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Let $A \in \mathbb{C}^{n\times n}$, is a hermitian matrix with guaranteed real and positive diagonal entries. Also $\det(A)>0$. Is $A$ positive-definite?

Positive-definite matrices have real and positive diagonal entries, but I have not found any resource whether the converse is true or not. I know that this is true in case of eigenvalues, but I wish not to calculate eigenvalues. This question solves it for a real matrix and I was wondering if the same can be done for complex matrix?

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    $\begingroup$ in that case $A=I + B B^H$ is positive definite, because $B B^H$ is positive semidefinite, and $I$ is positive definite. $\endgroup$ – orangeskid Jan 10 '20 at 15:15
  • $\begingroup$ @orangeskid thanks so much. I should have properly explained in my question. $\endgroup$ – impopularGuy Jan 10 '20 at 15:42
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The linked question only solves it for diagonally dominant matrices, and the same argument made there could be made for diagonally dominant complex matrices.


In general, the answer to your question is no, even for real matrices. For example, the matrix $$\begin{bmatrix}1&-3 &&-3\\ -3 &1&&2\\-3&&2&&1\end{bmatrix}$$ has two negative eigenvalues and one positive eigenvalue.

Note, of course, that the matrix I provide is not diagonally dominant.

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  • $\begingroup$ Thanks, I now understand. What else can be done for checking whether $A$ is positive-definite or not? $\endgroup$ – impopularGuy Jan 10 '20 at 14:19
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    $\begingroup$ @titusarmah99 The easiest way to show a matrix is positive definite is to calculate its Cholesky decomposition. en.wikipedia.org/wiki/Cholesky_decomposition $\endgroup$ – 5xum Jan 10 '20 at 14:20
  • $\begingroup$ In my case, $A$ is calculated as $A=BB^H$ where $B\in \mathbb{C}^{n\times m}$. How will it make sense to find Cholesky decomposition? $\endgroup$ – impopularGuy Jan 10 '20 at 14:34
  • $\begingroup$ Sorry, it is $A=I_n + BB^H$ where $I_n$ is identity matrix. $\endgroup$ – impopularGuy Jan 10 '20 at 14:42
  • $\begingroup$ @titusarmah99 A sum of positive definite matrices is positive definite. $\endgroup$ – 5xum Jan 10 '20 at 18:25

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