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})#.

Coordinate transformation
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#.

ExampleConsider 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:

Matrix of a coordinate transformation
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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