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!

• 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. Oct 19, 2016 at 8:42
• Sorry I'm still a bit lost :( would you mind elaborating a little bit? How do I diagonalise the linear transformation?
– user342661
Oct 19, 2016 at 9:30
• Do you know of the "spectral theorem"? Oct 19, 2016 at 10:13
• I certainly have been exposed to it, but I can't say I know how to utilise it
– user342661
Oct 19, 2016 at 10:14

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 $$[u \;d_g\;u^* ]^5 = u \; d_g^5 \; u^* = f$$
• To find $d_g^5$, we would just take the fifth root of the entries of $d_f$? Jan 5, 2021 at 15:29
• To find $d_g$, we would just take the fifth roots. There is no need to "find" $d_g^5$. Jan 5, 2021 at 15:30