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I have a matrix $A$, which satisfies :

  • $A$ is symmetric;

  • all the diagonal entries of $A$ are equal to $1$;

  • other entries of $A$ is between $0$ and $1$.

My question is, whether $A$ is a positive semi-definite matrix?

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Check whether it is diagonal dominant. If yes, then it is positive (semi) definite. see: – Shiyu Apr 21 '11 at 14:33
up vote 2 down vote accepted

Here's a counterexample for size $3 \times 3$ (there are probably simpler examples, but well...): $$\begin{pmatrix} 1 & 0.9 & 0.9 \\ 0.9 & 1 & 0.1 \\ 0.9 & 0.1 & 1 \end{pmatrix}$$ is indefinite, since the eigenvalues are $0.9$ and $(21 \pm \sqrt{649})/20$.

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okay, that is good. – user9889 Apr 21 '11 at 15:02

For the $3\times 3$ matrix $\begin{pmatrix}1&a&b\\a&1&c\\b&c&1\end{pmatrix}$, we have to check if $1-a^2-b^2-c^2+2abc$ and $1-a^2$ are positive.

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I think its not right , the correction is to check $1-a^2-b^2-c^2+2abc$ – user9889 Apr 21 '11 at 14:59

Well, it is positive semidefinite. If you take $\left(\begin{smallmatrix} 1 & a \\ a & 1 \end{smallmatrix}\right)$, then you see that the characteristic polynomial is $\lambda^2 -2\lambda +1-a^2$. Hence, $\lambda = 1\pm .5 \sqrt{4-4(1-a^2)}$, which is positive if $0\leq a<1$.

For bigger matrices this is not true anymore.

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The OP didn't ask about 2x2 matrices. – Brian Vandenberg Apr 21 '11 at 16:05

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