Looking for tighter bound on symmetric PSD matrices products

Yossi
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Homework Statement


Let K and L be symmetric PSD matrices of size N*N, with all entries in [0,1]. Let i be any number in 1...N and K’, L’ be two new symmetric PSD matrices, each with only row i and column i different from K and L. I would like to obtain an upper bound of the equation below: where .∗ is an element-wise multiply.

Homework Equations


|[sum(K' .∗ L') − sum(K .∗ L)] − (2/N)*[sum(K'L'') − sum(KL)] + (1/N*N)*[sum(K')sum(L') − sum(K)sum(L)]|

See attached for equation in LaTex/PDF.

The Attempt at a Solution


Using simple triangular inequality and bounding the three square brackets respectively,
I can bound the above with 12N − 13.
However, this is extremely loose.
Empirical experiments show that the constant coefficient should be much lower, probably closer to 1N.

Can you suggest any linear algebra properties connecting the equation's parts - which may help me get a tighter bound?
 

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Yossi said:

Homework Statement


Let K and L be symmetric PSD matrices of size N*N, with all entries in [0,1]. Let i be any number in 1...N and K’, L’ be two new symmetric PSD matrices, each with only row i and column i different from K and L. I would like to obtain an upper bound of the equation below: where .∗ is an element-wise multiply.

Homework Equations


|[sum(K' .∗ L') − sum(K .∗ L)] − (2/N)*[sum(K'L'') − sum(KL)] + (1/N*N)*[sum(K')sum(L') − sum(K)sum(L)]|

See attached for equation in LaTex/PDF.

The Attempt at a Solution


Using simple triangular inequality and bounding the three square brackets respectively,
I can bound the above with 12N − 13.
However, this is extremely loose.
Empirical experiments show that the constant coefficient should be much lower, probably closer to 1N.

Can you suggest any linear algebra properties connecting the equation's parts - which may help me get a tighter bound?

Does ##\sum(A)## mean ##\sum_i \sum_j a_{ij}## for matrix ##A = (a_{ij})##? When you say that each of ##K', L'##have only row ##i## and column ##i## different from ##K,L##, do you mean that ##K'## has both its row and column ##i## different from ##K## and that ##L'## has both its row and its column ##i## different from ##L##? I think that is what you mean, but it is worth checking. Are all the elements of ##K', L'## also in ##[0,1]##?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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