CHANGE OF BASIS PICTURES
LANCE D. DRAGER
1. Introduction
In these notes, I’ll try to draw some pictures to demonstrate change of basis.
The notes are best viewed in color.
2. Change of Basis
2
2.1. Change of Basis in R . To start with, let’s work in R2 so we can draw
pictures. To draw a picture, let’s start with the Euclidean plane and choose a
coordinate system, so the picture looks like Figure 2.1.
y
4
3
2
1
−4
−3
−2
−1
1
−1
2
3
4
x
−2
−3
−4
Figure 1. A Coordinate System on the Plane
If we have a point (v1 , v2 ), we can draw the arrow from the origin to (v1 , v2 ),
this arrow represents a geometric vector v (remember that any other arrow with the
same magnitude and direction also represents v). The pair of numbers (v1 , v2 ) are
called the components or coordinates of the vector v with respect to our coordinate
system.
From matrix calculations, it’s best to think of vectors in R2 as column vectors.
If we let
2
v=
,
3
1
2
LANCE D. DRAGER
the picture looks like Figure 2.
y
4
3
2
v
1
−4
−3
−2
−1
1
2
3
−1
4
x
−2
−3
−4
Figure 2. Coordinates of the Tip of v
For R2 we have the standard basis E = e1 e2 , where
1
0
e1 =
, e2 =
.
0
1
We can add the vectors e1 and e2 to our picture to get Figure 3. In terms of
matrices, we have
2
1
0
v=
=2
+3
= 2e1 + 3e2 .
3
0
1
In terms of matrix algebra, this seems trivial, but it’s not so trivial if we look at the
vector addition v = 2e1 + 3e2 in the geometric picture, Figure 4 We can summarize
the equation v = 2e1 + 3e2 as matrix multiplication
2
v = e1 e2
= E[v]E .
3
where we call
2
[v]E =
3
the coordinate vector of v with respect to the basis E.
For many purposes, it’s more convenient to use a basis different from the standard
basis. In our example, let’s introduce the basis U, where
−1
2
u1 =
, u2 =
.
1
−1
If we add these to the picture, we get Figure 5
Now, if we look at the world from the standpoint of the basis U, we are introducing new coordinate axes, where the directions are specified by the vectors u1 and
CHANGE OF BASIS PICTURES
y
3
4
3
2
v
1
e2
−4
−3
−2
−1
e1 1
2
3
−1
4
x
−2
−3
−4
Figure 3. The Standard Basis Vectors e1 and e2
y
4
3
2
v
3e2
1
e2
−4
−3
−2
−1
2e1
e1 1
2
−1
3
4
x
−2
−3
−4
Figure 4. v = 2e1 + 3e2
u2 and the scale on each axis is determined by taking multiples of the vector that
determines the axis. If we draw the picture in Figure 6, we see that the tip of v has
new coordinates. It looks like the u1 -coordinate of v is somewhat more that 2 and
the u2 -coordinate is near 1/2. Another way to look at it is that v = c1 u1 + c2 u2
where c1 is somewhat more that 2 and c2 is near 1/2, see Figure 7. Of course,
4
LANCE D. DRAGER
y
4
3
v
2
u1
1
−4
−3
−2
−1
1
2
−1
3
4
x
u2
−2
−3
−4
Figure 5. New Basis Vectors u1 and u2
3
2
−3
1
−2
u1
−1
v
u2
−1 1
2
−2
3
−3
Figure 6. The world from the Viewpoint of U.
v = c1 u1 + c2 u2 = u1
u2
c1
= U[v]U ,
c2
so
c1
= [v]U .
c2
The problem is, how do we calculate c1 and c2 , or to put it another way, how do
we calculate [v]U ?
CHANGE OF BASIS PICTURES
5
3
2
1
−3
−2
v
u1
−1
u2
−1 1
2
−2
3
−3
Figure 7. Expressing v in the Basis U
We need to go back to Figure 5 and express u1 and u2 in terms of the basis e1
and e2 . Of course we have
−1
u1 =
= (−1)e1 + 2e2
2
3
u2 =
= 3e1 + (−1)e2 .
−1
We can write this in matrix form as
u1 u2 = e1
e2
−1
2
3
−1
We introduced notation for the change of basis matrices, in the present setup we
have
U = ESEU .
Thus, in the present setup,
−1
3
SEU =
.
2 −1
To find [v]U , the coordinate vector of v with respect to U, we can use the change
of coordinates equation
[v]U = SU E [v]E .
Thus, we to know SU E , not SEU . But, we know
SU E = (SEU )−1 ,
so
SU E =
1/5
2/5
3/5
.
1/5
Since we know [v]E from above we get
1/5 3/5 2
11/5
[v]U = SU E [v]E =
=
.
2/5 1/5 3
7/5
Recall that this means
v = U[v]U = u1
11/5
7
11
u2
u1 + u2 .
=
7/5
5
5
6
LANCE D. DRAGER
Since 11/5 = 2.2 and 7/5 = 1.4, this is in accord with our estimates from Figure 7.
2.2. Change of Basis in a Space of Functions. Here is another example of a
change of basis problem. Let’s introduce the notation C(R) for the vector space of
all continuous functions on the real line.
You may not have studied how to solve this kind of differential equation yet, but
I can tell you that the general solution of the differential equation
d2 y
−x=0
dx2
(2.1)
is
(2.2)
y(x) = c1 ex + c2 e−x .
In other words, the space S of all solutions of (2.1) is the subspace of C(R) spanned
by ex and e−x . (I’m using the usual calculus abuse of notation here.) It’s easy to
check that the functions ex and e−x are independent. Since there are two of them,
and they are not the zero function, if they we dependent, we could write one as
a multiple of the other. If you look at the graphs, you can see this is impossible.
Thus, we have one basis of S, given by
U = ex e−x .
It’s often easier to use a different basis for S, namely the functions hyperbolic
cosine cosh(x) and hyperbolic sine sinh(x). You can find out more about these in
any calculus book. All we need here are the definitions
ex + e−x
2
ex − e−x
sinh(x) =
2
Since cosh(x) and sinh(x) are linear combinations of ex and e−x , they are in S. We
can write the definitions in matrix form as
1/2
1/2
cosh(x) sinh(x) = ex e−x
(2.3)
1/2 −1/2
cosh(x) =
Since the matrix
A=
1/2
1/2
1/2
,
−1/2
is invertible, we see that cosh(x) and sinh(x) also form a basis of S. If we set
V = cosh(x) sinh(x)
the V is an ordered basis of S. The change of basis matrix SU V is defined by
V = USU V
comparing this with (2.3), we see
SU V
1/2
=A=
1/2
The defining equation for SVU is
(2.4)
U = VSVU .
1/2
.
−1/2
CHANGE OF BASIS PICTURES
7
but we know that
−1
SVU = [SU V ]
1/2
=
1/2
−1 1/2
1
=
−1/2
1
1
.
−1
If we plug this into (2.4) and write things out in more detail we get
x
1
1
−x
e
e
= cosh(x) sinh(x)
.
1 −1
In other words,
ex = cosh(x) + sinh(x)
e−x = cosh(x) − sinh(x).
In differential equations, we often want to solve an initial value problem. So
consider the initial value problem
(2.6)
d2 y
−x=0
dx2
y(0) = α
(2.7)
y 0 (0) = β.
(2.5)
We might try to find the solution in terms of the basis U = ex
any solution of the differential equation can be written as
e−x . We know
y(x) = c1 ex + c2 e−x .
(2.8)
In other words,
y(x) = ex
e−x
c1
,
c2
so
[y(x)]U =
c1
.
c2
If we substitute x = 0 into (2.8) and use the first initial condition (2.6) we get
α = y(0) = c1 + c2 .
If we differentiate (2.8) we get
y 0 (x) = c1 ex − c2 ex .
Setting x = 0 is this equation and using the second initial condition (2.7) gives
β = y 0 (0) = c1 − c2 .
Thus, to find c1 and c2 we have to solve the system of equations
(2.9)
c1 + c2 = α
c1 − c2 = β.
8
LANCE D. DRAGER
That’s not hard, or course, but life is easier if we use the basis V = cosh(x)
because of the following easy to check identities
sinh(x)
cosh(0) = 1
sinh(0) = 0
d
cosh(x) = sinh(x)
dx
d
sinh(x) = cosh(x).
dx
So, let’s write our solution as a linear combination of the basis functions V, so
(2.10)
y(x) = a1 cosh(x) + a2 sinh(x).
In other words,
y(x) = cosh(x)
a1
sinh(x)
,
a2
so we have
a
[y(x)]V = 1
a2
If we plug x = 0 into (2.10) and use the first initial condition (2.6) we get
α = y(0) = a1 cosh(0) + a2 sinh(0) = a1 (1) + a2 (0) = a1 .
If we differentiate (2.10) we get
y 0 (x) = a1 sinh(x) + a2 cosh(x).
If we plug in x = 0 and use the second initial condition (2.7) we get
β = y 0 (0) = a1 sinh(0) + a2 cosh(0) = a2
Thus, our solution is
y(x) = α cosh(x) + β sinh(x),
our to put it another way,
α
[y(x)]V =
.
β
Now, if we want to find c1 and c2 in Equation (2.8), that’s the same thing as
finding [y(x)]U . We can use the basis change of coordinates equation
[y(x)]U = SU V [y(x)]V .
This gives us
c1
= [y(x)]U
c2
= SU V [y(x)]V
1/2
1/2 α
=
1/2 −1/2 β
α/2 + β/2
=
α/2 − β/2
CHANGE OF BASIS PICTURES
9
This, of course, is the same result you’ll get by solving the linear system (2.9).
Thus, we can write our solution in the basis U as
y(x) = U[y(x)]U
x
α/2 + β/2
−x
e
= e
α/2 − β/2
1
1
1
1
x
=
α+ β e +
α − β e−x .
2
2
2
2
Department of Mathematics and Statistics, Texas Tech University, Lubbock, TX
79409-1042
E-mail address: lance.drager@ttu.edu