Section 4.6

Change of Basis

In Section 4.4, we saw that both of the sets

B1 = {e1, e2} =

{(10

),

(01

)}and

B2 = {v1, v2} =

{(31

),

(22

)}form bases for R2.

In particular, this indicates to us that any vector in R2 can be written as a linear combinationof the vectors in B1, or as a linear combination of the vectors in B2. For example, we consideredthe vector whose tip is represented by the point in the graph below:

In terms of the first basis B1, the coordinates for the vector p are given by

(p)B1 = (−7, 3),

since we have to move −7e1 and 3e2 to get to p from the origin:

1

Section 4.6

However, in terms of the second basis B2, the coordinates for p are given by

(p)B2 = (−5, 4);

we must move −5v1 and 4v2 from the origin to get to p:

2

Section 4.6

The point of the discussion was that there are many ways to represent the location of the vectorp; the representation depends on which basis we choose to use.

As you may have guessed, there will be occasions on which we need to change the representationof a vector’s location from one basis to another; that is we may need to rewrite the coordinates ofa vector with respect to a new basis. In terms of the example above, how can we start with thecoordinates (−7, 3) for vector p relative to the basis B1, and rewrite them as coordinates for prelative to the basis B2?

The answer to our question will require us to go back to the idea of a linear transformation. InSection 1.8, we saw that

3

Section 4.6

Definition. A linear transformation f from Rn → Rm is any function from n dimensional Euclideanspace to m dimensional Euclidean space satisfying the following properties:

(1) f(u+ v) = f(u) + f(v)

(2) f(ku) = kf(u),

where u and v are any vectors in Rn and k is a scalar.

In addition, we learned that every linear transformation from Rn to Rm has a representation asan m× n matrix.

Under basis B1, we saw that vector p has coordinates (−7, 3), and under basis B2, its coordinatesare (−5, 4).

With this in mind, we can think of a “change of basis function” TB that sends the R2 coordinatesof a vector under the first basis to the R2 coordinates of the same vector under the second basis:

(−7, 3

) TB //(−5, 4

)It turns out that the function TB that sends the coordinates of a vector with respect to one

basis to its coordinates under a different basis is a linear transformation–thus has a representationas a matrix.

With this in mind, we are now going to think about coordinates such as (−7, 3) (which means−7 of the first basis element and 3 of the second) as vectors–so (−7, 3) is the vector(

−73

)in R2.

Now what we want to do is find the linear transformation TB (alternatively, the 2× 2 matrix Brepresenting the action of TB on R2) that sends the coordinates of a vector under its representationunder the first basis

B1 =

{(10

),

(01

)}to its coordinates under the second basis

B2 =

{(31

),

(22

)}.

For example, we know that this function TB (or matrix B) should satisfy

TB

((−73

))= B ·

(−73

)=

(−54

).

We will find the matrix B in the next example, and will use our solution as a model for a generalapproach to the change of basis problem.

4

Section 4.6

Example: Finding the Change of Basis Matrix

Our goal now is to find the linear function (alternatively, 2 × 2 matrix) that rewrites coordinatesin terms of

B1 = {e1, e2} =

{(10

),

(01

)}as coordinates in terms of

B2 = {v1, v2} =

{(31

),

(22

)}.

Since we plan to start out with vectors that are written in terms of the first basis B1, let’s writeboth of the vectors in our second basis in terms of e1 and e2: it is easy to see that

v1 = 3e1 + e2

v2 = 2e1 + 2e2.

Of course, we are planning to convert vectors into our new basis (consisting of v1 and v2)–solet’s rewrite the equations above so that e1 and e2 are written in terms of v1 and v2; solving fore1 and e2 in the first pair of equations, we have

e1 = 12v1 − 1

4v2

e2 = − 12v1 + 3

4v2.

In particular, this gives us the coordinates of e1 and e2 with respect to the new basis:

(e1)B2 =

1

2

−1

4

and (e2)B2 =

−1

2

3

4

.

Remember that a basis tells us everything that we need to know about the vector space. Thusall we need to know in order to get the linear transformation B (which we will now refer to as thetransition matrix) is the way that coordinates in the old basis ({e1, e2}) relate to coordinates inthe new basis ({v1,v2}). Since B is a transformation from R2 to R2, it has matrix representationas a 2× 2 matrix. I claim that the columns of this matrix B are made up of the coordinate vectorsof the old basis in terms of the new basis: in this case,

B =

| |(e1)B2 (e2)B2

| |

=

(12

− 12

− 14

34

).

Now we hope that B is the matrix describing the linear transformation of coordinates in R2 interms of the first basis to coordinates in R2 in terms of the second basis; if this is the case, thenwe can take any vector v whose coordinates are written in terms of the first basis, and calculateits coordinates in terms of the second basis as Bv.

Now we already have several vectors that we can test: we have seen that

5

Section 4.6

coordinates in terms of coordinates in terms offirst basis second basis

1

0

1

2

−1

4

0

1

−1

2

3

4

−7

3

−5

4

Let’s apply the transition matrix B to each of the vectors in the left-hand column; if we have

built B correctly, then we should get the corresponding vectors in the right-hand column:

B

(10

)=

(12

− 12

− 14

34

)(10

)

=

(12+ 0

− 14+ 0

)

=

(12

− 14

).

B

(01

)=

(12

− 12

− 14

34

)(01

)

=

(0− 1

2

0 + 34

)

=

(− 1

234

).

6

Section 4.6

B

(−73

)=

(12

− 12

− 14

34

)(−73

)

=

(− 7

2− 3

274+ 9

4

)

=

(− 10

2164

)

=

(−54

).

As you can see by comparing the results above with the table we built earlier, B successfullyrewrote each of the coordinates; indeed, B is the matrix that rewrites the coordinates of any vectorin R2 in terms of basis B1 as coordinates in terms of basis B2.

A General Process for Finding Transition Matrices

There will be many cases in which we wish to change bases in the coordinate representation of avector space; the following process will allow us to do so.

Key Point. LetB = {u1, u2, . . . , un}

andB′ = {u1

′, u2′, . . . , un

′}

be two different bases for a (finite-dimensional) vector space V . Suppose that vector v in V hascoordinates (v)B in terms of basis B. To calculate the coordinates (v)B′ of vector v in terms ofthe new basis B′:

1. Calculate the coordinates of the old basis vectors u1, u2, . . . , and un in terms of the newbasis vectors u1

′, u2′, . . . , and un

′; in other words, find the vectors

(u1)B′ , (u2)B′ , . . . , and (un)B′ .

2. Create the transition matrix

P =

| | |(u1)B′ (u2)B′ . . . (un)B′

| | |

whose columns are the coordinates vectors of basis elements from the old basis B in terms ofthe new basis B′.

7

Section 4.6

3. Calculate Pv, which will be the coordinates of v with respect to the new basis; i.e.,

(v)B′ = P (vB).

Example

The three-dimensional vector space U2 of 2 × 2 upper triangular matrices has two different basesgiven by

B = {e1, e2, e3},

where

e1 =

(1 00 0

), e2 =

(0 10 0

), and e3 =

(0 00 1

);

andB′ = {v1, v2, v3},

where

v1 =

(−1 −20 6

), v2 =

(−1 −30 6

), and e3 =

(1 20 −5

).

1. Find the coordinate representations of each of the six vectors above in terms of basis B.

2. Find the coordinate representations of each of the six vectors above in terms of basis B′.

3. Find the transition matrix P from B to B′.

4. Find the coordinate representation of the matrix

v =

(4 120 −5

)in terms of basis B.

5. Use the transition matrix P to calculate the coordinate representation of matrix v above interms of basis B′.

1. Find the coordinate representations of each of the six vectors above in terms of basis B:

To find the coordinate representations, we need to write each vector as a linear combinationof the vectors

e1 =

(1 00 0

), e2 =

(0 10 0

), and e3 =

(0 00 1

)

8

Section 4.6

from B. It is not too difficult to do this, especially with the first three vectors:

e1 =

(1 00 0

)= e1 + 0e2 + 0e3

e2 =

(0 10 0

)= 0e1 + e2 + 0e3

e3 =

(0 00 1

)= 0e1 + 0e2 + e3

v1 =

(−1 −20 6

)= −e1 − 2e2 + 6e3

v2 =

(−1 −30 6

)= −e1 − 3e2 + 6e3

v3 =

(1 20 −5

)= e1 + 2e2 − 5e3

Recall that the coordinates for a vector are just the coefficients of the basis elements in itsrepresentation as a linear combination; so we have coordinates

(e1)B =

100

(e2)B =

010

(e3)B =

001

(v1)B =

−1−26

(v2)B =

−1−36

(v3)B =

12−5

.

2. Find the coordinate representations of each of the six vectors above in terms of basis B′:

As with part 1, we need to write each vector as a linear combination of the vectors

v1 =

(−1 −20 6

), v2 =

(−1 −30 6

), and e3 =

(1 20 −5

)from B′.

9

Section 4.6

Now this is quite simple to do for v1, v2, and v3–you should check for yourself that

(v1)B′ =

100

, (v2)B′ =

010

, and (v3)B′ =

001

.

The easiest way to write each of e1, e2, and e3 as a linear combinations of v1, v2, and v3 isto start with the “opposite” linear combinations

v1 = −e1 − 2e2 + 6e3

v2 = −e1 − 3e2 + 6e3

v3 = e1 + 2e2 − 5e3

and solve for e1, e2, and e3.

Let’s start by adding the equations for v1 and v3:

v1 + v3 = −e1 − 2e2 + 6e3 + e1 + 2e2 − 5e3

= e3.

So we know that the linear combination for e3 is

e3 = v1 + v3.

Adding the equations for v2 and v3, and using the fact that e3 = v1 + v3, we have

v2 + v3 = −e1 − 3e2 + 6e3 + e1 + 2e2 − 5e3

= −e2 + e3

= −e2 + v1 + v3,

so thate2 = v1 − v2.

Finally, let’s replace e2 and e3 in the equation v3 = e1 + 2e2 − 5e3 with their linear combi-nations e2 = v1 − v2 and e3 = v1 + v3:

v3 = e1 + 2e2 − 5e3

= e1 + 2(v1 − v2)− 5(v1 + v3)

= e1 + 2v1 − 2v2 − 5v1 − 5v3

= e1 − 2v2 − 3v1 − 5v3

10

Section 4.6

so thate1 = 3v1 + 2v2 + 6v3.

With the linear combinations in place, we now see that the desired coordinate vectors are

(e1)B′ =

326

, (e2)B′ =

1−10

, and (e3)B′ =

101

.

3. Find the transition matrix P from B to B′:

The transition matrix P should be 3 × 3 (since U2 is a three dimensional space), and itscolumns should be the coordinate vectors of the elements of first basis B with respect to thesecond basis B′. Of course, we’ve already found these coordinate vectors:

(e1)B′ =

326

, (e2)B′ =

1−10

, and (e3)B′ =

101

.

So our transition matrix P is given by

P =

3 1 12 −1 06 0 1

.

4. Find the coordinate representation of the matrix

v =

(4 120 −5

)in terms of basis B:

Given that the basis B consists of matrices

e1 =

(1 00 0

), e2 =

(0 10 0

), and e3 =

(0 00 1

),

it is easy to see that the linear combination for v in terms of this basis is

v = 4e1 + 12e2 − 5e3,

so that the coordinate vector is

(v)B =

412−5

.

11

Section 4.6

5. Use the transition matrix P to calculate the coordinate representation of matrix v above interms of basis B′:

To get the coordinate matrix of v with respect to B′, all we need to do is calculate P (v)B:

P (v)B =

3 1 12 −1 06 0 1

412−5

=

12 + 12− 58− 12 + 024 + 0− 5

=

19−419

.

So the coordinate representation of the matrix

v =

(4 120 −5

)in terms of basis B′ is given by

(v)B′ =

19−419

.

We should check that these coordinates are correct; if they are, then

v = 19v1 − 4v2 + 19v3.

Let’s make the calculation:

19v1 − 4v2 + 19v3 = 19

(−1 −20 6

)− 4

(−1 −30 6

)+ 19

(1 20 −5

)

=

(−19 −380 6 · 19

)+

(4 120 −24

)+

(19 380 −5 · 19

)

=

(−19 + 4 + 19 −38 + 12 + 38

0 6 · 19− 24− 5 · 19

)

=

(4 120 −5

)= v.

12

Section 4.6

An Alternate Method for Computing Transition Matrices in Rn

If the vector space of interest is Rn, there is a more efficient way to find the transition matrix Pfrom an old basis B to the new one B′. Before we discuss it, we need a quick theorem:

Theorem 4.6.1. If P is the transition matrix from a basis B′ to a basis B, then P is invertibleand P−1 is the transition matrix from B to B′.

The theorem simply says that, if we know the transition matrix P for a change of basis, thenthe basis change in the opposite direction has transition matrix P−1:

(v)B′

P''

(v)BP−1gg

Key Point. Let B = {u1, u2, . . . , un} and B′ = {u1′, u2

′, . . . , un′} be two bases for Rn. The

following process will yield the transition matrix P from B to B′:

1. Form the matrix

(B | B′) =

| | | | | |u1 u2 . . . un u1

′ u2′ . . . un

′

| | | | | |

.

2. Use elementary row operations to reduce the first n columns of the matrix in step 1 to I.

3. The remaining matrix (I | P

)has P as the transition matrix from B to B′.

Example

Use the theorem to find the transition matrix from the basis

B = {(20

),

(−16

)}

for R2 to basis

B′ = {(41

),

(−11

)}.

We begin the process by building the matrix whose columns are made up of the vectors fromB, followed by the vectors from B′:

(B | B′) = (

2 −1 | 4 −10 6 | 1 1

).

13

Section 4.6

Next, we need to reduce the first half of the matrix to I using elementary row operations:(2 −1 | 4 −10 6 | 1 1

)→

(1 − 1

2| 2 − 1

2

0 6 | 1 1

)

→(1 − 1

2| 2 − 1

2

0 1 | 16

16

)

→(1 0 | 2 + 1

12− 1

2+ 1

12

0 1 | 16

16

)

→(1 0 | 25

12− 5

12

0 1 | 16

16

).

Thus the desired transition matrix is given by

P =

25

12− 5

12

1

6

1

6

14