Download Chapter 4 Isomorphism and Coordinates

Chapter 4
Isomorphism and Coordinates
Recall that a vector space isomorphism is a linear map that is both one-to-one and onto.
Such a map preserves every aspect of the vector space structure. In other words, if L : V →
W is an isomorphism, then any true statement you can say about V using abstract vector
notation, vector addition, and scalar multiplication, will transfer to a true statement
about W when L is applied to the entire statement. We make this more precise with
some examples.
Example. If L : V → W is an isomorphism, then the set {v1 , . . . , vn } is linearly independent in V if and only if the set { L(v1 ), . . . , L(vn )} is linearly independent in W . The
dimension of the subspace spanned by the first set equals the dimension of the subset
spanned by the second set. In particular, the dimension of V equals that of W .
This last statement about dimension is only one part of a more fundamental fact.
Theorem 4.0.1. Suppose V is a finite-dimensional vector space. Then V is isomorphic to W if
and only if dim V = dim W .
Proof. Suppose that V and W are isomorphic, and let L : V → W be an isomorphism.
Then L is one-to-one, so dim ker L = 0. Since L is onto, we also have dim imL = dim W .
Plugging these into the rank-nullity theorem for L shows then that dim V = dim W .
Now suppose that dim V = dim W = n, and choose bases {v1 , . . . , vn } and {w1 , . . . , wn }
for V and W , respectively. For any vector v in V , we write v = a1 v1 + · · · + an vn , and
define
L ( v ) = L ( a1 v1 + · · · + a n v n ) = a1 w 1 + · · · + a n w n .
We claim that L is linear, one-to-one, and onto. (Proof omitted.)
In particular, and 2–dimensional real vector space is necessarily isomorphic to R2 , for
example. This helps to explain why so many problems in these other spaces ended up
reducing to solving systems of equations just like those we saw in Rn .
Looking at the proof, we see that isomorphisms are constructed by sending bases to
bases. In particular, there is a different isomorphism V → W for each choice of basis for
V and for W .
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CHAPTER 4. ISOMORPHISM AND COORDINATES
One special case of this is when we look at isomorphisms V → V . Such an isomorphism is called a change of coordinates.
If S = {v1 , . . . , vn } is a basis for V , we say the n-tuple (a1 , . . . , an ) is the coordinate
vector of v with respect to S if v = a1 v + · · · + an vn . We denote this vector as [v]S .
Example. Find the coordinates for (1, 3) with respect to the basis S = {(1, 1), (−1, 1)}.
We set (1, 3) = a(1, 1) + b(−1, 1), which leads to the equations a − b = 1 and a + b = 3.
This system has solution a = 2, b = 1. Thus (1, 3) = 2(1, 1) + 1(−1, 1), so that [(1, 3)]S =
(2, 1).
Example. Find the coordinates for t2 + 3t + 2 with respect to the basis S = {t2 + 1, t +
1, t − 1}. We set t2 + 3t + 2 = a(t2 + 1) + b(t + 1) + c(t − 1). Collecting like terms gives
t2 + 3t + 2 = at2 + (b + c)t + (a + b − c). This leads to the system of equations


a = 1
b+c = 3


a+b−c = 2
The solution is a = 1, b = 2, c = 1. Thus we have t2 + 3t + 2 = 1(t2 + 1) + 2(t + 1) +
1(t − 1), so that [t2 + 3t + 2]S = (1, 2, 1).
Note that for any vector v in an n–dimensional vector space V and for any basis S for
V , the coordinate vector [v]S is an element of Rn .
Proposition 4.0.2. For any basis S for an n–dimensional vector space V , the correspondence
v #→ [v]S is an isomorphism from V to Rn .
Corollary 4.0.3. Every n–dimensional vector space over a R is isomorphic to Rn .
Chapter 5
Linear Maps Rn → Rm
Since every finite-dimensional vector space over R is isomorphic to Rn , any problem we
have in such a vector space that can be expressed entirely in terms of vector operations
can be tranferred to one in Rn . Since our ultimate goal is to understand linear maps
V → W , we will focus our efforts on understanding linear maps Rn → Rm , without
worrying about expressing things in abstract terms.
Remark. Unlike any previous section, we focus specifically on Rn in this chapter. To
emphasize the distinction, we use x to denote an arbitrary vector in Rn .
5.1 Linear maps from Rn to R
We’ve already seen above that the linear maps R → R are precisely those of the form
L( x ) = ax for some real number a. For the next step, we allow our domain to have
multiple dimensions, but insist that our target space be R. We will discover that linear
maps L : Rn → R are already familiar to us.
Theorem 5.1.1. If L : Rn → R is a linear map, then there is some vector m such that L(x) = a · x.
Proof. For j = 1, . . . , n, we set e j equal to the jth standard basis vector in Rn . Set a =
(a1 , . . . , an ), where each a j = L(e j ), and consider an arbitrary vector x = ( x1 , . . . , xn ) in
Rn . We compute
L(x) = L( x1 e1 + · · · + xn en ) = x1 L(e1 ) + · · · + xn L(en ) = x1 a1 + · · · + xn an = x · a.
Remark. Wait, didn’t we say that we weren’t going to think about dot products? Then
we would be studying inner product spaces rather than vector spaces! Yes, and that’s
still true. Within a given vector space, we will not be performing any dot products,
and so in particular will never speak of length or angle. And in fact our definition of
linear map did not use the notion of dot product; it used only vector addition and scalar
multiplication. What we’ve shown is that every linear map from Rn to R has the form
f ( x 1 , x 2 , . . . , x n ) = a1 x 1 + · · · + a n x n
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CHAPTER 5. LINEAR MAPS R N → R M
30
for some fixed real numbers a1 , . . . , an . It just so happens that we have a name for this
type of operation, and we call it the dot product, but this is just a convenient way to
explain what linear maps do; we’re not studying the algebraic or geometric properties
of the dot product in Rn .
5.2 Linear Maps Rn → Rm
One of the first things you learn in vector calculus is that functions with multiple outputs
can be thought of as a list of functions with one output. Thus given an arbitrary function f : R2 → R3 , say, we think of it as f ( x, y) = ( f 1 ( x, y), f 2 ( x, y), f 3 ( x, y)), where each
component function f j is a map R2 → R1 . We thus expect to find that linear maps from
Rn to Rm are those whose component functions are linear maps from Rn to R, which we
saw in the last section are just dot products. This is the content of the following.
Theorem 5.2.1. The function L : Rn → Rm is linear if and only if each component function
L j : Rn → R is linear.
Proof. Omitted.
Thus any linear map Rn → Rm is built up from a bunch of dot products in each
component. In the next section we will make use of this fact to come up with a nice way
to present linear maps.
5.3 Matrices
There are many ways to write vectors in Rn . For example, the same vector in R3 can be
represented as
3i + 2j − 4k,
$3, 2, −4%,
(3, 2, −4),
[3, 2, −4],


3
 2 .
−4
We will focus on these last two for the time being. In particular, whenever we have a
dot product x · y of two vectors x and y (in that order), we will write the first as a row
in square brackets and the second as a column in square brackets. Thus we have, for
example,
 
)
* 2
1 2 3  3  = 2 + 6 − 12 = −4.
−4
Note that we are also avoiding commas in the row vector.
5.3. MATRICES
31
Now suppose L is an arbitrary linear map from Rn to R. Then given input vector x,
L(x) is the dot product a · x for some fixed vector a. Thus we may write

x1
  x2   )
  
L   .   = a1 a2 · · ·
 .. 

xn


x1

*
 x2 
an  .  .
 .. 
xn
Now suppose L is a linear map from Rn to Rm , and the ith component functions is
the dot product with ai . The we can write


x1
  x2  
 
L  .  =
 .. 

a11
a
 21
 ..
 .
xn
a12
a22
..
.
am1 am2
···
···
 
a1n
x1
 x2 
a2n 
 
..   ..  =
.  . 
···
· · · amn
xn


a1 · x
 a2 · x 


 ..  .
 . 
am · x
Thus we can think of any linear map from Rn to Rm as multiplication by a matrix,
assuming we define multiplication in exactly this way.
Definition 5.3.1. If A = (aij ) is an m × n matrix and x is an n × 1 column vector, the
product Ax is defined to be the m × 1 column vector whose ith entry is the dot product
of the ith row of A with x.
Thus we are led to the fortuitous observation that every linear map L : Rn → Rm has
the form L(x) = Ax for some m × n matrix A. Thus linear maps from R to itself are just
multiplication by a 1 × 1 matrix; i.e., multiplication by a constant. This agrees with what
we saw earlier.
We now note an important fact about compositions of linear maps.
Theorem 5.3.2. Suppose L : Rn → Rm and T : Rm → R p are linear maps. Then the composition
T ◦ L : Rn → R p is a linear map.
Suppose L is represented by the m × n matrix A and T is represented by the p × m
matrix B. Because T ◦ L is also linear, it is represented by some p × n matrix C. We now
show how to construct C from A and B.
We begin with a motivating example. Suppose L maps from R2 to R2 , as does T, and
suppose L dots with a = (a1 , a2 , ) and b = (b1 , b2 ) while T dots with c = (c1 , c2 ) and
d = (d1 , d2 ). Then
T ◦ L(x ) = T
34
a·x
b·x
56
=T
34
4
a1 x 1 + a2 x 2
b1 x1 + b2 x2
56
4
c (a x + a2 x2 ) + c2 (b1 x1 + b2 x2 )
= 1 1 1
d1 (a1 x1 + a2 x2 ) + d2 (b1 x1 + b2 x2 )
c a + c2 b1 c1 a2 + c2 b2
= 1 1
d1 a1 + d2 b1 d1 a2 + d2 b2
54 5
x1
x2
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CHAPTER 5. LINEAR MAPS R N → R M
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4
5
4
5
a1 a2
c1 c2
Thus if L is multiplication by A =
and T is multiplication by B =
,
b1 b2
d1 d2
then T ◦ L is multiplication by C = (cij ), where cij is the dot product of the ith row of B
with the jth row of A. In other words, we have
4
54
5 4
5
c 1 c 2 a1 a2
c1 a1 + c2 b1 c1 a2 + c2 b2
=
.
d1 d2 b1 b2
d1 a1 + d2 b1 d1 a2 + d2 b2
This may seem a strange way to define the product of two matrices, but since we’re
thinking of matrices as representing linear maps, it only makes sense that the product of
two should be the matrix of the composition, so the definition is essentially forced upon
us.
Remark. According to this definition, we cannot just multiply any two matrices. Their
sizes have to match up in a nice way. In particular, for the dot products to make sense in
computing AB, the rows of A have to have just as many elements as the columns of B.
In short, the product AB is defined as long as A is m × p and B is p × n, in which case
the product is m × n.
Proposition 5.3.3. Matrix multiplication is associative when it is defined. In other words, for
any matrices A, B, and C we have A( BC) = ( AB)C, as long as all the individual products in
this identity are defined.
Proof. It is straightforward, though incredibly tedious, to prove this directly using our
algebraic definition of matrix multiplication. What is far easier, however, is simply to
note that function composition is always associative, when it’s defined. The result follows.
There are some particularly special linear maps: the zero map and the identity. It is
not to hard to see that the zero map Rn → Rm can be represented as multiplication by the
zero matrix 0m×n . The identity map Rn → Rm is represented by the aptly named identity
matrix Im×n , which has 1s on its main diagonal and 0s elsewhere. Note that it follows
that I A = AI = A for approriately sized I, while A0 = 0A = 0, for appropriately sized
0.