# Linear Algebra- Computing Inner Product

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.

• Please use \langle and \rangle to typeset the "$\langle$" and "$\rangle$" symbols. – zipirovich May 2 '18 at 14:31

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.

• Hi José. Will this be for axiom number 4a and 4b? – Jenny May 2 '18 at 13:22
• @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. – José Carlos Santos May 2 '18 at 13:23

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

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.