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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?

24n8
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3 Answers3

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

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

Will Jagy
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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.

p32fr4
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