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Berkeley problems in mathematics

problem 7.4.21: let $P$ be a linear operator on a finite dimensional vector space over a finite field. show that if $P$ is invertible, then $P^n$ = $I$ for some positive integer $n$.

I think there is a mistake in the above statement: it does not hold true if P is a scalar multiple of the identity operator.

But is the statement correct otherwise? i.e. is the following modified statement correct?

let $P$ be a linear operator on a finite dimensional vector space over a finite field. assume $P$ is not a scalar multiple of the identity operator. show that if $P$ is invertible, then $P^n$ = $I$ for some positive integer $n$.

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    $\begingroup$ The statement certainly holds for $P=aI$ because $a^{n-1}=1$ if $a\ne0$ and the finite field has $n$ elements. $\endgroup$ – lhf May 2 '19 at 13:04
  • $\begingroup$ @lhf thanks! I had difficulty understanding the notion of a finite field $\endgroup$ – Vinay Deshpande May 2 '19 at 13:07
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The statement is true even if $P$ is a scalar operator. Note that the field of scalars in question is finite.

Consider the group $G$ of invertible linear operators on the vector space in question. Since the space is finite-dimensional & the scalar field is finite, the vector space itself is finite, and the space of linear operators is finite. Hence $G$, being a subset of the space of all operators, is finite too. Now the result follows from general group theory: for any $P \in G$ we have $P^{|G|} = I$, the identity operator.

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I'll make a try:

Take $P,P^2,P^3,...$. Then these can't be all different because the have entries from a finite field. Then $P^t=P^s\Rightarrow P^n=I$ since $P$ is invertible

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