On $P_2$ define $\langle p(x),q(x)\rangle = p(1)q(1) + p(2)q(2)$ for $p(x),q(x)$ in $P_2$.
I am very confused regarding how to tell that this is an inner product.
I know that in order to show its an inner product the following must hold true:
1) Symmetry: $\langle f, g \rangle = \langle g, f \rangle$
2) Linearity: $\langle \alpha f,g\rangle = \alpha \langle f,g\rangle$
3) Positive-definite: $\langle f, f \rangle \geq 0$.
I have no idea how to prove it. How do I prove this is or is not an inner product?
Symmetry: Let $f, g \in \mathcal{P}2$ then $$\langle f,g \rangle = f(1)g(1) + f(2)g(2) = g(1)f(1) + g(2)f(2) = \langle g,f \rangle.$$
Linearity: Let $f,g \in \mathcal{P}2$ and $\alpha \in \mathbb{R}$ then $$\langle \alpha f,g \rangle = \alpha f(1) g(2) + \alpha f(2) g(2) = \alpha (f(1)g(1) + f(2)g(2)) = \alpha \langle f,g \rangle.$$
Semi positive-definite: Let $f \in \mathcal{P}2$ then $$\langle f,f \rangle = f(1)^2 + f(2)^2 \geq 0.$$ It is not positive-definite because the polynomial $f=(X-1)(X-2)$ is nonzero, has degree $2$ but $\langle f,f \rangle =0$. Positive definite means that for all $0 \neq f \in \mathcal{P}2$ we have $\langle f,f \rangle = 0$ and $\langle 0,0 \rangle=0$. So what you're describing is not a inner-product.
You can also show that for any $f,g,h \in \mathcal{P}2$ $$\langle f+g,h \rangle = \langle f,h \rangle + \langle g,h \rangle$$ which shows that this is a semi-inner-product.