Over what fields are finite order endomorphisms of vector spaces diagonalizable?

Question

The problem is as follows:

Let $V$ be a finite dimensional vector space over $F$. Let $T : V \to V$ be a linear endomorphism such that $T^{n} = I$ for some fixed $n$. What is a necessary and sufficient condition on $F$ for $T$ to be diagonalizable?

I have shown that $T$ will be diagonalizable if its minimal polynomial factors into distinct linear terms in $F$. I also know that the minimal polynomial must divide $x^n - 1$. From here, I am stuck, as I do not know how to gather any more information about the form of the minimal polynomial

Thank you in advance for any help.

Answer 1

A sufficient condition is that $F$ is algebraically closed. But of course, this is not necessary, for the identity operator is trivially diagonalizable over any field (indeed, it is already a diagonal operator).

Answer 2

If you want conditions on $F$ that are necessary and sufficient so that any endomorphism $T$ of any finite dimensional vector space with $T^k=I$ is diagonalizable, then...

1. For characteristic zero, a necessary and sufficient condition is that $F$ contain all $k$th roots of unity for each $k\geq 1$. To see the sufficiency, note that the minimal polynomial of such a $T$ divides $x^k-1$, and over such a field this splits into distinct linear factors. Conversely, the companion matrix of $x^k-1$ has minimal polynomial $x^k-1$, which you need to factor into linear terms in order to be diagonalizable. Thus, the condition (which is weaker than being algebraically closed) is both necessary and sufficient.

2. For positive characteristic, this is impossible. Let $p$ be the characteristic. The companion matrix of $x^p-1 = (x-1)^p$ has minimal polynomial $(x-1)^p$, and hence is not diagonalizable. So if $\mathrm{char}(F)\gt 0$, there is always an endomorphism of finite multiplicative order that is not diagonalizable.

If $V$ is fixed, of dimension $n$, the situation is slightly different. It is not hard to verify that if $T$ has finite order in $V$, then the order is at most $n$. So in this case:

1. If $F$ has characteristic $0$, or characteristic $p$, $2\leq p\leq n$, a necessary and sufficient condition is that the field contain all $k$th roots of unity, $1\leq k\leq n$. An argument as above works (use the companion matrix of $x^k-1$ and then complete it with $0$s to get an $n\times n$ matrix that has minimal polynomial $x(x^k-1)$).

2. If $F$ has positive characteristic $p\geq n$, then the same argument as above shows you cannot do it.

Per the comment, we actually have a third permutation: $k$ is fixed. What is required so that every endomorphism of order (dividing?) $k$ is diagonalizable?

1. If $F$ has characteristic $0$ or characteristic $p$ that does not divide $k$, then you need $F$ to contain (i) all $k$th roots of unity if you want order exactly $k$ only; and (ii) all $m$th roots of unity for all divisors $m$ of $k$ if you want it for any endomorphism such that $T^k=I$.

2. If the characteristic of $F$ is $p$ and $p$ divides $k$ and is no larger than $\dim(V)$, then you’re still out of luck. Writing $k=pm$, the companion matrix of $x^k -1 = (x^m)^p-1 - (x^m-1)^p$ has minimal polynomial $(x^m-1)^p$, and hence is not diagonalizable.

Answer 3

Since working in finite dimensional vector spaces one usually takes $n$ to stand for its dimension, I (like Arturo Magidin) will assume that the relation given is $T^k=I$. The question I will answer then is: for a fixed integer $k>0$, under what condition on $F$ is it the case that every endomorphism $T$ of any (finite dimensional) vector space $V$ over $F$ that satisfies $T^k=I$ is diagonalisable. (I cannot assume that $V$ is given, as the formulation in the question suggests, since that would automatically fix $F$; and it will turn out that a hypothesis of finite dimension does not simplify the answer.)

If the supposed annihilating polynomial $X^k-1$ is split over $F$ with simple roots, then $T$ will necessarily be diagonalisable (with spectrum contained in the set of roots of $X^k-1$; a theorem guarantees this, and it does not require $V$ to be finite dimensional). Since we can realise $X^k-1$ as the minimal polynomial of an appropriate $T$, for instance using a companion matrix, this sufficient condition is also necessary: the minimal polynomial of a diagonalisable endomorphism$~T$ is $\prod_{\lambda\in\operatorname{Spec}T}(X-\lambda)$, which is split and (by definition of the product) has simple roots.

If $F$ is of characteristic $0$ of of a characteristic not dividing $k$, then $X^k-1$ is relatively prime to its derivative $kX^{k-1}$, so it cannot have multiple roots, and therefore the condition is equivalent to $X^k-1$ being split over $F$, in other words to the existence in $F$ of all the $k$-th roots of unity ($F$ contains the $k$-th cyclotomic field). If the characteristic of $F$ does divide $k$, then multiplication in $F^2$ by $\left(1~1\atop0~1\right)$ provides an example of an endomorphism$~T$ satisfying $X^k=I$ but which is not diagonalisable, so no field of such characteristic can satisfy the condition.