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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 $f = 0$.

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What exactly is your question? What have you tried, and where are you stuck? – icurays1 Feb 6 '13 at 0:23

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.

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Why is it that $f(x,0) =0$? – user4593 Feb 8 '13 at 12:29

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