Hi I would like some help with the following exercise:
We have an orthonormal basis $B_1$ and the vectors $$u_1= \begin{bmatrix}\frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{bmatrix} u_2= \begin{bmatrix}\frac{2}{3} \\ \frac{1}{3}\\ -\frac{2}{3} \end{bmatrix} u_3=\begin{bmatrix}-\frac{2}{3} \\ \frac{2}{3} \\ -\frac{1}{3} \end{bmatrix}$$
Control that $B_2=\begin{bmatrix}u_1 \\ u_2 \\ u_3 \end{bmatrix}$ is an orthonormal basis by putting together a matrix and multiplying it with it's transponate.
Solution: the transponate of
$$ \begin{matrix} 1/3 & 2/3 & 2/3 \\ 2/3 & 1/3 & -2/3 \\ -2/3 & 2/3 & -1/3 \\ \end{matrix} $$
= $$ \begin{matrix} 1/3 & 2/3 & -2/3 \\ 2/3 & 1/3 & 2/3 \\ 2/3 & -2/3 & -1/3 \\ \end{matrix} $$
We have the vectors $v_1=\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}$ and $v_2=\begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}$ in $B_1$. Calculate them in $B_2$.
Solution: I think im supposed to take the transpose i got from the previous answer and multiply it with the $v_1$ to get it in base $B_2$ but i get the wrong answer.
For point a) we have that
$$AA^T=I$$
and thus $B_2$ is an orthonormal basis.
For point b) we need to solve the following systems
$au_1+bu_2+cu_3=v_1\implies A[a\,b\,c]^T=[1,0,1]^T\implies [a\,b\,c]^T=A^T[1,0,1]^T$
$du_1+eu_2+fu_3=v_2\implies A[d\,e\,f]^T=[0,1,1]^T\implies [d\,e\,f]^T=A^T[0,1,1]^T$