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.


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    $\begingroup$ 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.) $\endgroup$ – darij grinberg Sep 30 '15 at 22:54
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    $\begingroup$ Note: The converse direction holds. A symmetric matrix which is diagonally dominant is positive definite. I guess the same holds with "Hermitian" (no guarantees). $\endgroup$ – darij grinberg Sep 30 '15 at 22:56
  • $\begingroup$ Thanks that answers my question $\endgroup$ – muon Oct 1 '15 at 13:29
  • $\begingroup$ @darijgrinberg you wrote: 'A symmetric matrix which is diagonally dominant is positive definite.' This is only true if the values in the diagonal are positive. Diagonal dominance looks at the magnitude. For example $\begin{bmatrix} -3 & 1\\ 1 & 2 \end{bmatrix}$ is symmetric diagonally dominant but not positive definite. $\endgroup$ – Hjan Dec 10 '18 at 15:53
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    $\begingroup$ @Hjan: thanks for catching this. I was using a nonstandard definition of diagonal dominance, in which the diagonal entries were required to be positive. $\endgroup$ – darij grinberg Dec 10 '18 at 18:11

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