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

How to proceed ? Have no idea

Thanks

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closed as off-topic by Shailesh, Joey Zou, Henrik, wythagoras, Davide Giraudo Aug 5 '16 at 19:11

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

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

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

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