In the plane $V$ defined by the equation $x_1-2x_2+2x_3=0$, consider the basis $\mathcal{A} = (\vec{a_1}, \vec{a_2}) = \Bigg(\begin{bmatrix}2\\1\\0\end{bmatrix},\begin{bmatrix}-2\\0\\1\end{bmatrix}\Bigg)$.
a. Construct another basis $\mathcal{B}=(\vec{b_1}, \vec{b_2})$ of $V$, such that neither $\vec{b_1}$ nor $\vec{b_2}$ has any zero components.
b. Find the change of basis matrix $S$ from $\mathcal{B}$ to $\mathcal{A}$.
c. Find the change of basis matrix $S$ from $\mathcal{A}$ to $\mathcal{B}$.
d. Write an equation relating the matrices $[\vec{a_1}\,\,\,\vec{a_2}]$, $[\vec{b_1}\,\,\,\vec{b_2}]$, and $S = S_{\mathcal{B}\rightarrow\mathcal{A}}$.
I am utterly at a loss as to how to proceed here. I know that for changes between any two bases of a subspace, $AS=SB$ or $A=SBS^{-1}$ or $B=S^{-1}AS$, but the first part seems pretty arbitrary. Can I choose any two vectors $(\vec{b_1}, \vec{b_2})\in V$ that are linear functions of $(\vec{a_1}, \vec{a_2})$?
For (a) we wish to find a basis $\mathcal B=\{\vec b_1,\vec b_2\}$ for the subspace $V$ of $\Bbb R^3$ defined by the equation $$ x_1-2\,x_2+2\,x_3=0 $$ such that the components of $\vec b_1$ and $\vec b_2$ are all nonzero. Since we are given that $\{\vec a_1,\vec a_2\}$ are a basis, we can define \begin{align*} \vec b_1 &= 2\,\vec a_1+\vec a_2 & \vec b_2 &= 2\,\vec a_1-\vec a_2 \\ &= (2,2,1) & &= (6,2,-1) \end{align*}
The change of basis matrix $S$ from $\mathcal A$ to $\mathcal B$ is defined by $$ S=\begin{bmatrix}\alpha & \beta\\ \gamma&\delta\end{bmatrix} $$ where \begin{array}{rcccc} \vec b_1 & = & \alpha\,\vec a_1 & + & \gamma\,\vec a_2 \\ \vec b_2 & = & \beta\,\vec a_1 & + & \delta\,\vec a_2 \end{array} But here we clearly have $$ S=\begin{bmatrix}2 & 2\\ 1 & -1\end{bmatrix} $$
Finally, the change of basis matrix from $\mathcal B$ to $\mathcal A$ is $T=S^{-1}$. Can you compute $T$ in this case?