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Given a finite dimensional vector space $E$, let us consider a non-negative operator $P:E\rightarrow E$. The question is: besides its standard characterization, where $P$ is self-adjoint and $\langle Pv,v\rangle \geq 0$, is there any other criteria to identify them? It is worth mentioning here it is also supposed to be known that such condition is equivalent to: $P$ self-adjoint and $\lambda\geq 0$, for every eigenvalue. Thanks in advance.

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Any factorization of the form $P=M^{T}M$ is sufficient to show that $P$ is positive semidefinite.

If $P$ is given to you in matrix form and you're interested in a numerical computation then you can check for positive definiteness by computing the eigenvalues of $P$ or by computing a Cholesky factorization. For positive semidefiniteness, you'd have to use some kind of tolerance since any floating point computation of the eigenvalues or the Cholesky factorization would be subject to round-off errors.

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I will presume that $E$ is a complex vector space, or at least that $P$ is self-adjoint.

To determine that $P$ is positive definite, it suffices to apply Sylvester's criterion.

Now, let $P$ be a self-adjoint operator, and let $X$ be a matrix whose columns form an orthonormal basis for the image of $P$ (that is, the image of $X$ is the image of $P$ and $X^*X = I$). Then $P$ is positive (non-negative definite) if and only if the matrix $X^*PX$ is positive definite.

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  • $\begingroup$ A little nuance: @Brian Borchers asks for condition of semidefiniteness, connected to semidefinite positiveness. $\endgroup$ – Jean Marie Jun 26 '16 at 22:07
  • $\begingroup$ @JeanMarie see the entire last paragraph of my post $\endgroup$ – Omnomnomnom Jun 27 '16 at 0:49

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