1
$\begingroup$
  • 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?

$\endgroup$
3
$\begingroup$

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$$\begin{pmatrix}-1&0&1\end{pmatrix}A\begin{pmatrix}-1\\0\\1\end{pmatrix}=-2.$$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can you provide the specific page? Thank you. $\endgroup$ – Bach Jul 30 at 16:21
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.