Let $A$ be a complex matrix such that $A^n = I$, show that $A$ is diagonalisable.
How do I do this?
I would hazard a guess that since $A^n = I$ that means that $A^n$ is obvious diagonal, and I suppose that a diagonal matrix can only be formed by the square of another diagonal matrix, but I am not sure if that is the case. Also I think this neglects the complex cases.
Is there a Jordan block argument that I should use?
His airness Jordan does not play here. Let $k$ be a field and $A\in \mathbf{M}_n(k)$ a matrix. Then $A$ is a diagonalizable over $k$ if and only if its minimal polynomial is split with distincts roots. But in your case the minimal polynomial divides $X^n - 1$ who has distinct roots and all non constant polynomials are split, because you are over $\mathbf{C}$.
Detail for the split part. It uses the following well-known fact : over any algebraically closed field, any non constant polynomial is split (i.e. product of polynomials of degree $1$). Indeed : take a polynomial $P$ that is non constant : it has a root $z$ because $k$ is algebraically closed, so that you can write $P(X) = (X-z)Q(X)$ and continue with $Q$, which again has a root... Finally, $\mathbf{C}$ is algebraically closed, and this is a standard result.
Remark 1. $A$ is trigonalizable over $k$ if and only if its characteristic polynomial over $k$ is split.
Remark 2. I won't give details as they are in any good course. To prove the previously stated equivalences for diagonalizability or trigonalizability, you don't need Jordan normal form (JNF). For trigonalizability one direction is obvious and the other one is done by induction and uses the fact that a root of a characteristic polynomial is an eigenvalue. For diagonalizability one direction results from the kernel decomposition theorem (which does not use JNF and uses only Bézout) and the other is obvious. By the way, you don't need Schur's lemma, nor Cayley-Hamilton.
If $J$ is the Jordan canonical form of $A$, then $J^n=I$, so the same identity holds for every Jordan block. Thus you can work with an upper triangular matrix $A$ of order $m$ (where $m\le n$) with the same diagonal coefficients, and $A^n=I$.
Let $\varepsilon$ be the common coefficient on the diagonal, so $A=\varepsilon I+B$ where $B=[b_{ij}]$ is nilpotent and nonzero, if $m>1$. From $A^n=I$ we get $$ I=\sum_{k=0}^n \binom{n}{k}\varepsilon^{n-k}B^k $$ When you consider $B^k$ with $k>1$, the coefficients in positions $(i,i+1)$ (for $i=1,\dots,m-1$) are zero, so we conclude that $b_{i,i+1}=0$. Similarly, the coefficients in positions $(i,i+2)$ are zero in $B^k$ for $k>2$, so $b_{i,i+2}=0$ and we can go on. Thus $B$ is diagonal. Since $A$ is a Jordan block, assuming $m>1$ is a contradiction.
