1
$\begingroup$

A matrix $A$ is positive semi-definite IFF $x^TAx\geq 0$ for all non-zero $x\in\mathbb{R}^d$. If all elements of $A$ are non-negative, does this guarantee that $A$ is positive semi-definite?

$\endgroup$
1
$\begingroup$

No. The matrix $A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ is not psd, as you can check by seeing that $(1,-1)A(1,-1)^T=-2$.

$\endgroup$
0
$\begingroup$

$$ \left( \begin{array}{cc} 1 & 153 \\ 153 & 1 \end{array} \right) $$

$\endgroup$
0
$\begingroup$

In general no. One way of defining positive definiteness is through the leading principal minors of a matrix. The $k^{th}$ leading minor if found by computing the deturminant of the matrix after deleting the last $n-k$ colomns and rows in an $n \times n $ matrix. It is quite common to see a matrix with all positive entries that has a negative deturminnant, this therefore means this matrix would not be positive definite. For example, If you look at the leading principal minors of the following matrix $ A \in \mathbb{R}^{n\times n}$:

$$ A = \left( \begin{array}{cc} 1 & 1 \\ 2 & 1 \end{array}\right),$$

for $A$ to be positive definite $det(1)>0$ and $det(A)>0$. This is clearly not the case as $det(A)=-1$. In fact this particular matrix is indefinite.

$\endgroup$

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.