Inner product of the function

Question

Let $x, y \in \mathbb{R}^n$. Under which condition on $\alpha \in \mathbb{R}^n$ does the function $$ f(x, y)=\sum_{k=1}^n \alpha_k x_k y_k $$ define an inner product on $\mathbb{R}^n$ ?

My idea:

1. $\sum_{k=1}^n \alpha_k x_k (y_k+z_k) =\sum_{k=1}^n \alpha_k x_k y_k+\sum_{k=1}^n \alpha_k x_kz_k=f(x, y)+f(x,z)$ This will hold for all $\alpha \in \mathbb{R}^n$.
2. $\sum_{k=1}^n \alpha \alpha_k x_k y_k = \alpha \sum_{k=1}^n \alpha_k x_k y_k$. This will hold for all $\alpha \in \mathbb{R}^n$.
3. $\sum_{k=1}^n \alpha_k x_k^2 > 0$ This holds if $\alpha_k$ is positive.
4. $\sum_{k=1}^n \alpha_k x_ky_k = \sum_{k=1}^n \alpha_k y_kx_k$ I am not sure about this symmetry case, what should I do here? Is the rest ok? Did I forgot to include something?

Answer 1

Symmetry holds simply because $x_ky_k=y_kx_k$.

You need $\alpha_k >0$ for each $k$ to make $\sum \alpha_kx_kx_k \geq 0$ and $\sum \alpha_kx_kx_k =0$ if and only if $x_k=0$ for each $k$.

[If $\alpha_i=0$ for some $i$ then $x_k=1$ for $k=i$ and $0$ for $k \neq i$ gives a contradiction].