Given the following inner space over $\mathbb{R}^{2}$: $$ \left< \left(\begin{pmatrix} x_{1}\\ x_{2} \end{pmatrix} ,\begin{pmatrix} y_{1}\\ y_{2} \end{pmatrix}\right) \right> =x_{1} y_{1} -3x_{1} y_{2} \ -3x_{2} y_{1} +ax_{2} y_{2} \ $$ To which values of $a$ is it an inner space?
I figured it has to do with the positive/definite property since I already checked for the other twos and it's true for any values of $a$.
But I don't understand how am I supposed to solve such equation:
$x_{1} y_{1} -3x_{1} y_{2} \ -3x_{2} y_{1} +ax_{2} y_{2} \ \geqslant 0$
Thank you so much for the help!
In order to check positive definiteness, you should check the inner product of $x$ with itself, not with a general $y$. Hence, your inequality becomes
$$ \langle x_1,x_2\rangle_a:=x_1^2-6x_1x_2+a x_2^2\geq 0 $$ Now, this is a quadratic in $x_2$ with discriminant $36x_1^2-4ax_1^2=(36-4a)x_1^2$. Now, in order for the quadratic to always be strictly positive for $x_1\neq 0$, it must have no root, so the discriminant must be negative, which happens if and only if $a > \frac{36}{4}$ for $x_1\neq 0$.
Now, we must check that for such a choice of $a$, the constant sign of the inner product is positive, no matter the value of $x_1,x_2$. To this end, we see for $a>\frac{36}{4}$ that the above quadratic is convex and hence, for such an $a$ and $x_1\neq 0$, $$ \lim_{x_2\to\infty} \langle x_1,x_2\rangle_a=\infty, $$ implying that $\langle x_1,x_2\rangle_a\geq 0$ for all $x_2$, since the quadratic has no root and hence, constant sign. If $x_2\neq 0$ and $x_1=0$, then positivity is obvious.
We conclude that for $\langle \cdot,\cdot\rangle_a$ is positive definite if and only if $a>\frac{36}{4}$..