I am reading Representation Theory A First Course by William Fulton and Joe Harris. In Section 3, Lecture 1, they gave a method to find all the irreducible representations of the symmetric group $\mathfrak{S}_3$.

Suppose $W$ is an representation of $\mathfrak{S}_3$. ($W$ is a complex vector space). Let $\tau$ be a generator of the subgroup $\mathfrak{A}_3\subset \mathfrak{S}_3$, for example, $(123)$. Then the space $W$ is spanned by eigenvectors $v_i$ for the action of $\tau$.

I don't understand why $W$ is spanned by eigenvectors of $\tau$. Even in complex field, not all invertible matrix is diagonalizable. I guess it may have something to do with the fact that $\tau^3 = 1$, and so the eigenvalues of $\tau$ are all roots of unit.

  • 1
    $\begingroup$ A nondiagonalizable complex matrix cannot satisfy $\tau^3 = {\bf 1}$: To see this, observe that the $3$rd power (for that matter, the $n$th power for all $n \geq 1$) of any Jordan block of size $> 1$ is not the identity matrix. $\endgroup$ Sep 22, 2023 at 12:59
  • $\begingroup$ Thank you. :) So every action by a group element on a finite-dimensional vector space is diagonlizable? That is a little surprising to me. $\endgroup$ Sep 22, 2023 at 13:04
  • 1
    $\begingroup$ If the group is finite, yes, for that reason. But that's not always the case for infinite groups---just consider the standard action of a nondiagonalizable element of $\operatorname{GL}(\Bbb V)$ (where the underlying field has characteristic $0$). $\endgroup$ Sep 22, 2023 at 13:16

3 Answers 3


You can check this by hand, and then you will notice a pattern:

Say $\tau$ has order $3$, and acts on $W$. Let $\omega = \exp\frac{2 \pi i}{3}$. We can write every element $w$ in $W$ as

$$w = w_0 + w_1 + w_2$$

where $$w_k = \frac{1}{3} \cdot\sum_{l=0}^2 \omega^{-k l} \tau^l w$$


$$\tau w_k = \frac{1}{3}\sum_l\omega^{-k l} \tau^{l+1} w= \omega^k w_k$$

Similar to " every function is the sum of an even function and an odd function".


Every action $\rho(g)$ by an element $g$ of a complex representation $G \hookrightarrow GL(n, \Bbb C)$ of a finite group $G$ is diagonalizable. Suppose not, that is, that for some $g \in G$ the Jordan normal form of $\rho(g)$ contains a Jordan block of size $k > 1$, say, $J_k(\lambda)$. Now, for $n \geq 1$, $J_k(\lambda)^n \sim J_k(\lambda^n)$, so $\rho(g^n) = \rho(g)^n$ also contains a Jordan block of size $k > 1$. On the other hand, if we set $n$ to be the order of $g$, then $\rho(g^n) = \rho(1_G) = {\bf 1}$, whose Jordan blocks all have size $1$, a contradiction.

(In fact, computing gives that the superdiagonal entries of $J_k(\lambda)^n$ are $n \lambda^{n - 1} = n \lambda^{-1} \neq 0$, so if $k > 1$ then $J_k(\lambda)^n \not\sim {\bf 1}_k$.)


The general result is the following. Let $V$ be a finite-dimensional vector space over a field $k$ and $T : V \to V$ a linear map.

Proposition: Suppose $f(t) \in k[t]$ is a polynomial which splits over $k$ and has no repeated roots such that $f(T) = 0$. Then $T$ is diagonalizable.

Note that $f$ doesn't have to be either the minimal or the characteristic polynomial. If $G$ is a finite group, $\rho : G \to GL(V)$ is a representation of $G$ over an algebraically closed field $k$, and $g \in G$ has order $n$, then applying the above proposition, $\rho(g)^n = 1$ implies that $\rho(g)$ is diagonalizable as long as $\text{char}(k) \nmid n$.

Proof. Write $f(t) = \prod_{i=1}^n (t - \lambda_i)$. We are actually going to write down, explicitly, the projections onto each eigenspace of $T$. The idea is that the polynomial $\frac{f(t)}{t - \lambda_i} = \prod_{j \neq i} (t - \lambda_j)$ vanishes everywhere except at $t = \lambda_i$, so the polynomial

$$f_i(t) = \frac{f(t)}{(t - \lambda_i) \prod_{j \neq i} (\lambda_i - \lambda_j)}$$

vanishes everywhere except at $t = \lambda_i$ where it takes the value $1$. Now if $v \in V$ we compute that

$$(T - \lambda_i) f_i(T) v = \frac{f(T)}{\prod_{j \neq i} (\lambda_i - \lambda_j)} v = 0$$

so $f_i(T) v$ is an eigenvector with eigenvalue $\lambda_i$. Moreover, if $v_i \in V_i = \text{ker}(T - \lambda_i)$ is already such an eigenvector then

$$f_i(T) v_i = f_i(\lambda_i) v_i = v_i$$

so $f_i(T)$ fixes every eigenvector with eigenvalue $\lambda_i$. It follows that $f_i(T)$ is a projection onto the eigenspace $V_i$ as desired (which may be zero since $f$ is not necessarily the minimal polynomial).

From here we want to show that every $v \in V$ is a sum of eigenvectors. This would follow if we could show that $\sum_{i=1}^n f_i(T) = I$. Now, the polynomial $\sum_{i=1}^n f_i(t) - 1$, by construction, takes the value $0$ at each $\lambda_i$. But it also has degree $n-1$, and has $n$ roots. So it must be identically zero, and we have $\sum_{i=1}^n f_i(T) = I$ and hence every vector $v \in V$ can be written as a sum of eigenvectors

$$v = \sum_{i=1}^n f_i(T) v$$

as desired. (We can also show that $f_i(T) f_j(T) = 0$ for $i \neq j$ so the $f_i(T)$ are a system of orthogonal idempotents corresponding to the direct sum decomposition $V \cong \bigoplus_i V_i$ into eigenspaces but it's not totally necessary.) $\Box$

This can also be proven more abstractly by observing that $V$ has the structure of a module over $k[t]/f(t) \cong k^n$ but this argument can feel a little bloodless sometimes and it's nice to know that it can be made a lot more explicit. In any case it requires knowing a little commutative algebra whereas the above argument requires only a solid grasp of how polynomials behave. It's overkill bordering on circular to cite the Jordan normal form theorem here, because a generalization of the above result is used as a lemma in some proofs of that theorem.

In the special case that $f(t) = t^n - 1$ relevant to representations of finite groups the explicit direct sum decomposition above is essentially the discrete Fourier transform, or what is known sometimes as the "roots of unity filter."


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