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Given $||\mathbf{u}|| = 4, ||\mathbf{v}|| = 3 $ and the equation $$ \langle 2 \mathbf{u}-3 \mathbf{v},4 \mathbf{u}+5 \mathbf{v} \rangle=-10 $$

How do I partition this correctly so I can determine $\langle \mathbf{u,v}\rangle$?

Another question of mine already tackled transformation using linearity, but the fact that there's more than one element in the first component of the scalar product has dazed me a little.

My (wrong) transformation looked like the following:

$$ \langle 2 \mathbf{u},4 \mathbf{u}+5 \mathbf{v} \rangle + \langle 5 \mathbf{v},2 \mathbf{u}-3 \mathbf{v} \rangle =-10 $$ $$ \langle 2 \mathbf{u},4 \mathbf{u}\rangle + \langle 2 \mathbf{u},5 \mathbf{v}\rangle +\langle 5 \mathbf{v},2 \mathbf{u}\rangle - \langle 5 \mathbf{v},3 \mathbf{v}\rangle =-10 $$

After that, pull the scalars in front of the scalar products so we get $$ 8\langle \mathbf{u}, \mathbf{u}\rangle + 2(10\langle \mathbf{u}, \mathbf{v}\rangle) - 15\langle \mathbf{v}, \mathbf{v}\rangle =-10 $$

Going on with this, it yields a weird result for $\langle \mathbf{u,v}\rangle$ , which leads me to the suspicion that my transformation has gone horribly wrong.

What would be the correct way to transform the initial equation?

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2 Answers 2

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A bilinear function F satisfies $$F(a+b,x+y) \equiv F(a,x+y) + F(b,x+y) \equiv F(a,x)+F(a,y)+F(b,x)+F(b,y)$$ which you can check by letting $z=x+y$ first. You could also let $c=a+b$ instead and expand in the other order, but you'd get the same result.

If it helps, imagine the simple (1D) example $F(a,x)=ax$. You're essentially just multiplying out brackets.

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Hints:

$$-10=\langle 2u-3v\,,\,4u+5v\rangle=8\langle u,u\rangle-2\langle u,v\rangle-15\langle v,v\rangle=-2\langle u,v\rangle-7\;\ldots$$

Of course, assuming your inner product is real and not complex.

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