Let $\textbf{A} \in M_{n\times n}(\mathbb{C})$ be a Matrix and $\textbf{A}^m=\textbf{I}$ where $m\neq \infty$. Show (with Jordan normal form) that $\textbf{A}$ is diagonalizable. I have no clue how to do that...
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I wouldn't use Jordan normal form either. – May 16 '21 at 19:42
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1Hint: the polynomial $p(x) = x^m - 1$ factorises into distinct linear factors over $\mathbb{C}$. – Jose Avilez May 16 '21 at 19:43
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1What are the roots of $;x^m-1;$ ? How many of them are different? – DonAntonio May 16 '21 at 19:43
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To use Jordan: compute the powers of a Jordan block. – GEdgar May 16 '21 at 19:58
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Does this answer your question? Let $A$ be a complex matrix such that $A^n = I$, show that $A$ is diagonalisable. There are other versions of this question on math.SE with solutions of varying approaches and levels of detail. See also e.g. https://math.stackexchange.com/questions/2676557/prove-that-t-is-diagonalizable-if-and-only-if-the-minimal-polynomial-of-t-has-no ($T$ is diagonalizable if and only if the minimal polynomial of $T$ has no repeated roots) – leslie townes May 16 '21 at 22:04
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Yes. We use the following:
a matrix $M$ is diagonalisable iff its minimal polynomial is the product of distinct linear factors.
Now, can you see how to apply this to $A^m -I=0$?. Hint: the minimal polynomial will have to divide $t^m - 1$
riemann_lebesgue
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Using the Jordan normal form:
We have to show that all the blocks have size $1$. Now, if $A^m=I$, then it is easy to see that for the Jordan form $J$ we have $J^m = I$, so for every block $J_{d, \lambda}$ we have $$J_{d, \lambda}^m = I_d$$
However, if $d>1$, then the elements of $J^m_{d, \lambda}$ just above the diagonal have value $m$, so not $0$. Therefore, all the $d$'s must be $1$, all the blocks have size $1$, so $A$ is diagonalizable.
orangeskid
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