# showing that map $f$ is positive

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?

• 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 – rubikscube09 Feb 12 at 1:07

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.

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