# If $P$ is an invertible linear operator on a finite dimensional vector space over a finite field, then there is some $n>0$ such that $P^n=I$

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 $$n$$.

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

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.
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