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After some search here and on Google, I couldn't find a way to determine the definiteness of this matrix:

\begin{bmatrix}0&1\\1&0\end{bmatrix}

My understanding is that it should be negative semi-definite since all principal minors are $\leq 0$.

However, in the sample solutions of the book I am working through it stated that it is neither positive nor negative semi-definite.

A general procedure (for positive and negative semi-definiteness) would be really helpful.

Why is that?

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    $\begingroup$ A matrix is negative definite if its k-th order leading principal minor is negative when k is odd, and positive when k is even. The latter condition is not true. $\endgroup$
    – Dark
    Jun 14, 2016 at 8:28
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    $\begingroup$ Are all the principal minors of $\left(\begin{smallmatrix} -1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$ nonpositive? And yet the matrix is clearly negative definite... $\endgroup$ Jun 14, 2016 at 10:23

4 Answers 4

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(Not a general procedure) Going back to the original definition, a matrix $M$ is positive semi-definite if $v^T Mv\ge 0$ for all $v\in\mathbb R^n$ (similar for negative). But

$$ \begin{bmatrix}a&b\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix} = 2ab $$ The expression is not always $\ge 0$ or $\le 0$, so the matrix is indefinite.

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In general, the signs of the eigenvalues of a symmetric matrix $A$ determine whether the matrix is positive (semi)-definite or negative (semi)-definite. If all the eigenvalues are positive (non-negative), the matrix is positive (semi)-definite. If all the eigenvalues are negative (non-positive), the matrix is negative (semi)-definite. In your case, the characteristic polynomial of $A$ is

$$ p_A(\lambda) = \lambda^2 - \operatorname{tr}(A) \lambda + \det(A) = \lambda^2 - 1 = (\lambda - 1)(\lambda + 1). $$

Thus, $A$ has two eigenvalues $\pm 1$. Since one is positive and another is negative, the matrix $A$ is not positive nor negative.

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Let $M$ be your matrix. You have $M(1, 1) = (1, 1)$ and $M(1, -1) = (-1, 1)$, so it has the eigenvalues $1 > 0$ and $-1 < 0$, so it is neither positive- nor negative-semidefinite.

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A less complicated way of remembering/explaining this is to look at your first principle minor(H1). H1≤0 but technically H1≥0 as well, as H1=0. In this case, your optimal is at a saddle point, and the function is indeterminant at that point(none of the definites,semi or otherwise). That last conclusion might not be a perfect explanation, but if you are just worried about which semi-definite it is, just remember that anytime a principle minor=0, it is none of them.

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