Is a symmetric positive definite matrix always diagonally dominant?

A Hermitian diagonally dominant matrix $A$ with real non-negative diagonal entries is positive semidefinite.

Is it possible to have a Hermitian matrix be positive semidefinite/definite and not be diagonally dominant?

In other words, if I know that a matrix $M$ is symmetric positive definite then can I ensure $M - dI$, for a real number $d$, is positive definite only by ensuring $M - dI$ is diagonally dominant with non-negative diagonal entries?

I am aware that $d \le \lambda_{min}$, $\lambda$ being eigenvalue, for the matrix to remain semidefinite, but I need to avoid eigenvalue computation.

Thanks!

This was answered in the comments.

The matrix $\left(\begin{matrix} 1&-2 \\ -2&4\end{matrix}\right)$ is symmetric and positive semidefinite, but not diagonally dominant. You can change the "positive semidefinite" into "positive definite" by changing the $−2$'s to $−3/2$'s. Does this answer your question? (I am not totally sure what you are asking.) – darij grinberg Sep 30 '15 at 22:54