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Does the fact that a square zero matrix contains non-negative eigenvalues (zeros) make it proper to say it is positive semidefinite?

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The $n \times n$ zero matrix is positive semidefinite and negative semidefinite.

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"When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.

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    $\begingroup$ Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $\vec x^{\mathsf T}\!A\vec x \ge 0$ for all $\vec x \in \mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition. $\endgroup$ – Misha Lavrov Mar 19 at 3:56
  • $\begingroup$ @Misha Lavrov Nitpicking the nitpicking. Because both are equivalent, either can be taken as the definition, so some authors could choose the nonnegative eigenvalue criterion as the definition of positive-semidefinite. Anyhow, this is so presuming attention is restricted to symmetric matrices. $\endgroup$ – Mark L. Stone Mar 24 at 11:41

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