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Let $V$ be a finite-dimensional complex inner product space. Prove that given any self-adjoint linear transformation $f:V\rightarrow V$ there exists a self-adjoint linear transformation $g:V\rightarrow V$ such that $f=g^5$.

I'm not sure how to even begin so I would appreciate some guidance if it's not too much to ask. Thank you in advance!

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  • $\begingroup$ Start by diagonalizing the linear transformation. Then it should be clear how $g$ should look like. You need only to prove that $g$ is still self-adjoint. $\endgroup$
    – b00n heT
    Oct 19, 2016 at 8:42
  • $\begingroup$ Sorry I'm still a bit lost :( would you mind elaborating a little bit? How do I diagonalise the linear transformation? $\endgroup$
    – user342661
    Oct 19, 2016 at 9:30
  • $\begingroup$ Do you know of the "spectral theorem"? $\endgroup$
    – Ben Grossmann
    Oct 19, 2016 at 10:13
  • $\begingroup$ I certainly have been exposed to it, but I can't say I know how to utilise it $\endgroup$
    – user342661
    Oct 19, 2016 at 10:14

1 Answer 1

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Hint: We need to use the spectral theorem. In particular, we know that there exists a diagonal transformation and unitary $u$ such that $d_f:V \to V$ such that $f = u \; d_f \; u^*$.

Start by finding a diagonal $d_g$ such that $d_g^5 = d_f$. Then, it suffices to note that $$ [u \;d_g\;u^* ]^5 = u \; d_g^5 \; u^* = f $$

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  • $\begingroup$ To find $d_g^5$, we would just take the fifth root of the entries of $d_f$? $\endgroup$
    – Eugene
    Jan 5, 2021 at 15:29
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    $\begingroup$ To find $d_g$, we would just take the fifth roots. There is no need to "find" $d_g^5$. $\endgroup$
    – Ben Grossmann
    Jan 5, 2021 at 15:30

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