How to show that every element of $G$, where $G$ is a finite subgroup of $GL_n(\mathbb{k})$, the general linear group of square matrices of order $n$ over some algebraic closed field $\mathbb{k}$, is diagnonalizable if $\mathbb{k}$ is an algebraic closure of $\mathbb{Q}$?

I know that a matrix is diagonalizable if its minimal polynomial is seperable in the field on which the matrix is defined. Now, what if the minimal polynomial has repeated roots? How do we ensure the diginalizability? Any hints. Thanks beforehand.


2 Answers 2


Let $K$ be a field. As you have already mentioned, a matrix $A \in \operatorname{M}_n(K)$ is diagonalizable (over $K$) if and only if there exists a polynomial $f(t) \in K[t]$ with $f(A) = 0$ such that $f$ decomposes into pairwise different linear factors over $K$.

For $A \in G$ and $n := |G|$ we have that $A^n = I$, so that $A$ satisfies the polynomial $f(t) := t^n - 1 \in \mathbb{k}[t]$. The polynomial $f(t)$ decomposes into linear factors because $\mathbb{k}$ is algebraically closed. It follows from $\operatorname{char}(\mathbb{k}) = 0$ that the polynomial $f$ is seperable (because $f(t) = t^n - 1$ and $f'(t) = n t^{n-1}$ are coprime), so that $f(t)$ decomposes into pairwise different linear factors. Thus $A$ is diagonalizable.

(To see that $f(t)$ is seperable one can also embed $\mathbb{k}$ into $\mathbb{C}$ because $\mathbb{k}$ is an algebraic closure of $\mathbb{Q} \subseteq \mathbb{C}$. As the roots of unity in $\mathbb{C}$ are pairwise different, the same goes for $\mathbb{k}$.)

  • $\begingroup$ What an awesome answer. Just one thing: Is 'if and only if' in the first para correct? I am taught that if distinct eigenvalues then linear transformation diagonalizable. Converse may not hold, I think. $\endgroup$
    – Silent
    May 12, 2019 at 15:15
  • $\begingroup$ @Silent: If $A$ is diagonalizable with pairwise different eigenvalues $\lambda_1, \dotsc, \lambda_r$ then one can choose the polynomial $f(t) = (t - \lambda_1) \dotsm (t - \lambda_r)$. (The polynomial $f(t)$ does not need to be the characteristic polynomial of $A$, whose linear factors may not be pairwise different.) $\endgroup$ May 12, 2019 at 15:46

Let $A\in G$ be a matrix. After changing bases, $A$ is in JNF. Let us write $A=D+N$ with $D$ diagonal and $N$ the nilpotent part. Now since $G$ is finite, there is some $m\in\Bbb N$ with $A^m=I$, the identity matrix. Since $DN=ND$, we have $$I = A^m = (D+N)^m = \sum_{k=0}^m \binom mk N^k D^{m-k} = D^m + \sum_{k=1}^m \binom mk N^k D^{m-k} =: D^m+\tilde N. $$ Note that $\tilde N$ is a nilpotent, strictly upper triangular matrix. It follows that $\tilde N=0$ and $D^m=I$. Hence, $$ 0 = \tilde N = N\cdot\left(mD^{m-1}+\sum_{k=2}^{l} \binom mk N^{k-1} D^{m-k}\right) $$ and as the second factor is an invertible upper trianglar matrix (because we are in characteristic zero and $m$ is invertible), so we must have $N=0$. In other words, $A=D$ is already diagonal. In other words, $A$ is diagonal up to changing bases, which means that it is diagonalizable.

  • $\begingroup$ superb answer! by the way, the crux was the finiteness of the subgroup, right? $\endgroup$
    – vidyarthi
    Mar 10, 2018 at 16:31
  • $\begingroup$ but, the algebraic closure of $\mathbb{Q}$ is not at all utilized $\endgroup$
    – vidyarthi
    Mar 10, 2018 at 16:41
  • $\begingroup$ @vidyarthi; the comment by @JendrikStelzner was accurate and the relevance of the field characteristic is now highlighted. $\endgroup$ Mar 11, 2018 at 10:20

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