Removing non-diagonal entries from a PSD matrix keeps the PSD property?

Question

I was thinking about positive semidefinite (PSD) sparse matrices and I began to wonder whether the following is true:

Let $A$ be a PSD matrix and let $\bar{A}$ be obtained from $A$ by replacing some of its non-diagonal entries by zero, while keeping symmetry. Is $\bar{A}$ PSD as well?

Thanks!

Answer 1

The matrix $$ \begin{pmatrix} 1 & 1 & 1\\ 1 & 2 & 2\\ 1 & 2 & 2 \end{pmatrix} $$ is SPD, but $$ \begin{pmatrix} 1 & 1 & 0\\ 1 & 2 & 2\\ 0 & 2 & 2 \end{pmatrix} $$ has negative determinant, so it is not SPD.