# 4 Chapter 4 Lecture Notes. Vector Spaces and Subspaces

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Math 2040 Matrix Theory and Linear Algebra II
4 Chapter 4 Lecture Notes. Vector Spaces and Subspaces
4.1 Vector Spaces and Subspaces
1. Notation: The symbol ∅ means ”the empty set”.
The symbol ∈ means ”is an element of”. The symbol ⊆ means ”is a subset of”. The symbols {x|P (x)} mean ”the set of x such that x has the property P . R = ”the real numbers”. Elements of R are called scalars.
2. The deﬁnition of an abstract vector space and examples.
(a) There are 10 axioms for a vector space, given on page 217 of the text. The ﬁrst ﬁve axioms concern the operation of addition and may be named 1. Closure, 2. Commutivity, 3. Associativity, 4. Unit and 5. Inverse, respectively. These ﬁrst ﬁve axioms are the axioms of addition and alone deﬁne the idea of an abelian group. There are some strange and wonderful abelian groups, yet no vector space is strange. This is due to the second set of axioms which guarantee no strange abelian group can be made into a vector space. The axioms 5 through 10 concern scalar multiplication and may be named 6. Closure, 7. Right Distribution, 8. Left Distribution, 9. Associativity and 10. Unit, respectively.
(b) The idea of a vector space as given above gives our best guess of the objects to study for understanding linear algebra. We will abandon this idea if a better one is found.
(c) An important consequence of the axioms is the Cancellation Law, that u + v = u + w implies v = w.
(d) We denote the zero vector by 0. This can cause confusion so the student must take care to distinguish the zero vector from the zero scalar. Some practice is gained by reading the next item carefully.
(e) It follows from the axioms that 0v = 0 (here on the left of the equality appears the scalar zero while the vector zero appears on the right), and −v = (−1)v for any vector v and also that α0 = 0 for any scalar α ∈ R. It is also the case that 0 ∈ V is unique; that is, if v satisﬁes the property that v + u = u for all vectors u then v = 0. It is a worthwhile exercise for the student to try to prove these facts.
(f) Rn is a vector space. (g) Pn = ”the space of real polynomials of degree n or less” is a vector space.
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(h) C[0, 1] = ”the space of continuous real valued functions on the unit interval” is a vector space.

(i)

Let V

be R with

1 2

removed.

Compute v + w

in V

by computing v + w − 2vw

in R (e.g.

2 + 3 = −7 in V ). We will show in class that V with this addition is an abelian group which

cannot be made into a vector space.

3. Subspaces.

(a) A subspace of a vector space V is a subset W which is a vector space under the inherited operations from V . Thus, W ⊆ V is a subspace iﬀ 0 ∈ W and W nonempty and is closed under the operations of addition of vectors and multiplication of vectors by scalars.
(b) The trivial subspace of a vector space V is {0} ⊆ V . Another example of a subspace of V is V itself.
(c) Subspace Spanned By Vectors: Let V be a vector space and let S be a set of in V . The set Span(S) of all ﬁnite linear combinations of the vectors taken from S is a subspace of V . This subspace is called the subspace spanned by S.

4.2 Linear Transformation, Null Space, and Column Space

1. Linear transformations.

(a) A linear transformation T : V1 −→ V2 between two vector spaces is a function preserving all of the algebra; that is, T (αv + βu) = αT (v) + βT (u) for all scalars α, β ∈ R and vectors
v, u ∈ V1.

(b) V1 is called the domain of T and V2 is called the codomain of T .

(c)

Formal

diﬀerentiation

of

polynomials

d dx

:

Pn

−→

Pn−1

and

formal

integration

: Pn −→

Pn+1 (with 0 as the constant of integration) are examples of linear transformations.

(d) A linear transformation T : V1 −→ V2 is onto if for each u ∈ V2 there is at least one v ∈ V1 such that u = T (v). Formal diﬀerentiation is an onto linear transformation.

(e) When T : V1 −→ V2 and v ∈ V1, we write v → T (v).

(f) An m × n matrix transformation A is onto iﬀ the number of pivot columns is m.

(g) A linear transformation T : V1 −→ V2 is one to one if for each u ∈ V2 there is at most one v ∈ V1 such that u = T (v). Formal diﬀerentiation is not one to one.

(h) An m × n matrix transformation A is one to one iﬀ the number of pivot columns is n.

(i) A linear transformation T : V1 −→ V2 is an isomorphism if it is one to one and onto. When T is an isomorphism, we say V1 and V2 are isomorphic vector spaces and we write V1 V2.

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(j) If T : V1 −→ V2 is an isomorphism then for any u ∈ V2 there is exactly one v ∈ V1 such that u = T (v). Thus, we can deﬁne a function T −1 : V2 −→ V1 by taking T −1(u) to be the unique v ∈ V1 such that u = T (v). T −1 is linear whenever T is linear and T −1 is an isomorphism of vector spaces whenever T is an isomorphism.
(k) The identity linear transformation 1V : V −→ V associated with any vector space is deﬁned by 1V (v) = v. 1V is an isomorphism.
(l) If T1 : V1 −→ V2 and T2 : V2 −→ V3 are two linear transformations then the composition (or composite) of T2 with T1 is (T2 ◦ T1) : V1 −→ V3 deﬁned by (T2 ◦ T1)(v) = T2(T1(v)) for each v ∈ V1.
(m) If T1 and T2 are both linear (both one to one, both onto or both isomorphisms) then so is the composition T2 ◦ T1.
(n) If T : V1 −→ V2 is a linear transformation then T = T ◦ 1V1 and T = 1V2 ◦ T . (o) If T : V1 −→ V2 is an isomorphism then 1V1 = T −1 ◦ T and 1V2 = T ◦ T −1. 2. Kernel or Null space.
(a) The Kernel or Null Space of a linear transformation T : V1 −→ V2 is N ul(T ) = {u|0 = T (u)}.
(b) N ul(T ) is a subspace of V1. (c) When T is a matrix transformation, row reducing the augmented matrix [T |0] to reduced
echelon form produces a vector equation describing N ul(T ). We call this equation the kernel equation of the matrix. It provides us with a basis of N ul(T ) (see ”basis” below).
3. Range, Image and Column Space.
(a) The image or range of a linear transformation T : V1 −→ V2 is Image(T ), the set of all vectors u ∈ V2 such that u = T (v) for some vector v ∈ V1.
(b) The image of a linear transformation is a subspace of the codomain. (c) When T is a matrix transformation, its image is called the column space of T , denoted
Col(T ), and it is the subspace spanned by the columns of the matrix. Col(T ) ⊆ V2. (d) To ﬁnd a basis of the column space of a matrix, row reduce it to an echelon form. This
identiﬁes the pivot columns and these pivot columns (of the original matrix, not the columns of the echelon form) provide a basis of the column space (see ”basis” below).
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4.3 Linearly Independence; Basis
1. Linear independence.
(a) A subset B of a vector space is linearly independent if any ﬁnite linear combination α1b1 + ... + αnbn = 0 of elements b1, .., bn ∈ B implies α1 = 0,..., αn = 0.
(b) A set of vectors {b1, ..., bp} in Rn is linearly independent iﬀ the matrix [b1...bp] is a one to one linear transformation.
(c) A set of vectors {b1, ..., bp} in Rn is linearly independent iﬀ the matrix [b1...bp] has p pivot columns.
(d) The standard basis vectors {e1, ..., en} in Rn is a linearly independent set. (e) The set of vectors {1, t, t2, ..., tn} is linearly independent in Pn. (f) If B is a linearly independent set of vectors then 0 ∈/ B. (g) No vector in a linearly independent set of vectors is a linear combination of the other vectors
of the set. (h) Some vector in a linearly dependent set of more than one vector is a linear combination of
the other vectors of the set. (i) One to one linear transformations preserve and reﬂect linear independence. (j) In class problem: Decide if B = {1, 2t, −2 + 4t2} and C = {1, 1 − t, 2 − 4t + t2} are each
linearly independent sets in P2.
2. Basis.
(a) A basis of a vector space V is a set of linearly independent vectors spanning V . (b) {e1, ..., en} is a basis of Rn. It is called the standard basis of Rn. (c) The set of vectors {1, t, t2, ..., tn} is a basis of Pn. It is the standard basis of Pn. (d) One of the most important facts (deﬁnitely theorem material) about a basis is this: THE-
OREM: If {b1, ..., bn} is a basis of a vector space and if v is any vector in the space then there is exactly one set {α1, ..., αn} of scalars such that
v = α1b1 + ... + αnbn.
(e) The above theorem guarantees if B = {b1, ..., bn} is a basis of a vector space V then the linear transformation PB : Rn −→ V deﬁned at a standard basis vector ei ∈ Rn by PB(ei) = bi, i = 1, ..., n, is an isomorphism of vector spaces.
(f) If {b1, ..., bn} is a set of vectors in Rn then it is a basis of Rn iﬀ the n × n column matrix [b1...bn] is invertible.
(g) If any vector not in a basis is added to the basis then the resulting set is not a basis.
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(h) If any vector is deleted from a basis then the resulting set is not a basis. (i) These last two facts imply n = m if Rm Rn. This is very important. (j) The zero vector is never a basis vector. (k) Isomorphisms of vector spaces preserve and reﬂect basis. (l) In class problem: Show B = {1, 2t, −2 + 4t2} and C = {1, 1 − t, 2 − 4t + t2} are each basis
of P2.
4.4 Coordinate Systems
1. Coordinate Systems: Let B = {b1, ..., bn} be a basis of a vector space V . Let x ∈ V .
(a) From the notes of the last section above there are unique scalars c1, ..., cn ∈ Rn such that x = c1b1 + ... + cnbn. The vector [x]B ∈ Rn whose coordinates are c1, ..., cn, respectively, is called the coordinate vector of x relative to B.
(b) The assignment of [x]B ∈ Rn to x ∈ V is the inverse PB−1 : V −→ Rn to the isomorphism PB : Rn −→ V , called the coordinate mapping determined by B.
(c) In case V = Rn, PB is the column matrix [b1...bn] and it is called the change-ofcordinate matrix of B. Thus, the change-of-coordinate equations are [x]B = PB−1(x) and x = PB([x]B).
(d) The inverse PB : Rn −→ V of a coordinate mapping deﬁnes a representation of V . Generally it is easier to work with the representing space than with V .
(e) We use the standard basis {1, t, t2, ..., tn} to represent the vector space Pn as Rn+1. (f) In general, if B is a given ﬁnite basis of a vector space, then PB is easy to compute and PB−1
is hard to calculate. When B is a standard basis, both PB and PB−1 are easy to calculate. (g) In class problem: Let B = {1, 2t, −2 + 4t2} and C = {1, 1 − t, 2 − 4t + t2} in P2. Study the
B and C representations of quadratic polynomials.
4.5 Dimension
1. Dimension.
(a) Any two basis of a vector space have the same number of elements. In the case of ﬁnite basis, this follows from a count of the pivot elements of a matrix transformation constructed from the two basis. In the case of inﬁnite dimensional vector spaces this fact is more diﬃcult to prove and is outside the scope of the course.
(b) That every vector space has a basis is equivalent to an axiom of mathematics, called The Axiom Of Choice. Consequently, we will accept this fact at face value.
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(c) The dimension of a vector space is the number of elements in a basis of the vector space. That dimension is a well deﬁned concept follows from parts (a) and (b) just given.
(d) The dimension of Pn is n + 1. (e) If A is a matrix then the dimension of Col(A) is the number of pivot columns of A and
the dimension of N ul(A) is the number of free variables (i.e., the number of columns of A minus the number of pivot columns of A).
4.6 Rank
1. Rank. The rank of a matrix is the dimension of its column space. Thus, Rank(A) = dim(Col(A)).
2. Rank Theorem. If A : Rn −→ Rn is a linear transformation then n = Rank(A) + dim(N ul(A)).
3. Invertibility. An n × n matrix A is invertible iﬀ Rank(A) = n.
4.7 Change Of Basis
1. Change-of-coordinates matrix.
(a) Let B and C be two basis of an n dimensional vector space V . Both B and C contain n elements. For any x ∈ V we have x = PB([x]B) and [x]C = PC−1(x). Composing these gives [x]C = (PC−1 ◦ PB)([x]B) and this equation describes a linear transformation (PC−1 ◦ PB) : Rn −→ Rn. The standard matrix representation PC←B of PC−1 ◦PB is called the change-of-coordinates matrix from B to C.
(b) If x ∈ V then [x]B ∈ Rn is the representation of x relative to B and [x]C ∈ Rn is the representation of x relative to C. We have PC←B[x]B = (PC−1 ◦ PB ◦ PB−1)(x) = (PC−1 ◦ 1V )(x) = PC−1(x) = [x]C. Thus, PC←B maps B based representations to C based representations.
(c) Suppose B and C are both basis of Rn. Let B be the matrix of columns of B and let C be the matrix of columns of C. Then PB = B and PC = C, so PC←B = C−1B and consequently PC←B may be computed by row reducing the augmented matrix [C|B] to [I|PC←B].
2. In class problem: Let B = {1, 2t, −2 + 4t2} and C = {1, 1 − t, 2 − 4t + t2} in P2. Find PC←B.
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BasisVector SpaceVectorsColumnsVector