Coordinate transformation. Transition matrix.

Literature: Collection of Problems in Mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.

Let $L_n -$ be an arbitrary dimensional space, $B=(e_1, ..., e_n) -$ a fixed basis in it. Then each vector $x \in L_n$ corresponds uniquely to the column of its coordinates in this basis.

$$x=x_1e_1+...+x_ne_n\Leftrightarrow X=\begin{pmatrix}x_1\\ \vdots\\x_n\end{pmatrix}$$

In this case, linear combinations over vectors in coordinate form look as follows:

$z=x+y \Leftrightarrow Z=X+Y$

$y=\lambda x \Leftrightarrow Y=\lambda X.$

Let $B=(e_1, e_2, ..., e_n)$ and $B'=(e_1', e_2', ..., e_n') -$ be two different bases in $L_n.$ Decompose each vector of the basis $B'$ into the basis $B:$

$e_k'=t_{1k}e_1+...+t_{nk}e_n \Leftrightarrow E_k'=\begin{pmatrix}t_{1k}\ \vdots\ t_{nk}\end{pmatrix},\quad k=1, 2, ..., n.$

The transition matrix $T_{B \rightarrow B'}$ from the basis $B$ to the basis $B'$ is called the matrix

$T_{B \rightarrow B'} = \begin{pmatrix}t_{11}&...&t_{1n}\\...&...&...\\t_{n1}&...&t_{nn}\end{pmatrix}$ the $k$-th column of which is the column $E_k'$ of coordinates of the vector $e_k'$ in the basis $B.$ If $x -$ is an arbitrary vector from $L_n,$ $X$ and $X' -$ the columns of its coordinates in the bases $B$ and $B'$ respectively, then the equality holds $$X'=(T_{B\rightarrow B'})^{-1}X$$ (the formula for coordinate transformation when changing the basis).

Examples.

4.15. In the space $V_3,$ the vectors $e_1'=i+j,$ $e_2'=i-j,$ $e_3'=-i+2j-k$ are given. Prove that the system $B'=(e_1', e_2', e_3')$ is a basis in $R_3$ and write the transition matrix $T_{B \rightarrow B'},$ where $B=(e_1=i, e_2=j, e_3=k).$ Find the coordinates of the vector $x=i-2j+2k$ in the basis $B'.$

Solution.

To show that the system of vectors $B'=(e_1', e_2', e_3')$ is a basis in $R_3,$ it is sufficient to show that these vectors are non-coplanar.

From the condition, we have $e_1'=i+j=(1, 1, 0),$ $e_2'=i-j=(1, -1, 0),$ $e_3'=-i+2j-k=(-1, 2, -1).$ The vectors $e_1', e_2', e_3'$ are non-coplanar if $\begin{vmatrix}1&1&0\\1&-1&0\\-1&2&-1\end{vmatrix} \neq 0.$ Let's verify this:

$$\begin{vmatrix}1&1&0\\1&-1&0\\-1&2&-1\end{vmatrix}=\begin{vmatrix}-1&0\\2&-1\end{vmatrix}-\begin{vmatrix}1&0\\-1&-1\end{vmatrix}=1+1=2\neq 0.$$ Therefore, the system $B'=(e_1', e_2', e_3')$ is a basis in $R_3.$

Next, we write the transition matrix $T_{B \rightarrow B'}.$

$T_{B \rightarrow B'} = \begin{pmatrix}t_{11}&...&t_{1n}\\...&...&...\\t_{n1}&...&t_{nn}\end{pmatrix}$ the $k$-th column of which is the column $E'_k$ of coordinates of the vector $e'_k$ in the basis $B.$ That is, $$T_{B\rightarrow B'}=\begin{pmatrix}1&1&-1\\1&-1&2\\0&0&-1\end{pmatrix}.$$

Now, using the formula $X' = (T_{B \rightarrow B'})^{-1}X,$ find the coordinates of the vector $x = i - 2j + 2k$ in the basis $B'.$ Here $(T_{B \rightarrow B'})^{-1} = \begin{pmatrix}1&1&-1\\1&-1&2\\0&0&-1\end{pmatrix}^{-1},$ $X = \begin{pmatrix}1\\-2\\2\end{pmatrix}.$

Let's find the inverse matrix $(T_{B \rightarrow B'})^{-1}:$

Let $A = T_{B \rightarrow B'}.$ Then $det A = 2;$

$A_{11}=1;$ $A_{12}=1;$ $A_{13}=0;$

$A_{21}=1;$ $A_{22}=-1;$ $A_{23}=0;$

$A_{31}=1;$ $A_{32}=-3;$ $A_{33}=-2.$

From here $A^* = \begin{pmatrix}1&1&1\\1&-1&-3\\0&0&-2\end{pmatrix};$ $A^{-1} = \frac{1}{\det A}A^* = \frac{1}{2}\begin{pmatrix}1&1&1\\1&-1&-3\\0&0&-2\end{pmatrix}.$

Substituting this result into the formula $X' = (T_{B \rightarrow B'})^{-1}X,$ we get:

$X' = \frac{1}{2}\begin{pmatrix}1&1&1\\1&-1&-3\\0&0&-2\end{pmatrix}\begin{pmatrix}1\\-2\\2\end{pmatrix} = \frac{1}{2}\begin{pmatrix}1\\-3\\-4\end{pmatrix} = \begin{pmatrix}1/2\\-3/2\\-2\end{pmatrix}.$

Answer: $T_{B \rightarrow B'} = \begin{pmatrix}1&1&-1\\1&-1&2\\0&0&-1\end{pmatrix};$ $X' = \begin{pmatrix}1/2\\-3/2\\-2\end{pmatrix}.$

4.17. Let $B=(i, j, k)$ and $B'=(i', j', k')$ be orthogonal bases in $R_3.$ Write the transition matrix $T_{B \rightarrow B'},$ and list the column of coordinates for the vector $x=i-2j+k$ in the basis $B'.$

The basis $B'$ is obtained by the permutation $i'=j,$ $j'=k,$ $k'=i.$

Solution.

From the condition, we have $e_1=i, e_2=j, e_3=k;$ $e_1'=j=(0, 1, 0),$ $e_2'=k=(0, 0, 1),$ $e_3'=i=(1, 0, 0).$

Thus, the transition matrix is $$T_{B\rightarrow B'}=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}.$$

Now, using the formula $X'=(T_{B\rightarrow B'})^{-1}X,$ let's find the coordinates of the vector $x=i-2j+k$ in the basis $B'.$ Here $(T_{B\rightarrow B'})^{-1}=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}^{-1},$ $X=\begin{pmatrix}1\\-2\\1\end{pmatrix}.$

Let's find the inverse matrix $(T_{B\rightarrow B'})^{-1}:$

Designate $A=T_{B\rightarrow B'}.$ Then $det A=1;$

$A_{11}=0;$ $A_{12}=0;$ $A_{13}=1;$

$A_{21}=1;$ $A_{22}=0;$ $A_{23}=0;$

$A_{31}=0;$ $A_{32}=1;$ $A_{33}=0.$

From this, we have $A^* = \begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix};$ $A^{-1} = \frac{1}{\det A}A^* = \begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}.$

Substituting this result into the formula $X' = (T_{B \rightarrow B'})^{-1}X,$ we obtain:

$X' = \begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix} \begin{pmatrix}1\\-2\\1\end{pmatrix} = \begin{pmatrix}-2\\1\\1\end{pmatrix}.$

Answer: $T_{B \rightarrow B'} = \begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix};$ $X' = \begin{pmatrix}-2\\1\\1\end{pmatrix}.$

Homework:

Let $B=(i, j, k)$ and $B'=(i', j', k')$ be orthogonal bases in $R_3.$ Write the transition matrix $T_{B \rightarrow B'},$ and list the column of coordinates for the vector $x=i-2j+k$ in the basis $B'.$

4.16. The basis $B'$ is obtained by reversing the direction of all three basis vectors of $B.$

Answer: $T_{B \rightarrow B'} = \begin{pmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{pmatrix};$ $X' = \begin{pmatrix}-1\\2\\-1\end{pmatrix}.$

4.18. The basis $B'$ is obtained by rotating the basis $B$ by an angle $\varphi$ around the axis $i.$

Answer: $T_{B \rightarrow B'} = \begin{pmatrix}1&0&0\\0&\cos\varphi&-\sin\varphi\\0&\sin\varphi&\cos\varphi\end{pmatrix};$ $X' = \begin{pmatrix}1\\-2\cos\varphi+\sin\varphi\\2\sin\varphi+\cos\varphi\end{pmatrix}.$