# Positive definitness of infinite dimensional matrices

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.

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.