  • $\begingroup$ Others informations about it is that A = I - W - Wt + WtW, where I is identity and W is a weighting neighborhood matrix. I mean, as pixels have 8 neighborhood, this matrix aij element contains the value of the influence pixel i has on pixel j according to some weighting function w normalized by the sum of the neighbors elements. So, the rows (not the columns) of W sums 1. For this case, wij = exp -(Y(i) - Y(r))²/2*variance(i), where Y is the intensity of the pixel i and variance the variance in the neighborhood of i. $\endgroup$ – Joao Feb 21 '14 at 13:45
  • $\begingroup$ Oh... I just realize that it's obvious that A is semidefinite positve. $\endgroup$ – Joao Feb 21 '14 at 19:05
  • $\begingroup$ A = I - W - Wt + WtW = (W-I)t(W-I) xt(W-I)t(W-I)x = ((W-I)x)t((W-I)x) >= 0. Thanks! $\endgroup$ – Joao Feb 21 '14 at 19:07

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