How to decide if $Q(\underline u,\underline v) $ on $\mathbb R^2$ is inner product or not if $$ (\underline u, \underline v) = \underline u^T A\underline v$$ where $$A = \begin{pmatrix} -1 & -3 \\ 5& 8 \\ \end{pmatrix}$$
Can this be opposed by considering the symmetric property only? $$ (\underline v, \underline u) = \underline v^T A\underline u \neq\underline u^T A\underline v$$ How to prove this?
By definition, $Q$ will NOT be an inner product on $\mathbb{R}^2$ iff ANY of the properties of inner product does not hold:
https://en.wikipedia.org/wiki/Inner_product_space
(Where in this case, the conjugate symmetry is just symmetry)
So if you show that ANY of this properties doesn't hold, say the symmetry, then $Q$ is NOT an inner product; as you said.
If you don't know how to do that, it's like this:
Since the property says that $Q(u,v)=Q(v,u)$ for any pair $u,v\in\mathbb{R}^2$; it's enough to show that there exists a particular pair $u_0,v_0$ such that $Q(u_0,v_0)\neq Q(v_0,u_0)$ .
So just take $$ u_0= (1,0) \ \ v_0=(0,1)$$
Then $Q(u_0,v_0)=-3\neq 5 = Q(v_0,u_0)$
In general, any $Q$ of the form $ Q(u,v)=u^TAv$ is an inner product in $\mathbb{R}^n \Leftrightarrow A\ $ is symmetric and positive-definite(which in the $\mathbb{R}^2$ case is equivalent to $det(A)\geq 0 \wedge \ a_{11}\geq 0$ ) .