-1
$\begingroup$

If a matrix is both Hermitian and unitary show all its eigen values are ±1.

How to proceed ? Have no idea

Thanks

$\endgroup$
3
  • $\begingroup$ What are you allowed to know/use? Do you know about (left/right) singular values (or equivalently SVD)? $\endgroup$
    – Tom
    Commented Aug 5, 2016 at 14:42
  • $\begingroup$ Do you know what a Jordan normal form is? $\endgroup$
    – user322818
    Commented Aug 5, 2016 at 14:50
  • $\begingroup$ No I dont know about it $\endgroup$
    – Debjit Roy
    Commented Aug 5, 2016 at 16:09

2 Answers 2

4
$\begingroup$

If it is unitary, then $A^{\dagger}A=I$ and if it is Hermitian then $A=A^{\dagger}$. You can clearly see that $AA=A^2=I$ then. Since $\lambda$ of $I$ is $1$, you have $A^2x=\lambda x=x$. Can you see where to go from here?

$\endgroup$
2
  • $\begingroup$ I suspect you meant $\lambda^2 x$ above? $\endgroup$
    – copper.hat
    Commented Aug 5, 2016 at 16:20
  • $\begingroup$ Well $\lambda$ was the eigenvalue for $I$ but I could make that more clear $\endgroup$
    – m1cky22
    Commented Aug 5, 2016 at 16:21
4
$\begingroup$

If a matrix is unitary, then $\|Ux\| = \|x\|$ and so if $Ux= \lambda x$, we see that $|\lambda| = 1$.

If a matrix is Hermitian, its eigenvalues are real.

$\endgroup$
1
  • $\begingroup$ Thanks a lot ,you people almost did it for me $\endgroup$
    – Debjit Roy
    Commented Aug 5, 2016 at 16:08

Not the answer you're looking for? Browse other questions tagged .