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Prove that $f: x \to Tx$ is positive on $\mathbb{C}^n$ iff $T$ has ony non-negative eigen values, for a complex $n\times n$ Hermitian matrix $T$.

To prove that $f$ is positive I need to show that $\langle fx, x\rangle \geq 0, \forall x\in \mathbb{C^n}$. It is equivalent to show that $ \langle Tx, x\rangle \geq 0$, so I think that this means that $ x' Tx \geq 0$ ($x'$ is the transpose of $x$), but I am not sure how to argue it. Then it will be easy because I just need to show that $T$ is positive semi-definite. Can someone help me to get this conclusion?

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    $\begingroup$ Do you know how to do the case if $T$ is a diagonal operator? Once you can do that one, for general $T$ apply the spectral theorem $\endgroup$ – rubikscube09 Feb 12 at 1:07
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Let's do both directions separate.

Let us first note that every hermitian matrix has only real eigenvalues. If you don't know this fact, I will add a prove here.

If $f:x\mapsto Tx$ is positive, consider an eigenvalue $\lambda$ for an eigenvector $v$. We have $$0\leq\langle fv,v\rangle=\langle Tv,v\rangle=\langle \lambda v,v\rangle=\lambda \langle v,v\rangle.$$ Since $v\neq 0$, this shows that $\lambda \geq 0$.

Now let us assume that all eigenvalues are greater than or equal to zero. We can use the fact that every hermitian matrix is diagonalisable by a hermitian matrix. Thus take $A$ such that $A^{-1} T A=\overline{A}^T T A=D$, where $D$ is the diagonal matrix with entries the eigenvalues of $T$. $$\langle Dv,v\rangle=\langle \overline{A}^T T Av,v\rangle=\langle TAv,Av\rangle=\langle Tw,w\rangle $$ for $w=Av$. Since $A$ is invertible, it is bijective and we can write every vector $w$ as $Av$. So this shows that we can calculate the scalar product by using $D$ instead of $T$. But since $D$ has only real and positive eigenvalues, $\langle Dv,v\rangle \geq 0$.

If something remained unclear, feel free to comment.

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  • $\begingroup$ why do you assume in the second direction of the proof that all eigenvalues are positive? the statement requires that they are non-negative? $\endgroup$ – mandella Feb 12 at 14:14
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    $\begingroup$ With positive I mean $\geq 0$. I will edit to avoid confusion. $\endgroup$ – James Feb 12 at 14:18
  • $\begingroup$ Ah, ok! I saw that the proof worked out to use non-negative ones in the end. Thanks for the help $\endgroup$ – mandella Feb 12 at 14:19
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    $\begingroup$ You're welcome. $\endgroup$ – James Feb 12 at 14:20

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