Let $A$ be a complex $n \times n $ matrix.
a) Prove that if all the eigenvalues of $A$ are real, then $A$ is similar to a real matrix.
b) Classify up to similarity all the matrices $A$ such that $A^n=I$
My attempt:
a) Since the eigenvalues are real, the characteristic polynomial will split. So there exists a Jordan canonical form $J$ such that $A=Q J Q^{-1}$. Hence, part a) proved.
b) We want $I=A^n= (QJQ^{-1})^n= QJ^n Q^{-1}$.
So we must have $J^n=I$. We know that $J$ depends on the eigenvalues and their multiplicities. From here, I'm not sure how to continue.
Could someone please check my part a) and let me know how to continue with part b)?
Thank you!
For $b)$, because $A^n=I_n$, the minimal polynomial of the matrix divides $X^n-1$ and this polynomial has n distinct roots so the minimal polynomial will also have distinct roots, so the matrix is diagonalizable.
Now let $A=QDQ^{-1}$
Because $A^n=I_n$, $D^n=I_n$ but if we consider $D=diag(d_1,d_2,...,d_n)$, then $D^n=diag(d_1^n,d_2^n,...,d_n^n)$ so the matrices that you are looking for are the matrices that have their Jordan canonical form a diagonal matrix and that have their eignvalues roots of unity of order $n$, i.e. $d_i=e^{\frac{2k\pi i}{n}}$, where $k\in\{1,2,...,n\}$.