Could you please help me solve these quesitons??
Consider the bases S={$u_1, u_2, u_3$}, and T={$v_1, v_2, v_3$}, with
$u_1$=[-3, 0, -3], $u_2$=[-3, 2, -1], $u_3$=[1, 6, -1],
$v_1$=[-6, -6, 0],
$v_2$=[-2, -6, 4], $v_3$=[-2, -3, 7]
(a) Find the transition matrix from $S$ to $T$
(b) Using the result in (a), compute the coordinate vector $[w]_T$ where $w$=[-5, 8, -5]
On
As noted by @Emilio Novati, the key idea is to connect to the standard basis.
Vectors will be colored according to the basis membership, and named chromatically: $$ \color{blue}{\mathbf{B}\ (standard)}, \qquad \color{red}{\mathbf{R}\ (u)}, \qquad \color{green}{\mathbf{G}\ (v)}. $$ $$ \mathbf{R}_{\color{red}{R}\to \color{blue}{A}}= \color{black}{\left[ \begin{array}{rrr} -3 & -3 & 1 \\ 0 & 2 & 6 \\ -3 & -1 & -1 \\ \end{array} \right]} $$ Example: the second vector in the $\color{red}{\mathbf{S}}$ basis has the following coordinates in the $\color{blue}{standard}$ basis $$ \mathbf{R}_{\color{red}{S}\to \color{blue}{A}} \color{red}{\left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right]} = \color{blue}{\left[ \begin{array}{r} -3 \\ 2 \\ -1 \end{array} \right]} $$ You may think of the matrix $\mathbf{R}_{\color{red}{S}\to \color{blue}{A}}$ as an operator which takes a $\color{blue}{blue}$ vector and returns a $\color{red}{red}$ vector.
The inverse matrix is a map which connects vectors in the $\color{blue}{standard}$ basis to vectors in the $\color{red}{\mathbf{R}}$ basis: $$ \mathbf{R}^{-1}_{\color{blue}{B}\to \color{red}{U}} % \color{blue}{\left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right]} = % \frac{1}{24} \left[ \begin{array}{rrr} 2 & -2 & -10 \\ -9 & 3 & 9 \\ 3 & 3 & -3 \\ \end{array} \right] % \color{blue}{\left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right]} = % \color{red}{\frac{1}{24} \left[ \begin{array}{r} -10 \\ 9 \\ -3 \end{array} \right]} $$
For the $\color{green}{\mathbf{G}}$ basis, the maps are $$ \mathbf{G}_{\color{green}{G}\to \color{blue}{A}}= \color{black}{\left[ \begin{array}{rrr} -6 & -6 & 0 \\ -2 & -6 & 4 \\ -2 & -3 & 7 \\ \end{array} \right]}, \qquad % \mathbf{G}^{-1}_{\color{blue}{A}\to \color{green}{G}}= \color{black}{\left[ \begin{array}{rrr} -5 & 7 & -4 \\ 1 & -7 & 4 \\ -1 & -1 & 4 \\ \end{array} \right]} $$
Transition from $S$ to $T$
We connect all bases through the hub of the $\color{blue}{standard}$ basis. Start with a vector in the $\color{red}{\mathbf{R}}$ basis, map that to a vector in the $\color{blue}{standard}$ basis, then map that to a vector in the $\color{green}{\mathbf{G}}$ basis: $$ \color{red}{ \left[ \begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array} \right]} % \quad \Longrightarrow \quad % \color{blue}{ \left[ \begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array} \right]} % \quad \Longrightarrow \quad % \color{green}{ \left[ \begin{array}{c} x_{3} \\ y_{3} \\ z_{3} \end{array} \right]} % $$ The formal steps are $$ \begin{align} % \mathbf{R}_{\color{red}{R}\to \color{blue}{B}} \color{red}{ \left[ \begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array} \right]} % &= % \color{blue}{ \left[ \begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array} \right]} \\[3pt] % \mathbf{G}^{-1}_{\color{blue}{B}\to \color{green}{G}} \left( \mathbf{S}_{\color{red}{R}\to \color{blue}{B}} \color{red}{ \left[ \begin{array}{c} x_{1} \\ y_{1} \\ z_{1} \end{array} \right]} \right) % &= % \mathbf{G}^{-1}_{\color{blue}{B}\to \color{green}{G}} \color{blue}{ \left[ \begin{array}{c} x_{2} \\ y_{2} \\ z_{2} \end{array} \right]} % = % \color{green}{ \left[ \begin{array}{c} x_{3} \\ y_{3} \\ z_{3} \end{array} \right]} % \end{align} $$ The operator which maps $\color{red}{red}$ vectors to $\color{green}{green}$ is $$ \mathbf{X}_{\color{red}{R}\to \color{green}{G}} = \mathbf{G}^{-1}_{\color{blue}{B}\to \color{green}{G}} \mathbf{R}_{\color{red}{R}\to \color{blue}{B}} = % \frac{1}{24} \left( \begin{array}{rr} 27 & 33 & 41 \\ -15 & -21 & -45 \\ -9 & -3 & -11 \\ \end{array} \right)% $$
Computation
Turn a $\color{red}{red}$ vector into a $\color{green}{green}$ vector: $$ \mathbf{X}_{\color{red}{R}\to \color{green}{G}} \color{red}{ \left[ \begin{array}{r} -5 \\ 8 \\ -5 \end{array} \right]} = \color{green}{ \frac{1}{18} \left[ \begin{array}{r} -57 \\ 99 \\ 57 \end{array} \right]} $$
Hint:
If $U$ is the transition matrix from the standard basis to the basis $S$ ( i.e. the matrix that has as columns the vectors $u_i$), and $V$ is the transition matrix from the standard basis to the basis $T$, than the transition from $S$ to $T$ is given by the matrix $V^{-1}U$.