Be $f:\Bbb R^2 \times\Bbb R^2\to \Bbb R$ defined as $f\big((x_1,x_2),(y_1,y_2)\big)=x_1y_1-x_1y_2-x_2y_1+\alpha x_2y_2$. Find the values of $\alpha \in \Bbb R$ such that $f$ is an inner product.
I proved 3 of the 4 properties for $f$ to be an inner product, except the one that tells you that f is an inner product if and only if $f\big((x_1,x_2),(x_1,x_2)\big)=0 \iff (x_1,x_2)=(0,0)$. But here is where it lies my problem, that i don't understand completely how to do it, because I need to force the equation that I get by the definition "$x_1^2-2x_1 x_2+\alpha x_2=0$" to conclude that $(x_1,x_2)=(0,0)$,any tips? or something that I am missing in order to solve it?
UPDATE: For the other case: Let see when $\alpha<1$, so because of this case we have that $1-\alpha>0$ and with that $$\sqrt{1-\alpha}>0$$ $$ \Rightarrow(x_1 - x_2)^2 -(1-\alpha)x_2^2 = 0. $$ $$ \Rightarrow(x_1 - x_2)^2 -(\sqrt{(1-\alpha)}x_2)^2 = 0. $$ $$ \Rightarrow((x_1 - x_2) -\sqrt{(1-\alpha)}x_2)((x_1 - x_2)+\sqrt{(1-\alpha)}x_2) = 0. $$ $$ \Rightarrow((x_1 - x_2) -\sqrt{(1-\alpha)}x_2) = 0. $$ or $$ \Rightarrow((x_1 - x_2) +\sqrt{(1-\alpha)}x_2) = 0. $$ Finally $$ \Rightarrow x_1= x_2(1+ \sqrt{(1-\alpha)}). $$ or $$ \Rightarrow x_1= x_2(1- \sqrt{(1-\alpha)}). $$ And because there exist values different than zero for $x_1,x_2$ such that the inner product is zero, then it cannot happen that $\alpha \le 1$, and it would be an inner product if $\alpha>1$.
You want to check when $x_1^2 - 2x_1 x_2 + \alpha x_2^2 = 0$. Completing the square, you get the equation
$$ (x_1 - x_2)^2 + (\alpha - 1) x_2^2 = 0. $$
If $\alpha > 1$, then you can rewrite this expression as
$$ (x_1 - x_2)^2 + (\sqrt{\alpha - 1}x_2)^2 = 0 $$
and since both terms are non-negative, you get $x_1 = x_2$ and $\sqrt{\alpha - 1}x_2 = 0$ which implies that $x_1 = x_2 = 0$.
If $\alpha = 1$, then you get the equation $x_1 = x_2$ which has infinitely many non-trivial solutions. For example, taking $x_1 = x_2 = 1$, you get a non-zero vector $(1,1)$ for which $f((1,1),(1,1)) = 0$.
I'll leave the case $\alpha < 1$ to you with the hint to rewrite the equation as
$$ (x_1 - x_2)^2 - (\sqrt{1 - \alpha}x_2)^2 = 0. $$