Let $V$ be a non-zero vector space over a field $F$ with at least three elements. Prove that if $k \geq 2$ and $f : V^k = V \times V \times \cdots \times V \rightarrow F$ is a $k$-linear function which is also linear, then $f = 0$.
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
For simplicity, I assume $k=2$. Let $x, y$ be elements of $V$. Then because $f$ is $k$ linear we have $$f(2x, y) = 2 f(x,y)$$
And because $f$ is linear $$f(2x,y) = f(x,y) + f(x,0)$$
But $f$ is $k$-linear so $f(x,0) = 0$. And then $f(x,y) = 0$ for every $x$ and $y$ in $V$, which means $f$ is zero.