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Section 3.4 Change of Bases/Coordinates

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Subsection 3.4.1 Coordinate Systems

permalinkSuppose that we want to walk to the point (2,3) from the origin, how might we get there?

We can get from the origin to the point (2,3) by walking east 2 blocks and north 3.

2(10)+3(01)=(23)
Figure 3.4.1. Path one with the elementary basis E2.

We can get there by walking northeast(ish) 3 blocks and northwest(ish) 1.

3(11/2)+(13/2)=(23)
Figure 3.4.2. Path two with some basis B.

We can get there by walking northwest(ish) 2 blocks and northeast(ish) 4.

2(11)+4(11/4)=(23)
Figure 3.4.3. Path three with some basis D.

permalinkIn each case you get to the same place but using different paths. That is because in each case we are using different bases or coordinate systems.

permalinkIn Figure 3.4.1 we follow the vectors in the basis

E2={e1=(10),e2=(01)}.

permalinkWe say that the E2-coordinates for (2,3) are

RepE2(23)=(23)

permalinkbecause

2(10)+3(01)=(23).

permalinkIn Figure 3.4.2 we follow the vectors in the basis

B={β1=(11/2),β2=(13/2)}.

permalinkWe say that the B-coordinates for (2,3) are

RepB(23)=(31)

permalinkbecause

3(11/2)+1(13/2)=(23).

permalinkIn Figure 3.4.3 we follow the vectors in the basis

D={δ1=(11),δ2=(11/4)}.

permalinkWe say that the D-coordinates for (2,3) are

RepD(23)=(24)

permalinkbecause

2(11)+4(11/4)=(23).

permalinkIn each case the coordinates for the point (2,3) are the coefficients for a linear combination of basis vectors.

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Definition 3.4.4. B-Coordinates.

Given a basis

B={β1,β2,,βk}

the B-coordinates for a point or vector p are the coefficients b1,b2,,bk so that

p=b1β1+b2β2++bkβk

and we write

RepB(p)=[b1b2bk].

permalinkFinally, we can connect this to matrices by observing that if the columns of a matrix are the basis vectors then when we multiply that by a representation we get our point, for example

(β1 β2)RepB(23)=(111/23/2)(31)=(23).
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Subsection 3.4.2 Changing Bases

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Investigation 3.4.1. B to E2.

The elementary or standard basis is

E2={e1=(10),e2=(01)}.

Let B be the basis

B={β1=(11/2),β2=(13/2)}

as above. If we have a vector written in terms of B,

RepB(v)=(13)

and we want RepE2(v) then we multiply by the matrix

RepB,E2=(111/23/2).

So we get,

RepE2(v)=RepB,E2RepB(v)=(111/23/2)(13)=(44).

To change from any basis

B={β1,β2,,βk}

to Ek multiply by the matrix

RepB,Ek=(β1 β2  βk)

each of whose columns is an element of B.

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Investigation 3.4.2. E2 to B.

Let E2 and B be the same as before and suppose

RepE2(w)=(72)

to change from E2 to B multiply by the matrix

RepE2,B=(111/23/2)1=(3/41/21/41/2).

So we get,

RepB(w)=RepE2,BRepE2(w)=(3/41/21/41/2)(72)=(17/411/4).

To change from Ek to any basis

B={β1,β2,,βk}

multiply by the matrix

RepEk,B=(RepB,Ek)1=(β1 β2  βk)1.
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Investigation 3.4.3. B to D.

Finally, let

D={δ1=(11),δ2=(11/4)}.

and let B and E2 be as before. Suppose also that

RepB(v)=(13).

if we want to find RepD(v) then multiply by the matrix

RepB,D=RepE2,DRepB,E2=(1111/4)1(111/23/2)=(1/54/54/54/5)(111/23/2)=(1/57/56/52/5).

So that we get

RepD(v)=RepB,DRepB(v)=(1/57/56/52/5)(13)=(40).

We can verify that this is correct because

1β1+3β2=(44)

and

4δ1+0δ2=(44).

In general if

B={β1,β2,,βk}

and

D={δ1,δ2,,δk}

then

RepB,D=RepE2,DRepB,E2=(RepD,E2)1RepB,E2=(δ1 δ2  δk)1(β1 β2  βk).
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Investigation 3.4.4. Back to the Beginning.

We started above with

RepE2(23)=(23), RepB(23)=(31),  and RepD(23)=(24).

Now, we know that if we multiply the first one,

RepE2(23)=(23)

by

RepE2,B=(3/41/21/41/2)

or

RepE2,D=(1/54/54/54/5)

we get the other two. That is we have a reliable, algorithmic way to change from one basis to another in the same dimension.

permalinkSummary Diagram for Bases Changes:

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Figure 3.4.5. Changing from B to other bases.
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Subsection 3.4.3 Changing Morphisms

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Investigation 3.4.5. Change a transformation of E2 to one from B to D.

Let E2 be as bove and let

E3={e1=(100),e2=(010),e3=(001)}

Define a transformation P from R3 to R2 by

P(v)=(x,y).

From before the matrix for P is

A=(100010)

and it maps E3 to E2.

Now suppose that we want the transformation to change vectors with coordinates in the basis

B={β1=(120),β2=(031)β3=(101)}

to vectors in terms of the basis

D={δ1=(11),δ2=(34)}.

Do this by going from B to E3 to E2 to D,

(RepD,E2)1ARepB,E3=(4311)(100010)(101230011)=(4311)(101230)=(294131).

Using this we get

(xyz)(2x9y4zx+3y+z).

permalinkSummary Diagram for Transformation Changes:

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Figure 3.4.6. Changing transformations to new bases.