Change of Basis 2: Matrices
In this worksheet we will calculate what matrices look like in various bases. Whereas we need only one basis to write down a vector, we need two bases to write down a matrix. Loosely speaking we need a basis for the rows and a basis for the columns.
We will explain this more exactly in a minute.
For our calculations we will use the space of 3-D vectors R3, with the bases U, B, and C as follows.
U , the usual basis B =
1 0 2
,
2 1 0
,
0 1 2
C =
1 1 1
,
1 1 2
,
1 2 2
Remember that if I don’t specify a basis for a vector then the convention is that I am using the usual basis.
Last time we worked out the change of basis matrices for going back and forth between these different bases. Hopefully everyone ended up with the following answers.
1 2 0
0 1 1
2 0 2
U B
1 1 1
1 1 2
1 2 2
U C
1 6
2 −4 2
2 2 −1
−2 4 1
BU
2 0 −1
0 −1 1
−1 −1 0
CU
0 4 −2
2 −1 1
−1 −1 1
CB
1 6
0 2 −2
3 2 4
3 4 8
BC
What is a matrix? Sure it is a box full of numbers, but that isn’t a very useful answer.
It doesn’t give us any idea of what those numbers are supposed to mean or how they are related to the basis of our vector space.
What matrices do — what we really use them for (at least in this course), is to multiply column vectors and transform them into other column vectors. Just as a column vector is a way of describing a vectorvin terms of some basis, so too a matrix is just a way of describing a linear transformation. Remember that if we know what a linear transformation does to a basis, we know everything there is to know about it. Which basis? Any basis.
So before we do any more theory, let me tell you about a couple of linear transfor- mations and get you to do some calculations.
We define some linear transformations as follows.
1. T is the linear transformation that takes the basis elements {e1,e2,e3} of U respectively to the vectors
1 0 1
,
2 0
−1
,
0 1 0
2. S is the linear transformation that takes the respective basis elements of U to the vectors
0 1 1
C ,
0 1 0
C ,
1 0 0
C
3. R is the linear transformation that takes the respective basis elements of B to the vectors
0 0 6
U ,
0 6 0
U ,
6 0 0
U
4. M is the linear transformation that takes the respective basis elements ofB to the vectors
1 1 1
C ,
1 1 0
C ,
1 0 0
C
OK. Now you know everything there is to know about these linear transformations, at least in theory. So you should be able to work out the following.
Exercise 4: In each of the following cases work out what the linear transformation does to the given vector and write your answer in the specified basis.
a) T acting on
3
−1
−1
in basisU b) M acting on
0
−1 1
B
in basisC
c) S acting on
1 0 0
B
in basisC
d) R acting on
1 2
−3
B
in basisC
e) T acting on
1
−1 0
B
in basisC
f) M acting on
1 0
−1
in basisU
Once again the more observant among you will have noticed that I did not give you a particular method to solve these problems. This is so that you will have to think about what a linear transformation actually is. It is much too easy when you are studying linear algebra to just learn some routine for shuffling numbers around in boxes without really understanding what it is you are doing and why.
You will have found that some of the answers above were very easy to find. Others were not so easy. In parts a) and b), the calculation was fairly straightforward since all the bases matched up. In the other parts you will have found that it was neccessary to do some basis changing. This should not have been too difficult since all the change of basis matrices have already been worked out.
Let us look at part b) in more detail. The way that M is defined makes it easy to work out what M does to a vector if the vector is in base B and the answer is to be
written in base C. In particular if v=
a b c
B
=av1+bv2+cv3 where B={v1,v2,v3}, then
M(v) =aM(v1) +bM(v2) +cM(v3)
=a
1 1 1
C +b
1 1 0
C +c
1 0 0
C
=
1.a+ 1.b+ 1.c 1.a+ 1.b+ 0.c 1.a+ 0.b+ 0.c
C
so this problem is as easy as multiplying by a matrix. This matrix, whose columns are the images of respective elements ofBexpressed in the basisC is called thematrix of M fromB toC. When we multiply a column vector forvin basisBby this matrix, we obtain a column vector forM(v) in basisC. To remind us that this matrix should multiply a vector in baseB, and gives an answer which should be interpreted in base C, we write this matrix as
M =
1 1 1 1 1 0 1 0 0
CB
The general rule, when multiplying matrices, is that the bases which are written closest to one another must match up for the multiplication to make sense. If the bases don’t match up, you should change them until they do. We will see how to do this shortly. Now we can write
M(v) =
1 1 1 1 1 0 1 0 0
CB
a b c
B
Exercise 5: It should be possible to write down a matrix forM in any pair of bases.
Write down a matrix for M from C to U. You will have to work out what M does to the basis vectors inC, and express these in terms of basisU. These will be the columns of your matrix.
What you have just done in this exercise is change the bases for the matrix ofM. All we have to do now is tidy things up a bit.
Rather than working out the matrix for M in the new bases directly, we can get it from the old matrix for M using change of basis matrices. The matrix you were looking for in exercise 2, takes a vector v expressed in basis C, and computes the vector Mvin terms of the basis U. So suppose that
v=
a b c
C
In order to figure out what M does tov we first change to basisB by multiplying by the appropriate change of basis matrix.
v=
a b c
C
= 1 6
0 2 −2
3 2 4
3 4 8
BC
a b c
C
Now we have a vector in basisB, so we can figure out whatM does to it by multiplying by the matrix we worked out before.
Mv=
1 1 1
1 1 0
1 0 0
CB 1 6
0 2 −2
3 2 4
3 4 8
BC
a b c
C
This is great, but our answer is in base C, and we wanted it in base U. No problems!
Just multiply once again by the appropriate change of basis matrix to obtain M(v) =
1 1 1 1 1 2 1 2 2
U C
1 1 1 1 1 0 1 0 0
CB 1 6
0 2 −2
3 2 4
3 4 8
BC
a b c
C The product of the three matrices above thus does exactly what we wanted. So we can identify this product of three matrices as the matrix ofM fromC toU. In general to change a matrix from one set of bases to another, just whack the appropriate change of basis matrices on either side and multiply.
Exercise 6: Verify that this product of three matrices is the same as matrix you worked out in exercise five.
Exercise 7: Let’s slap a few bases around just to make sure we have it sussed. Write matrices for the following linear transformations in terms of the specified bases.
1. S as a matrix from U toU. 2. R as a matrix from B toB.
3. T as a matrix from C toC.
4. 1 as a matrix from B toU.
The last problem is particularly interesting. We see that our change of basis matrices are nothing more nor less than the identity transformation expressed from one basis to another. This explains why we can just stick change of basis matrices in wherever we need them to help make the bases match up. Inserting identity transformations doesn’t change anything. For example
If T =
1 2 0
0 0 1
1 −1 0
U U
and M =
1 1 1 1 1 0 1 0 0
CB
and u=
4 0 1
C Then
T Mu =
1 2 0
0 0 1
1 −1 0
U U
1 1 1 1 1 0 1 0 0
CB
4 0 1
C
Which we cannot multiply out since the bases don’t match up. But we also know that
1 =
1 1 1
1 1 2
1 2 2
U C
= 1 6
0 2 −2
3 2 4
3 4 8
BC
so
T Mu =T1M1u =
1 2 0
0 0 1
1 −1 0
U U
1 1 1 1 1 2 1 2 2
U C
1 1 1 1 1 0 1 0 0
CB
1 6
0 2 −2
3 2 4
3 4 8
BC
4 0 1
C and now the bases all match up, so we can go ahead and multiply out to compute the answer.
Recall from episode 1 that we have two equivalent notations for writing a vector in terms of a basis. For practical work we prefer the notation
v=
a b c
B
while for theoretical work it is usually easier to use the notation [v]B =
a b c
For linear transformations we also have two equivalent notations, one for practical work and one for theoretical work. So far we have used the practical notation where the basis labels are attached to the matrix. In the theoretical notation the basis labels are attached to the linear transformation. This gives us three equivalent ways of writing down the same facts.
1. T is the linear transformation that maps as follows v1 7→a.w1+d.w2+g.w3 v2 7→b.w1+e.w2+h.w3
v3 7→c.w1+f.w2+i.w3 Where B={v1,v2,v3} and C ={w1,w2,w3}are bases.
2.
T =
a b c d e f g h i
CB 3.
[T]CB=
a b c d e f g h i
In the last notation the matrix is just a box of numbers. We cannot interpret it as a linear transformation until someone tells us what bases to use. The notation on the left should be read as “the matrix of T from basis B to basis C”.
Let us end this handout by using the theoretical notation to sum up what we have discovered so far. Each of these facts is really a theorem, and we should prove these things properly. But if you read what is being said carefully and think about what it means in terms of practical examples, they are all pretty obvious.
a)
[T(v)]C = [T]CB[v]B b)
[T ◦S]CD = [T]CB[S]BD c)
([T]BC)−1 =hT−1i
CB
d)
[T]BC = [1]BF[T]F G[1]GC
Notice that without the theoretical notation, it would be very hard to write down these important facts (try it and see). This shows why we need a theoretical notation.
