If $A^4$ has an eigenvalue $x$, say, does $A$ have an eigenvalue $y$ s.t $y^4=x$?
I do not think this is true for all complex matrices but I cannot seem to find an example…
Also, how do I prove that $\mathrm{tr}(AB)^2 \leq \mathrm{tr}(A^2)\mathrm{tr}(B^2)$? Iam attempting Q5 from here: https://www.maths.cam.ac.uk/sites/www.maths.cam.ac.uk/files/pre2014/undergrad/pastpapers/2010/Part_IA/PaperIA_1.pdf
I have tried to apply the result proved in ii) to $A-B$ but I cannot get a strong enough inequality
Any complex matrix $A$ can be put into Jordan Normal Form: $Q^{-1}AQ = J$ for some $Q$, where $J$ is $0$ everywhere except the main and super diagonals, and has only $0$ or $1$ on the super diagonal. But for our purposes, it is enough that $J$ is upper-triangular.
An upper triangular matrix has its eigenvalues on the diagonal. Also, the product of two upper triangular matrices is also upper triangular, and the diagonal entries are the products of the diagonal entries of the multiplicands. Therefore $Q^{-1}A^4Q = J^4$, whose diagonal entries are the 4-th powers of the diagonal entries of $J$. But $J^4$ is upper-triangular, so its eigenvalues are the diagonal elements, which are the 4th powers of the diagonal elements of $J$, which are its eigenvalues.
But since $A$ is similar to $J$ and $A^4$ is similar to $J^4$, they have the same eigenvalues. So yes, it is true that every eigenvalue of $A^4$ is the 4-th power of an eigenvalue of $A$.