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$\langle u + v, w \rangle = \langle u, w \rangle + \langle v, w \rangle$ ; $u = (3, -2), v = (4, 5), w = (-1, 6)$

$\langle u, v \rangle = 4u_1v_1 + 5u_2v_2 $

When I tried to do this I got: $\langle u + v, w \rangle = \langle(48, -50), (-1, 6)\rangle = -348$ (probably the wrong answer)

Looking at it, it appears that:

$\langle u, v \rangle = v_1u_1v_1 + v_2u_2v_2$

but I don't know how to deal with the weighted inner product.

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1 Answer 1

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I got:
$$\langle u + v, w \rangle = \langle (3+4,-2+5),(-1,6)\rangle$$ $$=\langle (7,3),(-1,6)\rangle$$ $$=4( (7)(-1)) + 5((3)(6)) = -28+90=62$$

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So once you get it down to ⟨(7,3),(−1,6)⟩ you treat it like the ⟨u,v⟩? –  user8640 Mar 24 '11 at 1:25
    
Hey, I'm not an expert, but the weighting can only apply when you do the actual inner product. And <u,w> + <v,w> should get the same result. –  Carl Brannen Mar 24 '11 at 1:31

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