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I have seen it claimed multiple places (e.g. here and here) that a (not necessarily symmetric) matrix $M\in\mathbb{R}^{n\times n}$ is PD (i.e., $x^TMx>0$, $\forall x\in\mathbb{R}^n$) if and only if its symmetric part $\frac{1}{2}(M+M^T)$ is PD.

This is the case if $\frac{1}{2}x^T(M-M^T)x=0$, however I am only able to show this if $M$ is symmetric (in which case it is trivial).

Are the above linked claims true or false?

Bobson Dugnutt
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1 Answers1

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Note that $$\langle x, (M+M^{\mathrm{t}}) x\rangle = \langle x, M x\rangle + \langle x, M^{\mathrm{t}} x\rangle = \langle x, M x\rangle + \langle M x, x\rangle = 2\langle x, M x\rangle.$$

WimC
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  • Sorry, why is $\langle x, M x\rangle = \langle M x, x\rangle$? Wouldn't that require $M=M^T$? Or have I completely forgotten bra-ket notation? – Bobson Dugnutt Feb 25 '18 at 19:49
  • Ah, it is because they are each other's transpose, right? And the inner product is invariant under transposition (as it is a scalar)? – Bobson Dugnutt Feb 25 '18 at 19:55