I was given the following $\langle U, V \rangle = 3U_1V_1 + 2U_1V_2 + 2U_2V_1 + U_2V_2$ and I was asked to demonstrate if its a scalar product or not.
I've tested the first three axioms and they all worked out to be true for the given expression.
Now, I have trouble with the 4th Axiom ($\langle U, U \rangle \geq 0$ and $\langle U, U \rangle = 0\iff U = 0$ (positive definite) This is the definition from my lecture slides. )
If I try to find a counter-example, there seems to be two ways to do it.
1) Finding a vector u that gives $\langle u , u \rangle =$ a negative value.
2) Finding a vector $u$ that is negative but gives $\langle u, u\rangle = 0$.
I am not too sure if their valid and I dont know how to test it (How does the plug and chugg work in that case? i.e: how do I do I compute $U_1U_2$? Do I use the standard scalar product of $\mathbb{R}^2$?)
I know in advance that its not a scalar product because it doesn't verify the 4th axiom (I know this because I know the answer, I am interested in knowing how come).
For future problems like these, what if the expression was a scalar product. How would I go about proving it knowing that finding counter-examples didn't work to disprove it?
Note that\begin{align}\bigl\langle(u_1,u_2),(u_1,u_2)\bigr\rangle&=3{u_1}^2+4u_1u_2+{u_2}^2\\&=(2u_1+u_2)^2-{u_1}^2.\end{align}So, now it's easy: you take, for instance, $u_1=1$, and $u_2=-\frac12u_1=-\frac12$. Then$$\bigl\langle(u_1,u_2),(u_1,u_2)\bigr\rangle=0^2-1^2=-1<0.$$