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!



$$\langle x, A x \rangle = \langle x, B' B x \rangle = \langle Bx, Bx \rangle, \quad \forall x \in \mathbb C^n. $$

  • $\begingroup$ 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 $\endgroup$ – Sank Mar 28 '17 at 12:23
  • $\begingroup$ @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) $\endgroup$ – 5xum Mar 28 '17 at 12:25
  • $\begingroup$ I think I'm struggling to see the connection here $\endgroup$ – Sank Mar 28 '17 at 12:28
  • $\begingroup$ @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? $\endgroup$ – 5xum Mar 28 '17 at 12:31
  • $\begingroup$ $\lambda <x,x> = <Bx, Bx>$. It's getting super late/early I feel like I'm not seeing something very obvious $\endgroup$ – Sank Mar 28 '17 at 12:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.