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Problem: Let $V$ be a vector space over the field with three elements. Show that there exists an endomorphism $\alpha$ of V with $\alpha(v)+\alpha(v)=v$.

I have posted a solution attempt in the answers.

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up vote 2 down vote accepted

Identify the field with three elements with $\mathbb{Z}_3$.

The condition becomes $\alpha(2v)=v$.

Fix a basis $\{b_\alpha\}$ of the vector space. We can write any $v\in V$ in the form

$$v=\sum_1^n q_jb_j$$

for $q_j\in \mathbb{Z}_3$. Let our map send $b_\alpha$ to $2b_\alpha$ for each basis element and extend by linearity.


$$\alpha(2v)=\alpha\left(\sum_1^n 2q_jb_j\right)=\sum_1^n 2\cdot2q_jb_j=\sum_1^n q_jb_j=v$$

as desired.

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It works, but wouldn't it be simpler to do it without reference to a basis? Take $\alpha(v) = 2v$; then $\alpha(v)+\alpha(v) = 2v+2v = (2+2)v = v$ (because we are in characteristic $3$); and multiplication by a scalar is always linear. – Arturo Magidin Jun 2 '12 at 3:52
Yes, it would. It's just how I thought about it when solving the problem, I guess. – Potato Jun 2 '12 at 3:53

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