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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 $\left [ \vec{x} \right ]_E=\vec{x}$.

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  • $\begingroup$ Nothing because I don't know how to try. $\endgroup$ – Ninalol Apr 17 '16 at 17:08
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From what I understand, you are looking for the base transformation matrix from the base $E$ (standard basis) to $B$, i.e. a matrix $T$ such that $Te_1=b_1$, $Te_2=b_2$ and $Te_3=b_3$. This matrix is simply given by $T=(b_1\, b_2\, b_3)$.

Edit

There is nothing to do anymore. The matrix is \begin{align} T=\begin{pmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8 \end{pmatrix} \end{align} You can check that it indeed transforms the basis $E$ to the basis $B$.

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  • $\begingroup$ Yes, how to do it? $\endgroup$ – Ninalol Apr 17 '16 at 17:13
  • $\begingroup$ I have edited my answer. $\endgroup$ – B. Pasternak Apr 17 '16 at 17:17
  • $\begingroup$ and can you help how to do d)?? $\endgroup$ – Ninalol Apr 17 '16 at 17:19
  • $\begingroup$ But how do you know that matrix $T$ is the same like $B$?? $\endgroup$ – Ninalol Apr 17 '16 at 17:20
  • $\begingroup$ Edited my question. $\endgroup$ – Ninalol Apr 17 '16 at 18:01
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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: $$ M\mathbf{w}= \begin{bmatrix} 1& 0 & 2\\ 2 & -1 & 3\\ 4 & 1 & 8 \end{bmatrix} \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}= \begin{bmatrix} 7\\ 9\\ 30 \end{bmatrix} $$

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  • $\begingroup$ But how do you know that matrix $M$ is the same like $B$. I don't how can I substitute it and do this example $d)$ quesiton... $\endgroup$ – Ninalol Apr 17 '16 at 17:25
  • $\begingroup$ Simply calculate $M\mathbf{e_i}$ and you find $\mathbf{b_i}$. Or, in a more general way, think that the $\mathbf{b_i}$ are linear combinations of the $\mathbf{e_i}$ and the columns of the matrix $M$ represent such linear combinations. $\endgroup$ – Emilio Novati Apr 17 '16 at 17:35
  • $\begingroup$ Sorry can you help me because I don't know how to do it $\endgroup$ – Ninalol Apr 17 '16 at 17:45
  • $\begingroup$ I've added to may answer. I hope it's useful ( and that i've no mistake in the calculations). $\endgroup$ – Emilio Novati Apr 17 '16 at 19:42
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Let $E = (e_1, e_2, e_3)$ be the old coordinate basis, then any vector $x=(x_1, x_2, x_3)$ can be written as $x = x_1e_1+x_2e_2+x_3e_3$. Let $B = (b_1, b_2, b_3)$ be the new coordinate system. Let $\alpha_1, \alpha_2, \alpha_3$ be the coordinates of $x$ w.r.to the new basis. Then $x$ can be written as $x = \alpha_1b_1+\alpha_2b_2+\alpha_3b_3$. So

\begin{equation} \begin{split} \alpha_1b_1+\alpha_2b_2+\alpha_3b_3 &= x_1e_1+x_2e_2+x_3e_3\\ [b_1, b_2, b_3](\alpha_1, \alpha_2, \alpha_3)^{T} &= [e_1, e_2, e_3] (x_1, x_2, x_3)^{T}\\ (\alpha_1, \alpha_2, \alpha_3)^{T} &= [b_1, b_2, b_3]^{-1} [e_1, e_2, e_3] (x_1, x_2, x_3)^{T} \end{split} \end{equation}

Therefore the transformation matrix (from basis $E$ to $B$) is given by $T = [b_1, b_2, b_3]^{-1} [e_1, e_2, e_3]$

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