# Question regarding the diagonalizability of certain matrices

Let $A\in \mathbb{C}^{n^2}$ such that $A^m=I_n$, for some $m,n\in \mathbb{N}$.

Please prove that $A$ is diagonalizable.

Now let $B\in \mathbb{C}^{n^2}$ such that $B^m=B$, for some $m\in \mathbb{N}$ such that $m>1$.

Please prove that $B$ is diagonalizable.

-
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Dec 15 '12 at 14:26

## 1 Answer

Hints:
(1) A matrix $M$ is diagonalizable if and only if it's minimal polynomial $m_M(x)$ factors completely over the field (here $\mathbb{C}$) into distinct linear factors.The relevant theorem is here
(2) If $p(x)$ is a polynomial such that $p(M)=0$ then $m_M(x)|p(x)$.
(3) If we have two polynomials $f(x),g(x)$, $f(x)|g(x)$ and $g(x)$ factors completely into distinct linear factors, then so does $f(x)$.
(4) Observe that if $p(x)=x^m-1$ and $q(x)=x^m-x$ then $p(A)=0=q(B)$

-
The proof of (1) is ridiculously long and it's not on my notes, so I'm guessing I'm supposed to prove it in another way. My idea is to show that the matrices' Jordan Normal Form is a diagonal matrix. To do that it suffices to prove that $r(M-\lambda I)=r((M-\lambda I)^2)$ for every eigenvalue $\lambda$. –  Um burro Dec 15 '12 at 16:33
How do compute Jordan normal form? do you know that the size of the largest block corresponding to the eigenvalue $\lambda$ is equal to the power of $x-\lambda$ in $m_M(x)$? –  Dennis Gulko Dec 15 '12 at 16:57
I don't know that. In my notes the size of the largest block is the smallest natural $k$ such that $r((M-\lambda I)^k)=r((M-\lambda I)^{r+1})$, hence my comment above. –  Um burro Dec 15 '12 at 17:26
What is $r(\cdot)$? the rank? –  Dennis Gulko Dec 15 '12 at 17:36
Yes, it is the rank. Sorry. –  Um burro Dec 15 '12 at 17:41