I cant seem to solve the second part of this problem, which goes as follows:
Show that $\langle x,y \rangle = 2x_1y_1-2x_1y_2-2x_2y_1+5x_2y_2$ is an inner product on $\mathbb{R^2}$. Find an orthonormal basis for $\mathbb{R^2}$ with respect to this inner product.
The first part was straight-forward, but I cant seem to understand what to do in order to find an orthonormal basis. How can this be done?
Here's a way to construct a basis with the interesting property that it is orthogonal with respect to the given inner product and simultaneously orthogonal with respect to the dot product.
First, note that $\langle x, y \rangle = y^\top A x$, where $$A = \begin{pmatrix} 2 & -2 \\ -2 & 5 \end{pmatrix}.$$ Note that $A$ is symmetric. Indeed, $A$ is positive-definite if and only if $\langle \cdot, \cdot \rangle$ is an inner product. As $A$ is symmetric, it has real eigenvalues and orthogonal eigenvectors. Symmetry also tells us that $A$ is diagonalisable, and since it is $2 \times 2$, it has a repeated eigenvalue $\lambda$ if and only if $A = \lambda I$, which is clearly not true!
Thus, there will be distinct eigenvalues $\lambda_1, \lambda_2$ and corresponding eigenvectors $v_1, v_2$, which are orthogonal with respect to the dot product. That is, $v_2^\perp v_1 = 0$. Then, we have $$\langle v_1, v_2\rangle = v_2^\perp A v_1 = \lambda_1 v_2^\perp v_1 = 0.$$ That is, $v_1, v_2$ are not just orthogonal with respect to the dot product, but also with respect to $\langle \cdot, \cdot \rangle$.
If we want to normalise this basis, note that $$\langle v_i, v_i \rangle = v_i^\perp A v_i = \lambda_i v_i^\perp v_i = \lambda_i\|v_i\|^2_{\Bbb{R}^2},$$ where $\| \cdot \|^2_{\Bbb{R}^2}$ is the standard Euclidean norm. Thus, to normalise $v_i$, we simply let $$e_i = \frac{v_i}{\sqrt{\lambda_i}\|v_i\|_{\Bbb{R}^2}}.$$
As it turns out, $A$ has nice eigenvalues and eigenvectors. Specifically, we have \begin{align*} v_1 &= \begin{pmatrix} 2 \\ 1 \end{pmatrix} &\lambda_1 = 1 \\ v_2 &= \begin{pmatrix} 1 \\ -2 \end{pmatrix} &\lambda_1 = 6. \end{align*} You should verify that this basis is indeed orthogonal with respect to $\langle \cdot, \cdot \rangle$.
Note that $\|v_1\|_{\Bbb{R}^2} = \|v_2\|_{\Bbb{R}^2} = \sqrt{5}$, so our orthonormal basis (with respect to $\langle \cdot, \cdot \rangle$) is given by $$e_1 = \frac{1}{\sqrt{5}}\begin{pmatrix} 2 \\ 1 \end{pmatrix}, \quad e_2 = \frac{1}{\sqrt{30}}\begin{pmatrix} 1 \\ -2 \end{pmatrix}.$$