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I'm trying to check if $f(u, v) = (uv)(\alpha)$ (where $(u, v) \in F[x]$ - field of the rational numbers and $\alpha \in F$) is bilinear function and find it's matrix.

Am I right, that:

$f(u+x, v) = ((u+x)v)(\alpha) = (uv + xv)(\alpha)$

$f(u, v) + f(x, v) = (uv)(\alpha) + (xv)(\alpha) = (uv + xv)(\alpha)$

$f(u, v+x) = ((u(v+x))(\alpha) = (uv + xv)(\alpha)$

$f(u, v) + f(u, x) = (uv)(\alpha) + (ux)(\alpha) = (uv + ux)(\alpha)$ - oops... not bilinear, right?

Or have I interpreted it in a wrong way?

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    What is $F[x]$ here? – 2017-01-21
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    It's a field of rational numbers. – 2017-01-21

1 Answers 1

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Hint:

To show that the given function $f(u,v)$ is bilinear you have to prove that: $$ 1) \qquad f(u+x,v)=f(u,v)+f(x,v)\quad \mbox{and}\quad f(u,v+x)=f(u,v)+f(u,x) $$ Your work goes in this direction, but there is a mistake in the third line because $u(v+x)=uv+ux$

and you must also prove that: $$ 2) \qquad f(\beta u,v)=\beta f(u,v)\quad \mbox{and}\quad f( u,\beta v)=\beta f(u,v) $$ Can you do this?

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    Yes, It's bilinear, finally. But how to find the matrix for this function? – 2017-01-21
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    Is it $\begin{pmatrix}\alpha&0\\0&\alpha\end{pmatrix}$ ? – 2017-01-21