Verification of an Inner Product

[Parmenides]

My analysis instructor has posed an exercise to me in the following format:

"For {\bf{u}} = (u_1,u_2), {\bf{v}} = (v_1,v_2) \in R^2 define \left\langle{\bf{u}},{\bf{v}}\right\rangle = 3u_1u_2 - u_1v_2 - u_2v_1 + \frac{1}{2}u_2v_2..
Show that this is an inner product on $R^2$."

Not sure what the dots at the end represent, but I'm just posting it exactly as it appears. The textbook we're using (Advanced Calculus, Edwards) gives the criteria for an inner product as
1)\left\langle{\bf{x}},{\bf{x}}\right\rangle > 0, {\bf{x}} \neq 0
2) \left\langle{\bf{x}},{\bf{y}}\right\rangle = \left\langle{\bf{y}},{\bf{x}}\right\rangle
3) \left\langle{a}{\bf{x}} + b{\bf{y}},{\bf{z}}\right\rangle = a\left\langle{\bf{x}},{\bf{z}}\right\rangle + b\left\langle{\bf{y}},{\bf{z}}\right\rangle

I assume proving these would be sufficient, since this course doesn't deal with complex space. Thus, my attempt at a solution is:
1) \left\langle{\bf{u}},{\bf{u}}\right\rangle = 3u_1u_2 - u_1u_2 - u_2u_1 + \frac{1}{2}{u_2}^2 = u_1u_2 + \frac{1}{2}{u_2}^2
Which is non-zero if ${\bf{u}}$ is non-zero. Next,
2) \left\langle{\bf{v}},{\bf{u}}\right\rangle = 3v_1v_2 - v_1u_2 - v_2u_1 + \frac{1}{2}v_2u_2 = \left\langle{\bf{u}},{\bf{v}}\right\rangle
Proving property 3 seems to be giving me trouble. I believe I should begin by letting: {\bf{w}} = (w_1,w_2)
Then, \left\langle{a}{\bf{u}} + b{\bf{v}},{\bf{w}}\right\rangle = 3(au_1 + bv_1)(au_2 + bv_2) - (au_1 + bv_1)w_2 - (au_2 + bv_2)w_1 + \frac{1}{2}(au_2 + bv_2)w_2
The first term seems to pose a problem in rewriting in the desired form because I'm getting a quadratic-like form with the constants. Other than that, that's the only way I see how to show this. Any thoughts? Much appreciated.

 

[Parmenides]

Looking back, I think I've messed up proving property 2 and am now lost on that one, too.

 

[simba_lk]

I didn't even look at the third property because the firs and the second don't add up.
In the second as you sadi yourself what you wrote doesn't make sense.
but even more problematic is the fact that the first condition isn't met, and that there is no way to make it work.
As you wrote yourself condition is that an inner product of a vector with itself shouldn't only be non zero, it should also be definitely positive for each non-zero vector.

This term is interlinked with the way we define metrics, and as you perhaps know, due to this condition we can say that an inner product defines a norm over the space considered.

As you may know a norm is a non-negative operator and therefore your supposed innner product isn't an inner product at all.

 

[vela]

1) \left\langle{\bf{u}},{\bf{u}}\right\rangle = 3u_1u_2 - u_1u_2 - u_2u_1 + \frac{1}{2}{u_2}^2 = u_1u_2 + \frac{1}{2}{u_2}^2
Which is non-zero if ${\bf{u}}$ is non-zero.

That's not true. $\langle(1,0), (1,0)\rangle = 0$

Even if it were, you actually need to show that the inner product of a non-zero vector with itself is positive as simba_lk noted, but you have, for example, $\langle(-1,1), (-1,1)\rangle = -1/2$.

 

[Parmenides]

Correction; my instructor made a typo. The $u_2$ in the term $3u_1u_2$ should be a $v_1$ in which case I think all three properties come together easily:
\left\langle{\bf{u}},{\bf{v}}\right\rangle = 3u_1v_1 - u_1v_2 - u_2v_1 + \frac{1}{2}u_2v_2
So that:
1) \left\langle{\bf{u}},{\bf{u}}\right\rangle = 3{u_1}^2 - 2u_1u_2 + \frac{1}{2}{u_2}^2
2) \left\langle{\bf{v}},{\bf{u}}\right\rangle = 3v_1u_1 - v_1u_2 - v_2u_1 + \frac{1}{2}v_2u_2 = \left\langle{\bf{u}},{\bf{v}}\right\rangle
3) Let ${\bf{w}} = (w_1,w_2)$ so that:
\left\langle{a}{\bf{u}} + b{\bf{v}},{\bf{w}}\right\rangle = 3(au_1 + bv_1)w_1 -(au_1 + bv_1)w_2 - (au_2 + bv_2)w_1 + \frac{1}{2}(au_2 + bv_2)w_2 = (3au_1w_1 - au_1w_2 - au_2w_1 + \frac{1}{2}au_2w_2) + (3bv_1w_1 - bv_1w_2 - bv_2w_1 + \frac{1}{2}bv_2w2) = a\left\langle{\bf{u}},{\bf{w}}\right\rangle + b\left\langle{\bf{v}},{\bf{w}}\right\rangle

 

[vela]

Did you prove that the expression you got in #1 is positive?

 

[Parmenides]

I rested on the theorem of quadratic forms that it is positive-definite due to the positive coefficient, 3. The proof of *that* escapes me, however.