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Is it possible for a matrix in $M_n(\mathbb{C})$ to be both positive and negative semi-definite without it being the zero-matrix? My intuition suggests no, but I am finding it very hard to prove otherwise. Conversely, in trying to find such a matrix, no theorems really exist to generate one.

Pavan C.
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Michael Mao
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  • What definitions are you working with? – Mike F Apr 13 '21 at 03:26
  • only the definition of a positive / negative semi-definite matrix. I have also the quadratic form: Q(v) = v*Av for some matrix A, where v is a vector of length n from the complex vector field. – Michael Mao Apr 13 '21 at 03:33

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If $A$ is PSD, then $\langle Ax,x \rangle \geq 0$ for all $x$. On the other hand, if it is NSD, then $\langle Ax,x \rangle \leq 0$ for all $x$.

Overall, $\langle Ax,x \rangle =0$ for all $x$. Since $A$ is self adjoint (this is either part of the definition/a consequence of being positive/negative definite depending on your definition), you can conclude that $A=0$.

GSofer
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  • If A = A* and v*Av = 0, i unfortunately still really dont see how that implies then that A = 0. – Michael Mao Apr 13 '21 at 03:55
  • @MichaelMao See for instance: https://math.stackexchange.com/questions/312388/relation-between-t-0-and-tx-x-0/312397 This is a general property of Hilbert spaces. – GSofer Apr 13 '21 at 05:15
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    @MichaelMao Use Polarization Identity for $ \langle Ax, y \rangle$ to see that $ \langle Ax, y \rangle=0$ for all $x,y$ Then put $y=Ax$. https://en.wikipedia.org/wiki/Polarization_identity – Kavi Rama Murthy Apr 13 '21 at 05:16