Show that this $\mathbb{R}^2$ is an euclidean vector space

Question

Show that $\mathbb{R}^2$ with $\left \langle , \right \rangle: \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}, \left \langle \begin{pmatrix} x_u\\ y_u \end{pmatrix}, \begin{pmatrix} x_v\\ y_v \end{pmatrix} \right \rangle = x_ux_v + x_uy_v + y_ux_v + 2y_uy_v$ is an euclidean vector space.

If I understand the definition of euclidean space correctly, it's simply a vector space over the reals $\mathbb{R}$ with an inner product $\left \langle , \right \rangle$ and this inner product must be symmetric, bilinear and positive. Is this correct so far?

So I think I need to show that these 3 properties are satisfied on the given vector space. But how do you do this correctly?

It's symmetric because $x_vx_u + y_vx_u + x_vy_u + y_vy_u2 = x_ux_v + x_uy_v + y_ux_v + 2y_uy_v$, e.g. commutative also because we map from $\mathbb{R}^2$ to $\mathbb{R}$.

It's bilinear because it's distributive: $x_u(x_v+y_v) + y_u(x_v+2y_v) \Leftrightarrow x_ux_v + x_uy_v + y_ux_v + 2y_uy_v$

I'm not sure about the last property and if it's correct at all till here? :s

Answer 1

Up to here it is more or less correct.

Now note that\begin{align}\left \langle\begin{pmatrix} x_u\\ y_u \end{pmatrix}, \begin{pmatrix} x_u\\ y_u\end{pmatrix} \right \rangle&={x_u}^2+2x_uy_u+2{y_u}^2\\&=(x_u+y_u)^2+{y_u}^2\end{align}and that therefore $\left \langle\begin{pmatrix} x_u\\ y_u \end{pmatrix}, \begin{pmatrix} x_u\\ y_u\end{pmatrix} \right \rangle\geqslant0$ and that it is equal to $0$ if and only if $x_u=y_u=0$.