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Isomorphisms
of Vector
Spaces
4
...
Some of these generalizations are outlined in this section
...
Through
out this section, it is assumed that U, V, and W are vector spaces over JR and that
L : U --t V and M : V --t Ware linear mappings
...
One-to-One
Lemma 1
Lis one-to-one if and only if Null (L)
=
{0}
...
5
...
)
You are asked to prove Lemma 1 as Problem D 1
...
The fact that such
a transformation is one-to-one allows the definition of the inverse
...
5
...
The mapping
P : JR4 --t JR3 of Example 3
...
8 is not one-to-one
...
EXAMPLE2
Prove that L : JR2
--t
JR3 defined by Lx
( 1,x2)
=
x
( i,xi + x2,x2) is one-to-one
...
Then we have (x1,xi + X2,x2)
Y2, yz), and so x1
Y1 andx2 Y2· Thus, Lis one-to-one
...
2
Solution: Let p x
( ) = l+x andq(x) = l+x+x
...
EXERCISE 1
Suppose that {u1,
•
...
Prove
that {L( u1) ,
...
Definition
L : U --t V is said to be onto if for every v E V there exists some u E U such that
Onto
L(u) = v
...
EXAMPLE4
Invertible linear transformations of JR" are all onto mappings
...
5
...
5
...
EXAMPLES
2
Prove that L : JR
...
Solution: Let a+ bx be any polynomial in PI
...
,
such that UJ) = a+ bx
...
Therefore, we have
L(a, b - a) = a + bx, and so Lis onto
...
-t JR
...
Solution: If M is onto, then for every vector jl =
soch that L(X) = jl
...
3, there exists X =
[�:]
E
11!
...
Hence, M is not onto
...
, uk} is a spanning set for U and Lis onto
...
, L( uk)}
...
You are asked to prove Theorem 2 as problem D4
...
The word isomorphism comes from Greek words meaning "same form
...
It implies that the es
sential structure of the isomorphic vector spaces is the same, so that a vector space
statement that is true in one space is immediately true in any isomorphic space
...
EXAMPLE7
3
Prove that P2 and JR are isomorphic by constructing an explicit isomorphism L
...
Linear: Let any two elements of P2 be p(x)
and let t E R Then,
L(tp + q)
=
=
=
=
2
ao+a1x+a2x and q(x)
=
b 0+b1x+b2x
2
2
2
L(t(ao + a1x + a1x ) + (bo + b1x + b1x ))
2
L(tao + bo + (ta1 + b1)x + (ta2 + b1)x )
tL(p) +Lq
( )
Therefore, Lis linear
...
Then,
[�]
R3, we have La
( , + a1x + a2x')
=
Use Exercise 1 and Exercise 2 to prove that if L : 1I
...
,
2
L(a, + a1x + a2x )
=
[�]
{O} and thus Lis one-to-one by Lemma 1
...
EXERCISE 3
=
u,,} is a basis for 1I
...
[::J
...
V is an isomorphism and
, L(u,,)} is a basis for V
...
J and V are finite-dimensional vector spaces over R Then 1I
...
You are asked to prove Theorem 3 as Problem DS
...
The vector space M(m, n) is isomorphic to Rm"
...
The vector space P,, is isomorphic to R"+1
...
Every k-dimensional subspace of JR" is isomorphic to every k-dimensional sub
space of M(m, n)
...
However, even if we already know that
two vector spaces are isomorphic, we may need to construct an explicit isomorphism
between the two vector spaces
...
Theorem 4
If1!J and V are n-dimensional vector spaces over JR, then a linear mapping L : 1!J
--
V
is one-to-one if and only if it is onto
...
PROBLEMS 4
...
Prove that your map is an
isomorphism
...
Prove that your map is an
isomorphism
...
(Hint: Suppose that L is one-to
one
...
If L(u1) =
L(u2), then what is L(u1 - u2)?)
D2 (a) Prove that if Land Mare one-to-one, then Mo L
is one-to-one
...
(c) Is it possible to give an example where Lis not
one-to-one but M o Lis one-to-one? Explain
...
D4 Prove Theorem 2
...
To prove "isomorphic :: same
dimension," use Exercise 3
...
, u11} for
1!J, and a basis {v1 ,
, v11} for V
...
•
...
(You must prove that this is an iso
morphism
...
D7 Prove that any plane through the origin in JR3 is iso
2
morphic to JR
...
1
...
Prove that 1l
...
J x V that is isomorphic to 1!
...
D9 (a) Prove that JR2
(b) Prove that JRI!
JR is isomorphic to JR3
...
x
DlO Suppose that L
1l
...
Prove that L-1 o Mo Lis a linear mapping from 1l
...
J
...
:
CHAPTER REVIEW
Suggestions for Student Review
Remember that if you have understood the ideas of
Chapter 4, you should be able to give answers to these
questions without looking them up
...
1 State the essential properties of a vector space over
R Why is the empty set not a vector space? Describe
two or three examples of vector spaces that are not
subspaces of JR11• (Section 4
...
3)
3
(a) Explain the concept of dimension
...
(Section 4
...
(Section 4
...
4)
the standard procedure to determine its coordi
[ !]
nates with respect to :B
...
4)
(d) Pick any two vectors in JR5 and determine
whether they lie in your subspace
...
(Section 4
...
(Sections 4
...
4)
7 Give the definition of a linear mapping L : V � W
and show how this implies that L preserves linear
combinations
...
Describe how
to find a basis for the nullspace and a basis for the
S Invent and analyze an example as follows
...
(Section 4
...
(Don't make it too easy by choos
ing any standard basis vectors, but don't make
it too hard by choosing completely random
components
...
3)
(b) Determine the standard coordinates in JR5 of the
vecto' that hos cooniinate vectm
(c) Take the vector you found in (b) and carry out
[-!]
spect to your basis :B
...
4)
8 State how to determine the standard matrix and the
:B-matrix of a linear mapping L : V � V
...
s is determined in terms of [L]_s
...
6)
9 State the definition of an isomorphism of vector
spaces and give some examples
...
(Section 4
...
(a) The set of 4 x 3 matrices such that the sum of
the entries in the first row is zero (a11 + a12 +
a1 = 0) under standard addition and scalar
3
multiplication of matrices
...
(c) The set of 2 x 2 matrices such that all entries
are integers under standard addition and scalar
multiplication of matrices
...
E2 In each of the following cases, determine whether
the given set of vectors is a basis for M(2, 2)
...
[;
�] ' [;
�]}
-�]·[� �]}
�]}
-
E3 (a) Let § be the subspace spanned by v 1
1
0
1 ,
1
3
3
1
3
1
...
Determine the dimension of§
...
,
E4 (a) Find a basis for the plane in JR
...
(b) Extend the basis you found in (a) to a basis 13
for JR
...
(c) Let L : JR
...
3 be a reflection in the plane
from part (a)
...
(d) Using your result from part (c), determine the
standard matrix [L]s of the reflection
...
3
�
JR
...
{[i] [I] [ t]}
...
E6 Suppose that L : V � W is a linear mapping with
Null(L) = {O}
...
, vd is a linearly
independent set in V
...
•
•
...
If it is true, explain briefly; if it is false,
give an example to show that it is false
...
11 must have dimension less
than n
...
(c) If 13 is a basis for a subspace of JR
...
5 has
five components
...
11 � JR
...
11, the rank of the matrix [L]� is the
same as the rank of the matrix [L]s
...
(f) If L : V � W is one-to-one, then dim V =
dimW
...
They
may not be of interest to all students
...
(Hint: Begin by assuming that there are
two reduced row echelon forms R and
1
two matrices?)
A,
the three column sums of
A,
For example, if
A
=
[-� � -!]
...
0
(e)
(f)
0
and
l
3
...
=
(g) Find the coordinates of
A
=
[� � ;]1
2
respect to the basis 15
...
Exercises F4-F7 require the following definitions
...
(a) Show that MS 3 is a subspace of M(3, 3)
...
(c) Compute the nullspace of wt
...
, K1, K2} is a basis for MS 3
...
The subset of
0
Observe that!
...
)
weight k
...
+ PK1 + qK2
diagonal sums of A (a11 + a22 + a33 and a13 + a22 +
a31 ) all have the same value k
...
=
K2 form a
Null(wt), con
s in the
row sum is the sum of the entries in one row of
of
=
that have
F3 Magic Squares-An Exploration of
Their Vector Space Properties
We say that any matrix A E M(3, 3) is a 3 x 3
magic square if the three row sums (where each
A)
, v
a11K1 - a12K2
...
W hat can
you say about the columns with leading
A
(d) Let!
...
- 1 1 1l
[� 1 �
basis for Null(wt)
...
denote unknown entries
...
Prove that there exists a linear operator
L: V
a, b, c,
...
In general, given a subspace
S of V, one can choose a complement in many ways; the
2
complement of S is not unique
...
FS (a) If § is a k-dimensional subspace of JR
...
Let U be the
that any complement of§ must be ofdimension
subspace spanned by w and §
...
T but not in S, then v is in U
...
n that has a
unique complement
...
1?
either { O} or 1
F6 Suppose that v and w are vectors in a vector space
V
...
Let T
F7 Show that if § and T are finite-dimensional sub
spaces of V, then
dim § + dim T
=
dim(§ + T) + dim(§ n T)