Math 221: LINEAR ALGEBRA
Chapter 6. Vector Spaces
§6-2. Subspaces and Spanning Sets
Le Chen1
Emory University, 2020 Fall
(last updated on 11/03/2020)
Creative Commons License
(CC BY-NC-SA)
1
Slides are adapted from those by Karen Seyffarth from University of Calgary.
Subspaces and Spanning Sets
Linear Combinations and Spanning Sets
Subspaces and spanning Sets
Subspaces and spanning Sets
Definition (Subspaces of a Vector Space)
Let V be a vector space and let U be a subset of V. Then U is a subspace
of V if U is a vector space using the addition and scalar multiplication of V.
Subspaces and spanning Sets
Definition (Subspaces of a Vector Space)
Let V be a vector space and let U be a subset of V. Then U is a subspace
of V if U is a vector space using the addition and scalar multiplication of V.
Theorem (Subspace Test)
Let V be a vector space and U ⊆ V. Then U is a subspace of V if and only
if it satisfies the following three properties:
1. U contains the zero vector of V, i.e., 0 ∈ U where 0 is the zero vector of
V.
2. U is closed under addition, i.e., if u, v ∈ U, then u + v ∈ U.
3. U is closed under scalar multiplication, i.e., if u ∈ U and k ∈ R, then
ku ∈ U.
Subspaces and spanning Sets
Definition (Subspaces of a Vector Space)
Let V be a vector space and let U be a subset of V. Then U is a subspace
of V if U is a vector space using the addition and scalar multiplication of V.
Theorem (Subspace Test)
Let V be a vector space and U ⊆ V. Then U is a subspace of V if and only
if it satisfies the following three properties:
1. U contains the zero vector of V, i.e., 0 ∈ U where 0 is the zero vector of
V.
2. U is closed under addition, i.e., if u, v ∈ U, then u + v ∈ U.
3. U is closed under scalar multiplication, i.e., if u ∈ U and k ∈ R, then
ku ∈ U.
Remark
The proof of this theorem requires one to show that if the three properties
listed above hold, then all the vector space axioms hold.
Remark ( Important Note )
As a consequence of the proof, any subspace U of a vector space V has the
same zero vector as V, and each u ∈ U has the same additive inverse in U
as in V.
Remark ( Important Note )
As a consequence of the proof, any subspace U of a vector space V has the
same zero vector as V, and each u ∈ U has the same additive inverse in U
as in V.
Examples (Two extreme examples)
Let V be a vector space.
1. V is a subspace of V.
2. {0} is a subspace of V, where 0 denotes the zero vector of V.
Problem
Let A be a fixed (arbitrary) n × n real matrix, and define
U = {X ∈ Mnn | AX = XA},
i.e., U is the subset of matrices of Mnn that commute with A. Prove that U
is a subspace of Mnn .
Problem
Let A be a fixed (arbitrary) n × n real matrix, and define
U = {X ∈ Mnn | AX = XA},
i.e., U is the subset of matrices of Mnn that commute with A. Prove that U
is a subspace of Mnn .
Solution
I Let 0nn denote the n × n matrix of all zeros. Then A0nn = 0nn and
0nn A = 0nn , so A0nn = 0nn A. Thus 0nn ∈ U.
Problem
Let A be a fixed (arbitrary) n × n real matrix, and define
U = {X ∈ Mnn | AX = XA},
i.e., U is the subset of matrices of Mnn that commute with A. Prove that U
is a subspace of Mnn .
Solution
I Let 0nn denote the n × n matrix of all zeros. Then A0nn = 0nn and
0nn A = 0nn , so A0nn = 0nn A. Thus 0nn ∈ U.
I Suppose X, Y ∈ U. Then AX = XA and AY = YA, implying that
A(X + Y) = AX + AY = XA + YA = (X + Y)A,
and thus X + Y ∈ U, so U is closed under addition.
Problem
Let A be a fixed (arbitrary) n × n real matrix, and define
U = {X ∈ Mnn | AX = XA},
i.e., U is the subset of matrices of Mnn that commute with A. Prove that U
is a subspace of Mnn .
Solution
I Let 0nn denote the n × n matrix of all zeros. Then A0nn = 0nn and
0nn A = 0nn , so A0nn = 0nn A. Thus 0nn ∈ U.
I Suppose X, Y ∈ U. Then AX = XA and AY = YA, implying that
A(X + Y) = AX + AY = XA + YA = (X + Y)A,
and thus X + Y ∈ U, so U is closed under addition.
I Suppose X ∈ U and k ∈ R. Then AX = XA, implying that
A(kX) = k(AX) = k(XA) = (kX)A;
thus kX ∈ U, so U is closed under scalar multiplication.
By the subspace test, U is a subspace of Mnn .
Problem
Let t ∈ R, and let
U = {p ∈ P | p(t) = 0},
i.e., U is the subset of polynomials that have t as a root. Prove that U is a
vector space.
Problem
Let t ∈ R, and let
U = {p ∈ P | p(t) = 0},
i.e., U is the subset of polynomials that have t as a root. Prove that U is a
vector space.
Proof.
I Let 0 denote the zero polynomial. Then 0(t) = 0, and thus 0 ∈ U.
Problem
Let t ∈ R, and let
U = {p ∈ P | p(t) = 0},
i.e., U is the subset of polynomials that have t as a root. Prove that U is a
vector space.
Proof.
I Let 0 denote the zero polynomial. Then 0(t) = 0, and thus 0 ∈ U.
I Let q, r ∈ U. Then q(t) = 0, r(t) = 0, and
(q + r)(t) = q(t) + r(t) = 0 + 0 = 0.
Therefore, q + r ∈ U, so U is closed under addition.
Problem
Let t ∈ R, and let
U = {p ∈ P | p(t) = 0},
i.e., U is the subset of polynomials that have t as a root. Prove that U is a
vector space.
Proof.
I Let 0 denote the zero polynomial. Then 0(t) = 0, and thus 0 ∈ U.
I Let q, r ∈ U. Then q(t) = 0, r(t) = 0, and
(q + r)(t) = q(t) + r(t) = 0 + 0 = 0.
Therefore, q + r ∈ U, so U is closed under addition.
I Let q ∈ U and k ∈ R. Then q(t) = 0 and
(kq)(t) = k(q(t)) = k · 0 = 0.
Therefore, kq ∈ U, so U is closed under scalar multiplication.
By the subspace test, U is a subspace of P, and thus is a vector space.
Examples (more...)
1. It is routine to verify that Pn is a subspace of P for all n ≥ 0.
Examples (more...)
1. It is routine to verify that Pn is a subspace of P for all n ≥ 0.
2. U = A ∈ M22 | A2 = A is NOT a subspace of M22 .
To prove this, notice that I2 , the two by two identity matrix, is in U,
but 2I2 6∈ U since (2I2 )2 = 4I2 6= 2I2 , so U is not closed under scalar
multiplication.
Examples (more...)
1. It is routine to verify that Pn is a subspace of P for all n ≥ 0.
2. U = A ∈ M22 | A2 = A is NOT a subspace of M22 .
To prove this, notice that I2 , the two by two identity matrix, is in U,
but 2I2 6∈ U since (2I2 )2 = 4I2 6= 2I2 , so U is not closed under scalar
multiplication.
3. U = {p ∈ P2 | p(1) = 1} is NOT a subspace of P2 .
Because the zero polynomial is not in U: 0(1) = 0.
4. Cn ([0, 1]), n ≥ 1, is a subspace of C([0, 1]).
Linear Combinations and Spanning Sets
Linear Combinations and Spanning Sets
Definitions (Linear Combinations and Spanning)
Let V be a vector space and let {u1 , u2 , . . . , un } be a subset of V.
1. A vector u ∈ V is called a linear combination of u1 , u2 , . . . , un if there
exist scalars a1 , a2 , . . . , an ∈ R such that
u = a1 u1 + a2 u2 + · · · + an un .
2. The set of all linear combinations of u1 , u2 , . . . , un is called the span of
u1 , u2 , . . . , un , and is defined as
span{u1 , u2 , . . . , un } = {a1 u1 + a2 u2 + · · · + an un | a1 , a2 , . . . , an ∈ R}.
Linear Combinations and Spanning Sets
Definitions (Linear Combinations and Spanning)
Let V be a vector space and let {u1 , u2 , . . . , un } be a subset of V.
1. A vector u ∈ V is called a linear combination of u1 , u2 , . . . , un if there
exist scalars a1 , a2 , . . . , an ∈ R such that
u = a1 u1 + a2 u2 + · · · + an un .
2. The set of all linear combinations of u1 , u2 , . . . , un is called the span of
u1 , u2 , . . . , un , and is defined as
span{u1 , u2 , . . . , un } = {a1 u1 + a2 u2 + · · · + an un | a1 , a2 , . . . , an ∈ R}.
3. If U = span{u1 , u2 , . . . , un }, then {u1 , u2 , . . . , un } is called a spanning
set of U.
Problem
Is it possible to express x2 + 1 as a linear combination of
x + 1, x2 + x, and x2 + 2 ?
Equivalently, is x2 + 1 ∈ span{x + 1, x2 + x, x2 + 2}?
Problem
Is it possible to express x2 + 1 as a linear combination of
x + 1, x2 + x, and x2 + 2 ?
Equivalently, is x2 + 1 ∈ span{x + 1, x2 + x, x2 + 2}?
Solution
Suppose that there exist a, b, c ∈ R such that
x2 + 1 = a(x + 1) + b(x2 + x) + c(x2 + 2).
Then
x2 + 1 = (b + c)x2 + (a + b)x + (a + 2c),
implying that b + c = 1, a + b = 0, and a + 2c = 1.
Solution (continued)
Hence,
1. If this system is consistent, then we have found a way to express x2 + 1
as a linear combination of the other vectors; otherwise,
2. if the system is inconsistent and it is impossible to express x2 + 1 as a
linear combination of the other vectors.
Because
0 1 1 0 1 1
1 1
det 1 1 0 = det 0 1 −2 = det = −3 6= 0,
1 −2
1 0 2 1 0 2
Answer: Yes, i.e., x2 + 1 ∈ span{x + 1, x2 + x, x2 + 2}.
Remark
By solving the linear equation
b + c = 1
a + b + = 0
a + 2c = 1
we find that
1 1 2
a=− , b= , c= .
3 3 3
Hence,
1 1 2
x2 + 1 = − (x + 1) + (x2 + x) + (x2 + 2)
3 3 3
Problem
Let
1 −1 2 1 1 3
u= , v= and w= .
2 1 1 0 −1 1
Is w ∈ span{u, v}? Prove your answer.
Problem
Let
1 −1 2 1 1 3
u= , v= and w= .
2 1 1 0 −1 1
Is w ∈ span{u, v}? Prove your answer.
Solution (partial)
Suppose there exist a, b ∈ R such that
1 3 1 −1 2 1
=a +b .
−1 1 2 1 1 0
Then
a + 2b = 1
−a + b = 3
2a + b = −1
a + 0b = 1.
What remains is to determine whether or not this system is consistent.
Answer: No.
Example
The set of 3 × 2 real matrices,
1 0 0 1 0 0 0 0 0 0 0 0
M32 = span 0 0 , 0 0 , 1 0 , 0 1 , 0 0 , 0 0 .
0 0 0 0 0 0 0 0 1 0 0 1
Example
The set of 3 × 2 real matrices,
1 0 0 1 0 0 0 0 0 0 0 0
M32 = span 0 0 , 0 0 , 1 0 , 0 1 , 0 0 , 0 0 .
0 0 0 0 0 0 0 0 1 0 0 1
Remark ( A Spanning Set of Mmn )
In general, the set of mn m × n matrices that have a ‘1’ in position (i, j) and
zeros elsewhere, 1 ≤ i ≤ m, 1 ≤ j ≤ n, constitutes a spanning set of Mmn .
Example
Let p(x) ∈ P3 . Then p(x) = a0 + a1 x + a2 x2 + a3 x3 for some
a0 , a1 , a2 , a3 ∈ R. Therefore,
P3 = span{1, x, x2 , x3 }.
Example
Let p(x) ∈ P3 . Then p(x) = a0 + a1 x + a2 x2 + a3 x3 for some
a0 , a1 , a2 , a3 ∈ R. Therefore,
P3 = span{1, x, x2 , x3 }.
Remark ( A Spanning Set of Pn )
For all n ≥ 0,
Pn = span{x0 , x1 , x2 , . . . , xn } = span{1, x, x2 , . . . , xn }.
span{· · · } is a subspace and the smallest one.
Theorem
Let V be a vector space, let u1 , u2 , . . . , un ∈ V, and let
U = span{u1 , u2 , . . . , un }.
Then
1. U is a subspace of V containing u1 , u2 , . . . , un .
2. If W is a subspace of V and u1 , u2 , . . . , un ∈ W, then U ⊆ W. In other
words, U is the “smallest” subspace of V that contains u1 , u2 , . . . , un .
span{· · · } is a subspace and the smallest one.
Theorem
Let V be a vector space, let u1 , u2 , . . . , un ∈ V, and let
U = span{u1 , u2 , . . . , un }.
Then
1. U is a subspace of V containing u1 , u2 , . . . , un .
2. If W is a subspace of V and u1 , u2 , . . . , un ∈ W, then U ⊆ W. In other
words, U is the “smallest” subspace of V that contains u1 , u2 , . . . , un .
Remark
This theorem should be familiar as it was covered in the particular case
V = Rn . The proof of the result in Rn immediately generalizes to an
arbitrary vector space V.
Problem
Let
1 −1 0 1 1 −1 1 0
A1 = , A2 = , A3 = , A4 = .
−1 1 1 −1 −1 0 1 1
Show that M22 = span{A1 , A2 , A3 , A4 }.
Problem
Let
1 −1 0 1 1 −1 1 0
A1 = , A2 = , A3 = , A4 = .
−1 1 1 −1 −1 0 1 1
Show that M22 = span{A1 , A2 , A3 , A4 }.
Remark
We need to prove two inclusions
M22 ⊆ span{A1 , A2 , A3 , A4 }
and
span{A1 , A2 , A3 , A4 } ⊆ M22
Proof. ( First proof )
Let
1 0 0 1 0 0 0 0
E1 = , E2 = , E3 = , E4 = .
0 0 0 0 1 0 0 1
Since M22 = span{E1 , E2 , E3 , E4 } and A1 , A2 , A3 , A4 ∈ M22 , it follows from
the previous Theorem that
span{A1 , A2 , A3 , A4 } ⊆ M22 .
Proof. ( First proof )
Let
1 0 0 1 0 0 0 0
E1 = , E2 = , E3 = , E4 = .
0 0 0 0 1 0 0 1
Since M22 = span{E1 , E2 , E3 , E4 } and A1 , A2 , A3 , A4 ∈ M22 , it follows from
the previous Theorem that
span{A1 , A2 , A3 , A4 } ⊆ M22 .
Now show that Ei , 1 ≤ i ≤ 4, can be written as a linear combination of
A1 , A2 , A3 , A4 , i.e., Ei ∈ span{A1 , A2 , A3 , A4 } (lots of work to be done
here!), and apply the previous Theorem again to show that
M22 ⊆ span{A1 , A2 , A3 , A4 }.
Proof. ( Second proof )
(1) Since A1 , A2 , A3 , A4 ∈ M22 and M22 is a vector space,
span{A1 , A2 , A3 , A4 } ⊆ M22 .
Proof. ( Second proof )
(1) Since A1 , A2 , A3 , A4 ∈ M22 and M22 is a vector space,
span{A1 , A2 , A3 , A4 } ⊆ M22 .
a b
(2) For any ∈ M22 , we need to find x1 , · · · , x4 , such that
c d
a b
x1 A1 + x2 A2 + x3 A3 + x4 A4 =
c d
m
x1 + x3 + x4 = a
−x1 + x2 − x3 = b
−x1 + x2 − x3 + x4 = c
x1 − x2 + x4 = d
Since the coefficient matrix is invertible one can find unique solution and so
a b
∈ span{A1 , A2 , A3 , A4 }.
c d
Therefore, M22 ⊆ span{A1 , A2 , A3 , A4 }.
Problem
Let p(x) = x2 + 1, q(x) = x2 + x, and r(x) = x + 1. Prove that
P2 = span{p(x), q(x), r(x)}.
Problem
Let p(x) = x2 + 1, q(x) = x2 + x, and r(x) = x + 1. Prove that
P2 = span{p(x), q(x), r(x)}.
Solution (sketch)
(1) Since p(x), q(x), r(x) ∈ P2 and P2 is a vector space,
span{p(x), q(x), r(x)} ⊆ P2 .
Problem
Let p(x) = x2 + 1, q(x) = x2 + x, and r(x) = x + 1. Prove that
P2 = span{p(x), q(x), r(x)}.
Solution (sketch)
(1) Since p(x), q(x), r(x) ∈ P2 and P2 is a vector space,
span{p(x), q(x), r(x)} ⊆ P2 .
(2) As we’ve already observed, P2 = span{1, x, x2 }. To complete the proof,
show that each of 1, x and x2 can be written as a linear combination of
p(x), q(x) and r(x), i.e., show that
1, x, x2 ∈ span{p(x), q(x), r(x)}.
Then apply the previous Theorem.
