Logarithm of a Gram matrix Given a Gram matrix $K$, we are interested in calculating its matrix logarithm $\log(K)$, and in particular, to relate minus this logarithm to the Laplacian of a graph.
We have noticed that $-\log(K)$ always has positive diagonal entries. Which is a good thing, but we would like to prove it. Any ideas? 
 A: You cannot prove that without imposing further conditions, because the statement is not true in general. For instance, $K=I=I^TI$ is a Gram matrix, but $-\log K=0$ doesn't have any positive diagonal entries. In fact, $-\log(K)$ (assuming that you take real symmetric logarithm) has positive diagonal entries if and only if all eigenvalues of $K$ lie inside $(0,1)$.
A: Assume further that $K_{ii} =1$ for all $i$. These are the ones I'm interested in, since I want $K$ to be a Pearson correlation matrix.
Diagonalize $K = Q \Lambda Q^T$ where $\Lambda$ contains the positive eigenvalues $\{\lambda_j\}$ on the diagonal. The columns of $Q$ are orthonormal eigenvectors $\{v^j: j=1,\dots,n \}$, and $Q^T = Q^{-1}$. 
Working out the product, the diagonal of $K$ is given by $1 = K_{ii} = \sum_j \lambda_j (v^j_i)^2$. Also, since $QQ^T= I$, we have, $\sum_j (v^j_i)^2=1$.
Now, for each $\lambda_j$:
$$-\log(\lambda_j) \geq 1-\lambda_j$$
$$\Rightarrow -\sum_j (v^j_i)^2 \log(\lambda_j) \geq \sum_j (v^j_i)^2 (1-\lambda_j) = 0$$
