Basis transition

Linear maps: Matrices of Linear Maps

Basis transition

Let $V$ be an $n$ -dimensional vector space. Previously we discussed the coordinatization of $V$ using a basis. Now we look at the relationship between coordinatizations using two bases:

$\alpha=\basis{\vec{a}_1,\ldots ,\vec{a}_n} \phantom{xx}\hbox{ and }\phantom{xx}\beta =\basis{\vec{b}_1,\ldots ,\vec{b}_n}$ With each vector $\vec{x}\in V$ now correspond two sets of coordinates: $\alpha (\vec{x})$ with respect to the basis $\alpha$ and $\beta(\vec{x})$ with respect to the basis $\beta$ .

$\begin{array}{ccccccc} &&&V&&&\\ &\alpha&\swarrow&&\searrow&\beta&\\ \mathbb{R}^n&&&&&&\mathbb{R}^n\end{array}$ It is now clear what the relationship is between the $\alpha$ -coordinates of $\vec{x}$ and the $\beta$ -coordinates of $\vec{x}$ : we begin with the row of $\alpha$ -coordinates, apply the map $\alpha^{-1}$ to arrive at $\vec{x}\in V$ , and then apply the map $\beta$ in order to obtain the corresponding $\beta(\vec{x})$ .

Let $\alpha$ and $\beta$ be two bases for an $n$ -dimensional vector space $V$ . Then the linear map

$\beta\,\alpha^{-1}:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is called the coordinate transformation from $\alpha$ to $\beta$ .

Consider the bases $\alpha=\basis{1,x,x^2}$ and $\beta=\basis{x^2,x,1}$ for the vector space of all polynomials in $x$ of degree at most two. Then

$\begin{array}{rcl}\alpha(a_0+a_1x+a_2x^2) &=&\rv{a_0,a_1,a_2}\\ \beta(a_0+a_1x+a_2x^2) &=&\rv{a_2,a_1,a_0}\end{array}$

Clearly, the coordinate transformation from $\alpha$ to $\beta$ consists of interchanging the first and third coordinate. This is indeed the linear map $\beta\alpha^{-1}$ :

$\beta\alpha^{-1}\rv{v_1,v_2,v_3} = \beta\rv{v_1+v_2x+v_3x^2} = \rv{v_3,v_2,v_1}$

The coordinate transformation can be described by a matrix:

Let $\alpha$ and $\beta$ be bases for an $n$ -dimensional vector space $V$ and let

${}_\beta I_\alpha=\left(\beta\,\alpha^{-1}\right)_\varepsilon$ be the matrix of the linear map $\beta\,\alpha^{-1}$ .

If $\vec{x}$ is the $\alpha$ -coordinate vector of a vector $\vec{v}$ in $V$ , then the $\beta$ -coordinate vector of $\vec{v}$ is equal to ${}_\beta I_\alpha \,\vec{x}$ .

The matrix ${}_\beta I_\alpha$ is called the transition matrix of basis $\alpha$ to basis $\beta$ .

The coordinate transformation is a linear map from $\mathbb{R}^n$ to itself, so according to theorem Linear maps in coordinate spaces defined by matrices it is determined by an $(n\times n)$ -matrix ${}_\beta I_\alpha$ whose columns are the images of $\vec{e}_1,\ldots,\vec{e}_n$ under $\beta\,\alpha^{-1}$ . We have

$(\beta\alpha^{-1})(\vec{e}_i)=\beta(\alpha^{-1}(\vec{e}_i))=\beta(\vec{a}_i)$ The columns of this matrix are thus the $\beta$ -coordinate vectors of the $\alpha$ -basis vectors. By construction, the map $\beta\,\alpha^{-1}$ transforms $\alpha$ -coordinates into $\beta$ -coordinates. Such a conversion can be calculated by multiplying by the matrix ${}_\beta I_\alpha$ .

Two properties

Of course, we can also convert $\beta$ -coordinates into $\alpha$ -coordinates. This can be done with the matrix ${}_\alpha I_\beta$ . The product matrix ${}_\alpha I_\beta\ {}_\beta I_\alpha$ first transforms the $\alpha$ -coordinates of a vector $\vec{x}$ into the $\beta$ -coordinates of $\vec{x}$ and then transforms those back into the $\alpha$ -coordinates of $\vec{x}$ . So ${}_\alpha I_\beta\ {}_\beta I_\alpha = I$ , giving

${}_\alpha I_\beta= {}_\beta I_\alpha^{-1}$ If there is a third basis $\gamma$ in the game, we can convert $\alpha$ -coordinates into $\beta$ -coordinates and these into $\gamma$ -coordinates. Of course, the result is that we convert $\alpha$ -coordinates into $\gamma$ -coordinates, so

${}_\gamma I_\beta \, {}_\beta I_\alpha = {}_\gamma I_\alpha$ These two equalities also follow by determining compositions of maps:

$\begin{array}{rcl} \alpha\beta^{-1} &=& \left(\beta\alpha^{-1}\right)^{-1}\\ \left(\gamma \beta^{-1}\right) \,\left( \beta\alpha^{-1}\right) &=&\gamma \alpha^{-1}\end{array}$

Coordinate space

It is important to distinguish between calculations at the level of vectors, that is, with elements of the vector space $V$ , and calculations at the level of coordinates, that is, with rows of numbers. This difference is clear in the example below: the vectors are polynomials, while the coordinates are rows of numbers.

This distinction is more difficult when the vector space is $\mathbb{R}^n$ or $\mathbb{C}^n$ . For, the row of numbers representing a vector from $\mathbb{R}^n$ or $\mathbb{C}^n$ can also be understood as a row of coordinates with respect to the standard basis $\varepsilon=\basis{\vec{e}_1,\ldots,\vec{e}_n}$ :

$\rv{1,2,3}=1\vec{e}_1+2\vec{e}_2+3\vec{e}_3$

Consider the vector space $P_2$ of real polynomials of degree at most $2$ with bases

$\alpha=\basis{1,x,x^2}\phantom{xx}\text{ and }\phantom{xx} \beta=\basis{x-1,x^2-1,x^2+1}$ Determine the transition matrix ${}_\beta I_\alpha$ of the basis $\alpha$ to the basis $\beta$ and calculate the $\beta$ -coordinate vector of

$2 x^2-3 x+4$

transition matrix ${}_\beta I_\alpha=\matrix{0 & 1 & 0 \\ -{{1}\over{2}} & -{{1}\over{2}} & {{1}\over{2}} \\ {{1}\over{2}} & {{1}\over{2}} & {{1}\over{2}} \\ }$
$\beta$ -coordinate vector: $\left[ -3 , {{1}\over{2}} , {{3}\over{2}} \right]$

We can easily express the basis vectors of $\beta$ as linear combinations of the basis vectors of $\alpha$ :

$\begin{array}{rrr} x-1= & -1\cdot 1+1\cdot x+0\cdot x^2\\ x^2-1= & -1\cdot 1+0\cdot x+1\cdot x^2\\ x^2+1= & 1\cdot 1+0\cdot x+1\cdot x^2 & \end{array}$ Hence, we know the $\alpha$ -coordinates of the vectors of $\beta$ and therefore the transition matrix of $\beta$ to $\alpha$ :

${}_\alpha I_\beta = \matrix{-1 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \\ }$ The transition matrix ${}_\beta I_\alpha$ is the inverse of this matrix. We find

${}_\beta I_\alpha = \matrix{0 & 1 & 0 \\ -{{1}\over{2}} & -{{1}\over{2}} & {{1}\over{2}} \\ {{1}\over{2}} & {{1}\over{2}} & {{1}\over{2}} \\ }$ How do we determine the $\beta$ -coordinates of the vector $2 x^2-3 x+4$ ? The $\alpha$ -coordinates of this vector are $\rv{4,-3,2}$ . We can convert these into $\beta$ coordinates using the matrix ${}_\beta I_\alpha$ :

${}_\beta I_\alpha\ \left(\,\begin{array}{r} 4\\ -3\\ 2 \end{array}\,\right) = \frac{1}{2}\left(\,\begin{array}{rrr} 0 & 2 & 0\\ -1 & -1 & 1\\ 1 & 1 & 1 \end{array}\,\right) \left(\,\begin{array}{r} 4\\ -3\\ 2 \end{array}\,\right)\ =\matrix{-3 \\ {{1}\over{2}} \\ {{3}\over{2}} \\ }$

We verify the results:

The first column of ${}_\beta I_\alpha$ should contain the $\beta$ -coordinates of the first basis vector of $\alpha$ . Those $\beta$ -coordinates are

$\rv{0,-\frac12,\frac12}$ and these correspond to the vector

$0\, (x-1)-\frac12 (x^2-1)+\frac12(x^2+1)=1$ The first basis vector $\alpha$ is indeed equal to $1$ . Verify for yourself that the second column contains the $\beta$ -coordinates of $x$ and the third column those of $x^2$ .

Finally, we show that $\left[ -3 , {{1}\over{2}} , {{3}\over{2}} \right]$ is indeed the $\beta$ -coordinate vector of $2 x^2-3 x+4$ :

$-3(x-1)+\frac12(x^2-1)+\frac32 (x^2+1)=2x^2-3x+4$

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