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Chapter 3
Section 3.3
Examples of Subspaces
Span of a Set of Vectors
If 𝑆 = v1 , ⋯ , v𝑟 is a set of vectors in ℝ𝑛 then
all of the vectors that are linear combinations
of the vectors in S is called the span of S and is
denoted Sp 𝑆 = Sp v1 , ⋯ , v𝑟
Sp 𝑆 = Sp 𝐯1 , ⋯ , 𝐯𝐫 =
= x: x = 𝑥1 v1 + ⋯ + 𝑥𝑟 𝐯𝑟 , 𝑥1 , ⋯ , 𝑥𝑟 ∈ ℝ
Span of a Set of Vectors is a Subspace
For any finite set of vectors 𝑆 = v1 , ⋯ , v𝑟 in ℝ𝑛 the set Sp 𝑆 is a subspace of ℝ𝑛 . To show
this let x,y ∈ Sp 𝑆 and 𝑎 ∈ ℝ.
1. Show (a3): 0v1 + ⋯ + 0v𝑟 = 𝜽 which means 𝜽 ∈ Sp 𝑆
2. Show (c1): Let x = 𝑥1 𝐯1 + ⋯ + 𝑥𝑟 𝐯𝑟 and y = 𝑦1 𝐯1 + ⋯ + 𝑦𝑟 𝐯𝑟
then 𝐱 + 𝐲 = 𝑥1 𝐯1 + ⋯ + 𝑥𝑟 𝐯𝑟 + 𝑦1 𝐯1 + ⋯ + 𝑦𝑟 𝐯𝑟
= 𝑥1 + 𝑦1 𝐯1 + ⋯ + 𝑥𝑟 + 𝑦𝑟 𝐯𝑟 which means 𝐱 + 𝐲 ∈ Sp 𝑆
3. Show (c2): Let x = 𝑥1 𝐯1 + ⋯ + 𝑥𝑟 𝐯𝑟
then 𝑎𝐱 = 𝑎 𝑥1 𝐯1 + ⋯ + 𝑥𝑟 𝐯𝑟 = 𝑎𝑥1 𝐯1 + ⋯ + 𝑎𝑥𝑟 𝐯𝑟 ∈ Sp 𝑆
Example
The line 𝑙 𝑡 to the right can be written as a
span of a nonzero vector. In fact a line through
the origin in higher dimensions is the span of
a single nonzero vector.
3𝑡
3
3
𝑙 𝑡 = −4𝑡 = 𝑡 −4 = Sp −4
7𝑡
7
7
Example
The span of the two vectors to the right is a plane passing through
the origin. We can find the equation of this plane in 2 ways.
−1
1
Sp −2 , 3
0
1
Method 1 (Geometric)
Compute the normal vector using the cross product and plug into the equation for a plane.
0−3
𝐢
𝐣 𝐤
−1
1
−3
= −1
−2 × 3 = 1 −2 1 = − 0 − −1
0
1
1
−1 3 0
3−2
This plane has equation −3𝑥 − 𝑦 + 𝑧 = 0. This method will only work for 3 dimensions.
Method 2 (Algebraic)
𝑏1
Let the vector 𝐛 = 𝑏2 be a
𝑏3
linear combination in the
span. Set the linear
combination equal to b and
row reduce the matrix to
echelon form.
𝑏1
−1
1
𝑥1 −2 + 𝑥2 3 = 𝑏2
𝑏3
0
1
1 −1 𝑏1
−2 3 𝑏2
1
0 𝑏3
2R1+R2
-R1+R3
𝑏1
1 −1
0 1 2𝑏1 + 𝑏2
0 1 −𝑏1 + 𝑏3
-R2+R3
𝑏1
1 −1
2𝑏1 + 𝑏2
0 1
0 0 −3𝑏1 − 𝑏2 + 𝑏3
In order for this system to
be consistent it can not
have a nonzero number
in the augmented
column. This means:
−3𝑏1 − 𝑏2 + 𝑏3 = 0
𝑏1
𝑥
Setting 𝑏2 = 𝑦 we
𝑧
𝑏3
get:
−3𝑥 − 𝑦 + 𝑧 = 0
Example (Continued)
In general a plane is the span of any two nonzero vectors that point in different directions in
any dimension. This example showed that a subspace is always a solution to a homogeneous
system of equations.
The homogeneous system of
equations that a subspace of vectors
satisfies is called and algebraic
specification of the subspace.
Example
Give an algebraic
specification for the line
𝑙 𝑡 given below.
𝑡
𝑙 𝑡 = −4𝑡
3𝑡
Now,
1
𝑙 𝑡 = Sp −4
3
Form the augmented
matrix with the vector:
𝑥1 𝑥2 𝑥3 𝑇
1 𝑥1
−4 𝑥2
3 𝑥3
−1
1
Sp −2 , 3
0
1
4R1+R2
-3R1+R3
𝑥1
1
0 4𝑥1 + 𝑥2
0 −3𝑥1 + 𝑥3
Again to be consistent
both of the last rows
must be zero this gives
two equations
=
𝑥1
𝑥2 : −3𝑥1 − 𝑥2 + 𝑥3 = 0
𝑥3
𝑡
𝑙 𝑡 = −4𝑡
3𝑡
1
= Sp −4
3
𝑥1 4𝑥1 + 𝑥2 = 0
and
= 𝑥2 :
𝑥3 −3𝑥1 + 𝑥3 = 0
The intersection of two planes
is a line. Both of the planes
above contain the line 𝑙 𝑡 , so
this is the line of intersection.
Matrix Row & Column Vectors
Let A be an 𝑚 × 𝑛 matrix. Let
𝐚𝑖 be the row vector that is the
𝑖 𝑡ℎ row of matrix A. Let 𝐀𝑗 be
the column vector that is the 𝑗𝑡ℎ
column of matrix A.
𝑎11
𝐴= ⋮
𝑎𝑚1
𝐚1 = 𝑎11
⋯
⋯
⋱
⋯
𝑎1𝑛
𝐚𝑚 = 𝑎𝑚1
⋯ 𝑎𝑚𝑛
Row Space B = Sp 𝐛1 , 𝐛2 , 𝐛3 = Sp 1 4 3
1
𝐵= 2
−3
4 3
0 1
1 0
2
5
1
1
3
4
2
2 , 𝐁2 = 0 , 𝐁3 = 1 , 𝐁4 = 5
−3
1
0
1
Column
:𝐁 =
Vectors 1
Row Space of a Matrix
The row space of a matrix A is the
span of the row vectors of A. This is
a subspace since it is the span of a
set of vectors.
⋯ 𝐀𝑛
𝑎11
𝑎1𝑛
𝐀1 = ⋮ , ⋯ , 𝐀 𝑛 = ⋮
𝑎𝑚1
𝑎𝑚𝑛
⋮
Example
Find the row and column vectors for the matrix B given
to the right.
𝐛1 = 1 4 3 2
Row
: 𝐛 = 2 0 1 5
Vectors 2
𝐛3 = −3 1 0 1
𝑎1𝑛
𝐚1
⋮ = ⋮ = 𝐀1
𝑎𝑚𝑛
𝐚𝑚
𝑎11
𝐴= ⋮
𝑎𝑚1
⋯
⋱
⋯
𝑎1𝑛
𝐚1
⋮ = ⋮
𝑎𝑚𝑛
𝐚𝑚
Row Space A = Sp 𝐚1 , ⋯ , 𝐚𝑚
2, 2
0 1 5 , −3
1 0 1
The Column Space of a Matrix
The column space of a matrix A is the
span of the column vectors of A. Again
this is a subspace since the span of any
set of vectors is a subspace
1 4
If 𝐵 = 2 0
−3 1
𝑎11
𝐴= ⋮
𝑎𝑚1
⋯
⋱
⋯
𝑎1𝑛
⋮ = 𝐀1
𝑎𝑚𝑛
⋯ 𝐀𝑛
Column Space A = Sp 𝐀1 , ⋯ , 𝐀𝑛
3 2
1
4 3 2
1 5 then Column Space B = Sp 𝐁1 , 𝐁2 , 𝐁3 , 𝐁4 = Sp 2 , 0 , 1 , 5
0 1
−3 1 0 1
The Null Space (Kernel) of a Matrix
If A is a 𝑚 × 𝑛 matrix and x is a vector in ℝ𝑛 then 𝐴x is a
vector in ℝ𝑚 . The null space of A (denoted 𝒩 𝐴 ) are all
vectors x ∈ ℝ𝑛 such that 𝐴x = 𝜽 (the zero vector) in ℝ𝑚 .
In a previous example we showed the kernel of a matrix
was a subspace.
1. Show (a3): 𝐴𝜽 = 𝜽 which means 𝜽 ∈ 𝒩 𝐴
2. Show (c1): If x,y ∈ 𝒩 𝐴 then 𝐴x = 𝜽 and 𝐴y = 𝜽
then 𝐴 x + y = 𝐴x + 𝐴y = 𝜽 + 𝜽 = 𝜽 which
means x + y ∈ 𝒩 𝐴
3. Show (c2): If x ∈ 𝒩 𝐴 and 𝑎 ∈ ℝ then 𝐴x = 𝜽, then
𝐴 𝑎x = 𝑎 𝐴x = 𝑎𝜽 = 𝜽 which means 𝑎x ∈ 𝒩 𝐴
𝒩 𝐴 =
𝑥1
𝑥1
0
⋮ :𝐴 ⋮ = ⋮
𝑥𝑛
𝑥𝑛
0
The null space of a matrix A
are the set of solutions to the
homogeneous system of
equations with coefficient
matrix A. So every subspace is
the null space of some matrix.
The Range of a Matrix
If A is a 𝑚 × 𝑛 matrix and x is a vector in
ℝ𝑛 then y = 𝐴x is a vector in ℝ𝑚 . The range
of A (denoted ℛ 𝐴 ) are all vectors y ∈ ℝ𝑚
such that 𝐴x = 𝐲 for some vector x in ℝ𝑛 .
ℛ 𝐴 =
𝑦1
𝑦1
𝑥1
𝑥1
⋮ : ⋮ = 𝐴 ⋮ , ⋮ ∈ ℝ𝑛
𝑦𝑚 𝑦𝑚
𝑥𝑛
𝑥𝑛
Because of the way matrix multiplication is defined (row  column) the matrix product can
again be rewritten as a linear combination of the columns of the matrix.
𝑥1
𝐴 ⋮ = 𝑥1 𝐀1 + ⋯ + 𝑥𝑛 𝐀𝑛 ∈ Column Space of A
𝑥𝑛
Since this is the case for any combination of 𝑥1 , ⋯ , 𝑥𝑛 the range of A will be the span of the
columns of the matrix and we see that the range and the column space are equal.
ℛ 𝐴 = Sp 𝐀1 , ⋯ , 𝐀𝑛 = Column Space of 𝐴
In particular this shows that the range is a subspace.
This gives us several different ways to characterize a subspace, all of them in terms of a
matrix. Not only are subspaces solutions of homogeneous systems of equations but they are
various combinations of parts of matrices!