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Consider a matrix M over complex numbers of size m×m such that $M^n = I$ for some natural number $n$. Then prove or disprove that $M$ is diagonalisable.

My try:

I imitated the proof of projection linear transformation. That is basis vectors of null space and image space will be Eigen vectors but I am not sure if it is correct way here.

Suppose $M^2 = I$ then consider a vector $x$ in null space. Then $x =M^2x = 0$ so null space will have only $0$. Then I am not sure if a basis for image space has some Eigen vectors.

I tried finding counter examples also but not successfully.

please give me some help. Anyother approach also is fine.

Infinity_hunter
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 04 '21 at 19:22
  • To community: I am asking help – Infinity_hunter Sep 04 '21 at 19:23
  • I'm sure we have done this somewhere on the site. Hint: Can you show that every vector is a sum of eigenvectors of $M$? If $M^2=I$, then to each $x$ both $\dfrac12(x+Mx)$ and $\dfrac12(x-Mx)$ are eigenvectors of $M$. In general you need more terms (and $n$th roots of unity). – Jyrki Lahtonen Sep 04 '21 at 19:24
  • If $M^n=I$ then $M$ is diagonalizable over $\mathbb{C}$ since $M$ is a zero of $X^n-1$ which is a product of linear terms with only simple roots – Evariste Sep 04 '21 at 19:26
  • Related. Many proofs on the site rely on the concept of a minimal polynomial and the fact (follows from Jordan canonical form) that if the minimal polynomial has no multiple roots, then the matrix is diagonalizable. In the present case there is a constructible proof also. Looking.... – Jyrki Lahtonen Sep 04 '21 at 19:37
  • Another. Leaving this as a link for I'm contemplating writing an answer that does not need Jordan canonical forms. – Jyrki Lahtonen Sep 04 '21 at 19:43
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    To JyrkiLahtonen:I will definitely try the hint(which you have given in your first comment) and generalising to nth power by finding n Eigen vectors. – Infinity_hunter Sep 04 '21 at 19:49
  • @quantum_spin For what it's worth, I wrote the argument as an answer to this question that is, in my opinion, a better duplicate target. – Jyrki Lahtonen Sep 04 '21 at 20:44

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