So my given problem is:
$Let\,\, A \in \mathbb{C}^{n \,\times \,n}\,\, be\,\, such\,\, that\,\,\\ \forall x\in \mathbb{C}^n : \langle\,Ax,x\rangle \geq 0 \\ where \,\, \langle\,\cdot,\cdot\rangle \,\, is \,\, the \,\, standard \,\, inner \,\, product \,\, of \,\, \mathbb{C}^n.\,\, Show \,\, that\,\,\\ spec(A) \subset [0, \infty).$
Know I don't have any approach how to tackle the problem.
Thank you in advance
$\textbf{EDIT:}$
I came this far:
Let $\lambda$ be e-value of A and x the corresponding e-vector.
$Ax = \lambda x \,\,\, multiply \,\, x^T\\ x^TAx = \lambda x^Tx$
The left hand side of the equation is greater or equal to zero since
$x^TAx=\langle\,Ax,x\rangle \geq 0$
Then
$ = \lambda \sum_{i=1}^{n} x^{2}_i $
Since $x^Tx=\sum_{i=1}^{n} x^{2}_i$
The sum is positive therefore $\lambda$ has to be greater or equal to 0 so that the whole term is greater or equal to zero.
But now I dont know how to show that all eigenvalues are real.
If
$\lambda \in \text{spec}(A), \tag 0$
we have a vector $x \in \Bbb C^n$ such that
$Ax = \lambda x, \; x \ne 0; \tag 1$
then
$\bar \lambda \langle x, x \rangle = \langle \lambda x, x \rangle = \langle Ax, x \rangle \ge 0; \tag 2$
we recall that the assertion $\langle Ax, x \rangle \ge 0$ (tacitly, by definition) implies $\langle Ax, x \rangle \in [0, \infty) \subset \Bbb R$, and since
$0 \ne \langle x, x \rangle \in (0, \infty) \subset \Bbb R, \tag 3$
(2) yields
$0 \le \bar \lambda \in [0, \infty) \subset \Bbb R; \tag 4$
now
$\bar \lambda \in [0, \infty) \subset \Bbb R \Longleftrightarrow \lambda \in [0, \infty) \subset \Bbb R; \tag 5$
therefore
$\text{spec}(A) \subset [0, \infty), \tag 6$
as was to be shown. $OE\Delta$.