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I'm having a little trouble with this question of determining whether the operation is an inner product:

Verify that the operation $<x, y> = x_1y_1 − x_1y_2 − x_2y_1 + 3x_2y_2$ where $x= (x_1, x_2)$ and $y = (y_1, y_2)$ is an inner product in $R^2$.

Now I do understand for the above to be an inner product, it has to satisfy 4 axioms:

  1. $<u,v>$ = $<v,u>$
  2. $\alpha$$<u,v>$= <$\alpha$ u, v>
  3. $<u+v,w>$= $< u, w>+<v,w>$

4a. $(u,u)$≥ 0

4b. $(u,u)$ = 0 $\implies$$u$=0

I have managed to proof axiom 1 & 2 but I'm having trouble with proofing axiom 3, 4a and 4b.

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  • $\begingroup$ Please use \langle and \rangle to typeset the "$\langle$" and "$\rangle$" symbols. $\endgroup$ – zipirovich May 2 '18 at 14:31
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Since we have$$\langle(x,y),(x,y)\rangle=x^2-2xy+3y^2=(x-y)^2+2y^2,$$it should be clear how to prove that the remaining properties hold.

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  • $\begingroup$ Hi José. Will this be for axiom number 4a and 4b? $\endgroup$ – Jenny May 2 '18 at 13:22
  • $\begingroup$ @Jenny Yes. What's the problem with axiom 3? It's just a matter of computing both sides and proving that they give the same thing. $\endgroup$ – José Carlos Santos May 2 '18 at 13:23
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Note that $$<u, w>+ <v, w> =$$

$$ u_1w_1 − u_1w_2 − u_2w_1 + 3u_2w_2+$$

$$ v_1w_1 − v_1w_2 − v_2w_1 + 3v_2w_2=$$

$$ (u_1+v_1)w_1 − (u_1+v_1)w_2 − (u_2+v_2)w_1 + 3(u_2+v_2)w_2=<u+v, w> $$

Also, $$<u, u> = u_1u_1 − u_1u_2 − u_2u_1 + 3u_2u_2= (u_1-u_2)^2+ u_2^2\ge 0$$ $$<u, u> = 0 \iff (u_1-u_2)^2+ u_2^2 =0 \iff u=0$$

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I tried attempting axiom 3 but something weird is going on. This is what I tried:

Let $w$=$(w_1,w_2)$

$<x+y,w> = (x_1+y_1)w_1 - (x_1+y_2)w_2-(x_2+y_1)w_1+3(x_2+y_2)w_2$

= $x_1w_1+y_1w_1-x_1w_2-y_2w_2-x_2w_1-y_1w_1+3x_2w_2+3y_2w_2$

= $x_1w_1-x_1w_2-x_2w_1+3x_2w_2+y_1w_1+y_2w_2-y_1w_1+3y_2w_2$

I managed to get $<x,w>$ but struggling to get $<y,w>$ since $y_1w_1-y_1w_1$ on my third step will cancel each other out.

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