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This time I brought a different problem.

I'm starting the study of bilinear forms and I came across this question. This is probably simple, however still confuses me a bit. I must prove that

$f(u,v) = 2u_1 \cdot v_2 - 3u_2\cdot v_1$ is bilinear.

Now, I've seen two possible definitions, one with $4$ steps and the other with $2$. I'll stick with the second one as it is more straightforward.

$$f(au_1+b\cdot u_2, v) = af(u_1,v) + bf(u_2,v)$$ $$f(u, a v_1+bv_2) = af(u,v_1) + bf(u,v_2).$$

While I think I understand the concept, I'm not sure what to plug in $v$ when I decompose $u$ into $u_1$ and $u_2.$ Can anyone help?

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2 Answers

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I think you're getting confused about what is a vector and what is a scalar. Here $\mathbf{u}$ and $\mathbf{v}$ are vectors. So, for example, $\mathbf{u}_1 = \langle u_{11}, u_{12}\rangle$ and $\mathbf{u}_2 = \langle u_{21},u_{22}\rangle$. Then \begin{align} f(a\mathbf{u}_1 + b\mathbf{u}_2, \mathbf{v}) &= f(\langle au_{11}+bu_{21}, au_{12}+bu_{21}\rangle, \langle v_1, v_2\rangle) \\ &= 2(au_{11}+bu_{21})v_2 - 3(au_{12}+bu_{21})v_1 \\ &= a(2u_{11}v2 - 3u_{12}v_1) + b(2u_{21}v_2 - 3u_{21}v_1) \\ &= af(\mathbf{u}_1\mathbf{v}) + bf(\mathbf{u}_2\mathbf{v}). \end{align} The other case is similar.

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$$ (u_1, u_2) \left( \begin{array}{rr} 0 & 2 \\ -3 & 0 \end{array} \right) \left( \begin{array}{r} v_1 \\ v_2 \end{array} \right) = 2 u_1 v_2 - 3 u_2 v_1 $$

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