Hermitian Matrices and Eigenvalues

I'm working on a linear algebra problem that I just cannot wrap my head around.

1) Show that any $n × n$ matrix $A$ of the form $$A = B'B$$ has eigenvalues that are all real and positive. Here, $B$ is any invertible $n × n$ matrix with real or complex entries, and $B'$ is its conjugate transpose

Now I know, any Hermitian matrix $A$ has real eigenvalues, and any matrix $A$ of form $$A = B' B$$ is Hermitian, thus its eigenvalues are real. But I just cannot find out why they should be positive as well.

Any help would be great!

$$\langle x, A x \rangle = \langle x, B' B x \rangle = \langle Bx, Bx \rangle, \quad \forall x \in \mathbb C^n.$$
• This gives rise to $\lambda <x,x> = x'Ax$, right? Am I on the right direction? - edit I feel that this leads to just a trivial result – Sank Mar 28 '17 at 12:23
• @user282639 You didn't use Stefano's hint yet. You can write $x'Ax$ (or $\langle x, Ax\rangle$, which is the same thing) as something else (and here you use the hint) – 5xum Mar 28 '17 at 12:25
• @user282639 You already know that $\lambda \langle x, x\rangle = \langle x, Ax\rangle$ (which is what you wrote in your first comment). Now, use the hint which tells you that one of the sides of the equation $\lambda \langle x, x\rangle = \langle x, Ax\rangle$ is also equal to something else... What equation do you get? – 5xum Mar 28 '17 at 12:31
• $\lambda <x,x> = <Bx, Bx>$. It's getting super late/early I feel like I'm not seeing something very obvious – Sank Mar 28 '17 at 12:34