# Sufficient conditions for positive semidefinite matrices

• Is a symmetric matrix with positive terms (i.e., $a_{ij} > 0$) and positive determinant positive semidefinite?

• Is a symmetric matrix with positive terms, positive determinant, and terms that satisfy $a_{ij}^2 \leq a_{ii} a_{jj}$ positive semidefinite?

The answer to your first question is negative. Consider$$A=\begin{pmatrix}1&2&2\\2&2&2\\2&2&1\end{pmatrix}.$$Each entry is greater than $0$ and the determinant is $2$. However$$\begin{pmatrix}-1&0&1\end{pmatrix}A\begin{pmatrix}-1\\0\\1\end{pmatrix}=-2.$$

b) The following statements are equivalent:

• The symmetric matrix $A$ is positive semidefinite
• All eigenvalues are non negative
• All the principal minors of $A$ are non negative.
• There exists a $B$ such that $A = B^{T}B$ where $B$ can be thought of as a square root of a positive operator.

[1] Kreyszig, Erwin. Introductory functional analysis with applications. Vol. 1. New York: wiley, 1989.

• Can you provide the specific page? Thank you. – Bach Jul 30 at 16:21

a) The answer is no. Look at the matrix $$A = \begin{pmatrix} 1&2&1&1\\ 2&1&1&1\\ 1&1&1&2\\ 1&1&2&1 \end{pmatrix}$$ All its entries are positive. Its determinant is 5, and thus positive. But one of its eigenvalues is $-1$, and thus it is not positive semidefinite.

b) I could not find a counter-example at this stage.