# Linear Algebra: diagonalization problem

Let $T: \mathbb{C^2}\rightarrow \mathbb{C^2}$ be a linear map. If exist only one eigenvalue of $T$ and exist $k \neq 0$ where $T^k = I$ then T is diagonalizable.

My solution: If $\ T^k = I$ then the minimal polynomial $m_T(x)$ divides $x^k -1$. As existe $k$ distinct roots of $x^k -1$ and $T$ has only one eigenvalue $m_T (x) = x- \lambda$ where $\lambda ^k = 1$. So for all vectors $v \in \mathbb{C^2}$ $m_T(T(v)) = T(v) - \lambda v = 0$, in other words; for all $v \in \mathbb{C^2} \ \Rightarrow v \in Aut_T(\lambda)$. As $Aut_T(\lambda) \subseteq \mathbb{C^2}$ then $Aut_T(\lambda) = \mathbb{C^2}$. Therefore $T$ is diagonalizable.

My problem is: I don't know if it's correct, because I could change $\mathbb{C^2}$ for any $\mathbb{C}^n$ for $n \in \mathbb{N}$. Or this hypothesis is important and I'm not seeing?

-
Absolutely. You can prove that directly for $\mathbb{C}^n$. –  1015 Apr 2 '13 at 22:47

I think you should directly prove a more general result.

Let $A$ be a complex $n\times n$ matrix (or an operator from a finite dimensional complex vector space to itself).

Claim: If there exists $k\geq 1$ such that $A^k=I_n$. Then $A$ is diagonalizable.

Proof: the polynomial $X^k-1=\prod_{j=1}^k(X-\omega_j)$ annihilates $A$. Since the $\omega_j$ here are the $k$-roots of untity and are pairwise distinct, the factors are relatively prime and therefore $$\mathbb{C}^n=\bigoplus_{j=1}^k\mbox{Ker} (A-\omega_j I_n).$$ Each summand is either trivial or an eigenspace of $A$. Hence $A$ is diagonalizable. QED.

Note: of course, the real generalization, which is true over any field, is: $A$ is diagonalizable if and only if it is annihilated by a polynomial which splits and has only simple roots.

-
Thank you julien. You helped me a lot. –  jon jones Apr 5 '13 at 21:14
@jonjones You're welcome, jon. –  1015 Apr 5 '13 at 21:22