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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.

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1 Answer

(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 $p(A)=0=q(B)$

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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
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