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Assume $M=(b_{i,j})_{ i,j=1}^{\infty}$ is an infinite dimensional matrix such that $b_{i,i}>0$ and $b_{i,j}=b_{j,i}$ for all $i,j\in\mathbb{N}$ (i.e., $M$ is symmetric with positive diagonal entries). Let $\boldsymbol{a}=(a_i)\in \ell^2(\mathbb N)$. Assuming $\boldsymbol{a}M\boldsymbol{a}^T$ converges and makes sense can we conclude $\boldsymbol{a}M\boldsymbol{a}^T>0$?

It is known from the theory of matrices that a symmetric matrix with positive diagonal is indeed positive definite. I am trying so see weather the same is true for infinite dimensional matrices.

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It is not true that a symmetric (real, for example) matrix with positive diagonal is positive definite. This is manifest already in two dimensions: $S=\pmatrix{1 & -2 \cr -2 & 1}$ is not positive-definite because $\pmatrix{1&1}S\pmatrix{1\cr 1}=1+1-2-2<0$. For two-by-twos, the determinant must be positive. Etc.

Perhaps the question is not quite what was meant to be asked? Certainly notions of positivity for infinite matrices make sense, but the premise of the present one isn't right already in finite dimensions...

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  • $\begingroup$ Yes sir.Well that statement was too good to be true.Regardless I still think my matrix ,say its finite dimensional one is negative definite.It has the properties that is symmetric, all entries including off diagonal are negative and also their values are near -1/2. $\endgroup$ – BigM Jan 29 '14 at 1:38

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