# If the diagonal entries are real and positive, is the matrix positive-definite?

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?

• in that case $A=I + B B^H$ is positive definite, because $B B^H$ is positive semidefinite, and $I$ is positive definite. – orangeskid Jan 10 '20 at 15:15
• @orangeskid thanks so much. I should have properly explained in my question. – impopularGuy Jan 10 '20 at 15:42

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.
• Thanks, I now understand. What else can be done for checking whether $A$ is positive-definite or not? – impopularGuy Jan 10 '20 at 14:19
• 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? – impopularGuy Jan 10 '20 at 14:34
• Sorry, it is $A=I_n + BB^H$ where $I_n$ is identity matrix. – impopularGuy Jan 10 '20 at 14:42