Assuming $A \in R^{n \times n}$ and is PSD and hermitian, so it has $k$ non-zero eigenvalues and $k<n$. Also its entries are all non-zero values.
Then i create $B$ as the principal submatrix of $A$ obtained by selecting same rows and columns of $A$ indicated by $I$, a subset of $\{1,2,…n\}$.
I noticed for a wide range of random matrices of $A$, as long as having $size(I)<k$ the resulted $B$ will be positive-definite meaning it has only positive eigenvalues. To generate $A$ i chose each entry as a random number in the range [-100 100] and doing $A^T*A$ and afterwards making some of the eigenvalues equal to zero using eigenvalue decomposition of $A$.
So although it is possible to create specific matrices $A$ which don't obey the above rule, i think there should be some pre-conditions on $A$ to let the above happens! and i like to find the mathematical way to explain that!
Best Answer
It is not true.
Let $A = \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}$, then $k=1$, $n=2$.
Let $I=\{2\}$, then $B=0$ which is not positive definite.
Edit:
Let $A = \begin{bmatrix} 8 & 0.1 & 0.1 & 0.1 \\ 0.1 & 6 & 0.1 & 0.1 \\ 0.1 & 0.1 & 0.1 & 0.1 \\ 0.1 & 0.1 & 0.1 & 0.1\end{bmatrix}$. It has $k=3$ non-zero eigenvalues. Let $I = \{ 3,4\}$, the matrix $B$ is singular.