In $\mathbb{R}^3$ the canonical basis $E=\left (\mathbf{e_1},\mathbf{e_2},\mathbf{e_3} \right )$ and $B=\left (\mathbf{b_1},\mathbf{b_2},\mathbf{b_3} \right )$ with
$\mathbf{b_1}=(1,2,4)^T$, $\mathbf{b_2}=(0,-1,1)^T$ and $\mathbf{b_3}=(2,3,8)^T$.
How do I determine for vector $\mathbf{v}=2\mathbf{e_1}+\mathbf{e_2}+2\mathbf{e_3}$ coordinates $\left \lfloor \mathbf{v} \right \rfloor_E$ and $\left \lfloor \mathbf{v} \right \rfloor_b$ ?
a) $\left \lfloor \mathbf{v} \right \rfloor_E$ =$\begin{pmatrix} 2\\ 1\\
2\end{pmatrix}$
$\left \lfloor \mathbf{v} \right \rfloor_b$ =$\begin{pmatrix} -16\\ 6\\
9\end{pmatrix}$
How do I determine for vector $\mathbf{w}=\mathbf{b_1}+2\mathbf{b_2}+3\mathbf{b_3}$ coordinates $\left \lfloor \mathbf{w} \right \rfloor_E$ and $\left \lfloor \mathbf{w} \right \rfloor_b$ ?
b)
$\left \lfloor \mathbf{w} \right \rfloor_E$ =$\begin{pmatrix} 7\\ 9\\
30\end{pmatrix}$
$\left \lfloor \mathbf{v} \right \rfloor_b$ =$\begin{pmatrix} 1\\ 2\\
3\end{pmatrix}$
Where I need help?
Determine the transformation matrix T of the coordinate transformation from the base E to the base B, where the old coordinates on E and the new coordinates refer to B. Note: Determine the matrix T such that applies
$\left [ \vec{x} \right ]_E=T\left [ \vec{x} \right ]_B $
I really don't understand how to do it?
d) How can I calculate $\left [ \vec{x} \right ]_b$ from $\left [ \vec{x} \right ]_E$. Take care that $\left [ \vec{x} \right ]_E=\vec{x}$.
Best Answer
Hint:
The matrix $$ M= \begin{bmatrix} 1 & 0 & 2\\ 2 & -1 & 3\\ 4 & 1 & 8 \end{bmatrix} $$ represents the transformation: $$ \mathbf{e_1}\to \mathbf{b_1} \qquad \mathbf{e_2}\to \mathbf{b_2} \qquad \mathbf{e_3}\to \mathbf{b_3} $$ and its inverse: $$M^{-1} \begin{bmatrix} -11 & 2 & 2\\ -4 & 0 & 1\\ 6 & -1 & -1 \end{bmatrix} $$ represents the transformation: $$ \mathbf{b_1}\to \mathbf{e_1} \qquad \mathbf{b_2}\to \mathbf{e_2} \qquad \mathbf{b_3}\to \mathbf{e_3} $$
Use $M^{-1}$ to substitute $\mathbf{e_i}$ in the vector $\mathbf{v}$ and $M$ to substitute $\mathbf{b_i}$ in the vector $\mathbf{w}$
Note that the columns of $M$ are the vectors $\mathbf{b_i}$ in the standard basis, so $M\mathbf{e_i}=\mathbf{b_i}$. In the same way the columns of $M^{-1}$ are the vectors of the standard basis expressed in the basis $\mathbf{b_i}$. So, by linearity, your vector $\mathbf{v}$ that in the standard basis is $\mathbf{v}=2\mathbf{e_1}+\mathbf{e_2}+2\mathbf{e_3}$, in the basis $B$ is: $$ M^{-1}\mathbf{v}= \begin{bmatrix} -11 & 2 & 2\\ -4 & 0 & 1\\ 6 & -1 & -1 \end{bmatrix} \begin{bmatrix} 2\\ 1\\ 2 \end{bmatrix}= \begin{bmatrix} -16\\ -6\\ 9 \end{bmatrix} $$ and the vector $\mathbf{w}$ that in the basis $B$ is $[1,2,3]^T$ , in the standard basis is: $$ M\mathbf{w}= \begin{bmatrix} 1& 0 & 2\\ 2 & -1 & 3\\ 4 & 1 & 8 \end{bmatrix} \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}= \begin{bmatrix} 7\\ 9\\ 30 \end{bmatrix} $$