Let
$A=\left(\begin{array}{lll} 2 & 0 & 1 \\\ 0 & 2 & 0 \\\ 1 & 0 & 2 \end{array}\right) \in \mathrm{M}_{3}(\mathbb{R})$
(a) Using the matrix $A$, we define $\langle x, x\rangle:=x^{t} \cdot A \cdot x \geq 0$, where $x, y \in \mathbb{R}^{3}$. Show that this defines an inner product on $\mathbb{R}^{3}$.
The problem for me in (a) is just showing $\langle x, x\rangle:=x^{t} \cdot A \cdot x \geq 0$
So let $ x = \left(\begin{array}{l} x_1 \\\ x_2 \\\ x_3 \end{array} \right) \in \mathbb{R}^{3}$
I end up with $2x_1^2 + 2x_2^2 + 2x_3^2 + 2x_1x_3 \geq 0 $ What's the argument that it's greater or equal to 0. I tried applying some binomial formulas for 2 and 3 variables but I ended up nowhere.
(b) Starting from the standard basis $\left(e_{1}, e_{2}, e_{3} \right)$, determine an orthonormal basis $\left(f_{1}, f_{2}, f_{3}\right)$ for $\mathbb{R}^{3}$ with respect to the inner product defined in part (a).
Note that
$\langle e_1, e_1 \rangle =e_1^{t} \cdot A \cdot e_1 = 2 \\\ \langle e_1, e_2 \rangle =e_1^{t} \cdot A \cdot e_2 = 0 \\\ \langle e_2, e_3 \rangle =e_2^{t} \cdot A \cdot e_3 = 0 \\\ \langle e_1, e_3 \rangle =e_1^{t} \cdot A \cdot e_3 = 1 \\\ \langle e_2, e_2 \rangle =e_2^{t} \cdot A \cdot e_2 = 2$
for future calculations.
Now I want to solve this with Gram-Schmidt-Orthonormalization.
The first vector already has length $1$ so $e_1 = f_1$.
Since $f_2 = e_2 - \frac{\langle e_1, e_2 \rangle}{\langle e_1, e_1 \rangle} \cdot e_1$ and $\langle e_1, e_2 \rangle = 0$, so $f_2 = e_2.$
The third vector
$f_3= e_3 - \frac{\langle e_1, e_3 \rangle}{\langle e_1, e_1 \rangle} \cdot e_1 - \frac{\langle e_2, e_3 \rangle}{\langle e_2, e_2 \rangle} \cdot e_2 = e_1 - 0.5 \cdot e_1 = \left( \begin{array}{l} -0.5 \\\ 0 \\\ 1 \end{array} \right)$
Normalizing:
$f_3^{*} = \frac{2\sqrt{5}}{5}\left( \begin{array}{l} -0.5 \\\ 0 \\\ 1 \end{array} \right)$
But $f_3^{*} $ isn't orthogonal to the first basis vector. I tried finding the error for hours now.
Thanks for any help!