Change of basis for coordinates
The standard bases for some of our examples are very useful, because as
we have seen, it is very easy to find coordinates of a vector, often “by
inspection”, i.e., without calculations.
Of course, we can always do this in two steps:
2 3
c1
6 7
1. Given the B-coordinates [v ]B = 4 ... 5, we can calculate v directly:
But often it is necessary or desirable to use a di↵erent basis, one that is
more suited to a particular problem.
So it will be useful to be able to convert coordinates of a vector from one
basis to another.
cn
v = c 1 b1 + · · · + c n bn
Problem
0
2. Then find the B coordinates of v.
Given a vector space V , a vector v, and two di↵erent bases, B and B 0 , of
V , if we know the B-coordinates of v, namely [v]B , how can we
(efficiently) calculate the B 0 coordinates, [v]B 0 :
change-of-basis
! [v]B 0
[v]B
1
2

⇢ 
2
1
0 1
,
and B 0 =
,
.
1
2
1 1
(You can verify that these are bases.) Now suppose we know that the
3
B-coordinates of some vector v are [v]B =
. Then
4



2
1
10
1. v = 3b1 4b2 = 3
4
=
.
1
2
5
For example, let V = R2 , B =
⇢
For another example, let V = P2 , B = { 1 2x + x 2 , x + x 2 , 1 x 2 },
and B 0 = { 1 + x, x, 1 + x + x 2 }. You can easily check that these are
both bases of V . Now suppose we are told that p is a vector whose
B-coordinates are
2 3
3
[p]B = 4 2 5,
1
and are asked to find the B 0 -coordinates. First, we calculate p:
2. Now we must find the coordinates of v with respect to B 0 . If
(v)B 0 = (d1 , d2 ), we set v = d1 b01 + d2 b02 :



10
0
1
= d1
+ d2
5
1
1
p = 3b1 + 2b2
=2
15 and d2 = 10. So

15
=
.
10
3
2x + x 2 ) + 2(x + x 2 )
4x + 6x
(1
x 2)
2
Next, we need to find the B 0 -coordinates of p, so we set
p = c1 b01 + c2 b02 + c3 b03 , or
This is easily solved to give d1 =
[v]B 0
1b3 = 3(1
2
4x + 6x 2 = c1 (1 + x) + c2 (x) + c3 (1 + x + x 2 )
and solve for c1 , c2 , c3 .
4
This two-step procedure always works, but if we have a lot of vectors to
convert, we would have to set up and solve a set of equations for each
one. Next, we will develop a method that, once set up, will convert
coordinates for any vector with just a matrix multiplication.
This leads to
2
1
41
0
So we have c1 =
0
1
0
1 |
1 |
1 |
4, c2 =
3
2
2
1
45 · · ! 40
6
0
6, c3 = 6, and
2 3
4
[p]B 0 = 4 65
6
0
1
0
0 |
0 |
1 |
3
4
65.
6
Let V be any space, with bases B and B 0 . For v 2 V , let
2 3
c1
6 7
[v]B = 4 ... 5
which means that
v = c 1 b1 + · · · + c n b n .
cn
Now remember that the coordinate isomorphism preserves linear
combinations, so, switching to B 0 -coordinates, we have
[v]B 0 = [c1 b1 + · · · + cn bn ]B 0 = c1 [b1 ]B 0 + · · · + cn [bn ]B 0 .
5
6
Theorem (Change of basis formula)
[v]B 0 = P[v]B ,
This linear combination can be expressed in matrix form. Let P be the
n ⇥ n matrix whose j-th column is [bj ]B 0 . Then
2 3
c1
h
i6 c 2 7
6
7
P[v]B = [b1 ]B 0 [b2 ]B 0 · · · [bn ]B 0 6 . 7 = c1 [b1 ]B 0 + · · · + cn [bn ]B 0 .
4 .. 5
cn
where P is the transition matrix from B to B 0 :
h
i
P = [b1 ]B 0 [b2 ]B 0 · · · [bn ]B 0
whose j-th column is [bj ]B 0 .
For clarity, we will use the notation
But this is just [v]B 0 . Therefore we have proved the following theorem.
P = PB!B 0
to emphasize that the transition is from B to B 0 . So
h
i
PB!B 0 = [b1 ]B 0 [b2 ]B 0 · · · [bn ]B 0
7
8
In order to compute the transition matrix PB!B 0 , we need to compute
B 0 -coordinates for all the basis vectors in B, but once this is done, the
coordinates of any vector can be found by matrix multiplication.
⇢ 
⇢ 
2
1
0 1
In the first example: V = R2 , B =
,
B0 =
,
, we
1
2
1 1
need to find [b1 ]B 0 and [b2 ]B 0 , so we will row-reduce two augmented
matrices:



0 1 | 2 R1 $R2 1 1 | 1 R1 R2 1 0 |
1
[B 0 |b1 ] =
!
!
1 1 | 1
0 1 | 2
0 1 | 2
[B 0 |b2 ] =
So [b1 ]B 0


0 1 |
1 R1 $R2 1 1
!
1 1 | 2
0 1


1
3
=
, [b2 ]B 0 =
, and
2
1
|
|
2
1
R1 R2
!
h
i  1
P = PB!B 0 = [b1 ]B 0 [b2 ]B 0 =
2

|
|
1 0
0 1
In the example earlier, we were given [v]B =
find [v]B 0 using P:
[v]B 0 = PB!B 0 [v]B =
3
1
1
2
3
1

3
=
4
3
. Now we can easily
4

3 12
=
6+4

15
,
10
as before.
3
.
1
9
10
Notice that when finding P, we did


0 1 | 2 R1 $R2 1
[B 0 |b1 ] =
!
1 1 | 1
0

0
[B |b2 ] =
1
0
1
1
|
|
1
2

1
!
0
R1 $R2
1
1
|
|
1
1
|
|
1
2
R 1 R2
!
2
1

1 0
0 1

1 0
!
0 1
R1 R2
|
|
There are a few theorems that can help simplify finding transition
matrices, as well as being of theoretical importance later. The first
concerns a change of basis followed by a second change. These two
changes can be rolled together into a single one:
1
2
|
|
Theorem
3
1
Let B, B 0 , and B 00 be bases of a vector space V . Then
(PB 0 !B 00 )(PB!B 0 ) = PB!B 00
and both systems have the same left-hand parts, i.e., the same coefficient
matrices, and so the row operations involved are the same. Therefore we
can use the technique mentioned in Section 1.6 of the text (“Linear
Systems with a Common Coefficient Matrix”, p. 61, and Example 2
following). So we apply elimination to the “doubly augmented matrix”:



0 1 | 2
1 R1 $R2 1 1 | 1 2 R1 R2 1 0 | 1 3
!
!
,
1 1 | 1 2
0 1 | 2
1
0 1 | 2
1
and immediately get PB!B 0 =
11



1
2
Proof.
Recall that the j-th column of a product AB is A times the j-th column
of B (see text, page 31). So the j-th column of P = (PB 0 !B 00 )(PB!B 0 ) is
PB 0 !B 00 [bj ]B 0
3
.
1
But this is the formula for the B 00 coordinates of bj . So the j-column of
P is [bj ]B 00 , which is precisely the j-th column of PB!B 00 .
12
In the case where the two bases B and B 0 are the same, then “change of
basis” is no change at all. And the matrix that does this is the identity
matrix.So we have
Finally, by combining these two theorems we get the following.
Theorem
Theorem
PB!B = I
For any bases B and B 0 ,
PB 0 !B = (PB!B 0 )
Proof.
The j-th column of PB!B is [bj ]B . But clearly
bj = 0b12+3· · · + 1bj + · · · + 0bn , which is equivalent to
0
6 .. 7
6.7
6 7
7
[bj ]B = 6
617 = ej . And ej is the j-th column of the identity matrix.
6 .. 7
4.5
0
Since PB!B and I have the same j-th column for every j, they are equal.
13
Proof.
(PB 0 !B )(PB!B 0 ) = PB!B = I
14
For “standard” bases, finding coordinates is easy, so finding a transition
to a standard basis is also easy:
x + x 2 , 1 + 2x
Example: Let V = P2 , B = { 2
B 0 = { 1, x, x 2 }. Find PB!B 0
x 2, 1 + x
Then to get the reverse transition matrix PB 0 !B , by the previous
theorem, we only need to find the inverse.
x 2 } and
We only need to read o↵ the coefficients of the polynomials in B to get
the columns of P:
2
3
2
1
1
15
PB!B 0 = 4 1 2
1
1
1
15
PB 0 !B = (PB!B 0 )
1
2
2
=4 1
1
1
2
1
3
1
15
1
1
2
1/3
= ··· = 4 0
1/3
0
1
1
3
1/3
1 5
5/3
16
We can also use these two theorems together, when neither basis is
standard.
⇢ 
⇢ 
2
1
0
1
Consider again V = R2 with B =
,
B0 =
,
.
1
2
1
1
Then if S is the standard matrix of V , we easily get


2
1
0
PB!S =
and PB 0 !S =
1 2
1
This calculation can be simplified by the following observation.
Theorem
Let A be an invertible n ⇥ n matrix and let B be any n ⇥ k matrix. Then
A is row-equivalent to the identity matrix. Suppose, by elementary row
operations, that
[ A | B ] · · ! [ I | C ].
1
1
Then by the “repeated change” theorem:
Then C = A
PB!B 0 = PS!B 0 PB!S
But we also have PS!B 0 = (PB 0 !S )
PB!B 0 = (PB 0 !S )
17
1
1
1
PB!S =
0
1
1
1
B.
Proof.
, so

1
1
2
1
This is because the “multisystem” AX = B is equivalent to IX = C . But
the solution to AX = B is X = A 1 B, so C = X = A 1 B.
1
2
18
We can apply this to finding a transition matrix by way of a standard
basis, as in the previous example.
Applying this method to the previous example, we had


2
1
0 1
PB!S =
and PB 0 !S =
1 2
1 1
“An Efficient Method . . . p. 220”
Let B and B 0 be bases of V , and let S be the “standard” basis of V .
Then
PB!B 0 = (PS!B 0 )(PB!S ) = (PB 0 !S ) 1 (PB!S )
So we reduce
⇥
So to calculate this, (1) form the matrix
⇥
⇤
PB 0 !S | PB!S .

⇤
0
PB 0 !S | PB!S =
1
1 | 2
1 | 1
And
Then (2) convert this to reduced row-echelon form. The result will be
⇥
⇤
⇥
PB 0 !S | PB!S · · ! I | PB!B 0 ].
PB!B 0 =

1
2
1
2
··!

1
0
0 |
1 |
1
2
3
1
3
.
1
(3) Extract the right-hand side.
19
20
In our second example above, V = P2 , B = { 1 2x + x 2 , x + x 2 , 1 x 2 },
and B 0 = { 1 + x, x, 1 + x + x 2 }. Then if S = { 1, x, x 2 }, we have
2
3
2
3
1 0 1
1 0 1
PB!S = 4 2 1 0 5, PB 0 !S = 41 1 15,
1 1
1
0 0 1
2 3
3
And we can use this to find [p]B 0 where we were given [p]B = 4 2 5:
1
So we reduce
⇥
PB 0 !S
2
1
⇤
| PB!S =41
0
2
1
·· !40
0
2
0
And we have PB!B 0 = 4 3
1
1
1
1
0
1
0
0
1
0
1 |
1 |
1 |
0 |
0 |
1 |
1
2
1
0
1
1
0
3
1
1
1
1
3
1
05
1
[p]B 0
3
2
15
1
3
2
15 .
1
2
0
= PB!B 0 [p]B = 4 3
1
in agreement with our earlier answer.
21
1
1
1
32 3 2 3
2
3
4
15 4 2 5 = 4 65 ,
1
1
6
22
Here is another example, from the text (#16, p. 224).
(b) If w = ( 5, 8, 5), find [w]B . Then use (a) to find [w]B 0 .
Let V = R3 , B = { ( 3, 0, 3), ( 3, 2, 1), (1, 6, 1) }, and
B 0 = { ( 6, 6, 0), ( 2, 6, 4), ( 2, 3, 7) }.
To express w as a linear combination of B, we reduce [B
2
3
2
3
3 1 | 5
1 0 0 |
40
2
6 | 8 5 · · ! 40 1 0 |
3
1
1 | 5
0 0 1 |
(a) Find PB!B 0 .
Using the standard basis S, we form [ PB 0 !S | PB!S ] and reduce:
2
6
4 6
0
So
23
2
6
4
2 |
3 |
7 |
3
0
3
3
2
1
3
2
1
1
5
6 ·· ! 40
1
0
2
3/4
PB!B 0 = 4-3/4
0
3/4
-17/12
2/3
0
1
0
0 |
0 |
1 |
3
3/4
3/4
-3/4
-17/12
0
2/3
1/12
2 3
1
So [w]B = 415.
1
3
-17/125
2/3
(Check:
1/12
-17/125
2/3
24
1b01
+
1b02
+
1b03
2
| w]:
3
1
15
1
3 2 3 2 3 2 3
3
3
1
5
= 4 0 5 + 4 2 5 + 4 6 5 = 4 8 5 = w.)
3
1
1
5
Now to find [w]B 0 , we have
[w]B 0
2
3/4
= PB!B 0 [w]B = 4-3/4
0
3/4
-17/12
2/3
(c) Check by computing [w]B2 directly.
32 3
1/12
1
-17/125415
2/3
1
2
3
19/12
-43
4
/125
=
4/3
Comment: The instructions said to check by “computing [w]B2 directly”.
But there is an easier way. Check by using the definition of coordinates,
i.e., show that
19 0
43 0
4
b +
b + b0 = w
12 1
12 2 3 3
We can do this the same way we found [w]B1 :
Do this.
[PB2 !S
25
2
6
| [w]S ] = 4 6
0
2
6
4
2 |
3 |
7 |
3
2
5
1
5
8
· · ! 40
5
0
0 0 |
1 0 |
0 1 |
19/12
3
-43/125
4/3
26