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Vector Spaces
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•
Definition: A vector space is a nonempty set V of objects,
called vectors, on which are defined two operations, called
addition and multiplication by scalars (real numbers), subject to
the ten axioms (or rules) listed below. The axioms must hold for
all vectors u, v, and w in V and for all scalars c and d.
1. The sum of u and v, denoted by u+v , is in V.
2. u+v=v+u.
3. (u+v)+w=u+(v+w) .
4. There is a zero vector 0 in V such that u+(-u)=0.
5. For each u in V, there is a vector -u in V such that
u+(-u)=0.
6. The scalar multiple of u by c, denoted by cu, is in V.
7. c(u+v)=cu+cv.
8. (c+d)u=cu+du.
9. c(du)=(cd)u.
10. 1u=u.
The most important of these axioms are the closure properties (1) and (6).
In many cases we will be dealing with vectors that are a subset of a familiar
vector space (such as ℝ2), and if we can prove that the set is closed, it will be
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a subspace of the familiar vector space.
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Definition: A subspace of a vector space V is a subset H of V that
has three properties:
(a) The zero vector of V is in H.
(b)
H is closed under vector addition. That is, for each u and v
in H, the sum u+v is in H.
(c) H is closed under multiplication by scalars. That is, for each u
in H and each scalar c, the vector cu is in H.
•Properties (a), (b), and (c) guarantee that a subspace H of V is
itself a vector space, under the vector space operations already
defined in V.
•Every subspace is a vector space.
•Conversely, every vector space is a subspace (of itself and possibly
of other larger spaces).
•The set consisting of only the zero vector in a vector space V is a
subspace of V, called the zero subspace and written as {0}.
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A simple example.
Here is a homogeneous linear system:
It represents a linear transformation from ℝ3↦ℝ3.
Let’s look at the matrix of this linear transformation.
x1  2x3  0
x 2  x3  0
x1  x2  x3  0
1 0 2 
A  0 1  1
1 1 1 
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A simple example.
Here is a homogeneous linear system:
It represents a linear transformation from ℝ3↦ℝ3.
Let’s look at the matrix of this linear transformation.
The matrix will take a vector from ℝ3 (the domain) and
transform it into a different vector from ℝ3 (the co-domain).
x1  2x3  0
x 2  x3  0
x1  x2  x3  0
1 0 2 
A  0 1  1
1 1 1 
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A simple example.
Here is a homogeneous linear system:
It represents a linear transformation from ℝ3↦ℝ3.
Let’s look at the matrix of this linear transformation.
The matrix will take a vector from ℝ3 (the domain) and
transform it into a different vector from ℝ3 (the co-domain).
x1  2x3  0
x 2  x3  0
x1  x2  x3  0
1 0 2 
A  0 1  1
1 1 1 
If we solve the system in the usual way, we will obtain what
is called the Null Space of A (also called the kernel of the
linear transformation).
• Nul(A) is the set of vectors in the domain that get
mapped to the zero vector in the co-domain.
Let’s walk through the calculation:
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A simple example.
x1  2x3  0
Here is a homogeneous linear system:
It represents a linear transformation from ℝ3↦ℝ3.
Let’s look at the matrix of this linear transformation.
The matrix will take a vector from ℝ3 (the domain) and
transform it into a different vector from ℝ3 (the co-domain).
x 2  x3  0
x1  x2  x3  0
1 0 2 
A  0 1  1
1 1 1 
If we solve the system in the usual way, we will obtain what
is called the Null Space of A (also called the kernel of the
linear transformation).
• Nul(A) is the set of vectors in the domain that get
mapped to the zero vector in the co-domain.
Let’s walk through the calculation:
1 0 2 0
1 0 2 0
1 0 2 0
x1  2x3
 x1   2






   
0 1  1 0  0 1  1 0  0 1  1 0  x2  x3  x2    1   t
1 1 1 0
0 1  1 0
0 0 0 0
x3   1 
x3  x3






On the next slide is a graph of the situation.
t is the parameter of our explicit solution. Our
solution space represents a line in ℝ3.
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0
5
5
5
1.0
0.5
0.5
0.0
1.0
1.0
Domain (ℝ3)
Co-Domain (ℝ3)
0.5
This line is
Nul(A)
0
0.0
(0,0,0)
0.5
10
5
1.0
0.5
0.0
0
5
10
0.5
5
1.0
1.0
This line in the domain is the Null Space of A. It is a 1-dimensional subspace of ℝ3.
Every point in Nul(A) is mapped to the origin in the co-domain.
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A simple example.
Here is a homogeneous linear system:
It represents a linear transformation from ℝ3↦ℝ3.
Let’s look at the matrix of this linear transformation.
The matrix will take a vector from ℝ3 (the domain) and
transform it into a different vector from ℝ3 (the co-domain).
x1  2x3  0
x 2  x3  0
x1  x2  x3  0
1 0 2 
A  0 1  1
1 1 1 
Next we will find the Column Space of A.
By definition, Col(A) is the span of the columns of A.
We have 3 columns, but as we will see, the column space
will be a 2-dimensional subspace of ℝ3.
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A simple example.
Here is a homogeneous linear system:
It represents a linear transformation from ℝ3↦ℝ3.
Let’s look at the matrix of this linear transformation.
The matrix will take a vector from ℝ3 (the domain) and
transform it into a different vector from ℝ3 (the co-domain).
x1  2x3  0
x 2  x3  0
x1  x2  x3  0
1 0 2 
A  0 1  1
1 1 1 
Next we will find the Column Space of A.
By definition, Col(A) is the span of the columns of A.
We have 3 columns, but as we will see, the column space
will be a 2-dimensional subspace of ℝ3.
1 0  2  


Col(A)  span0, 1,  1 
1 1  1  
      
This set of vectors is linearly dependent (we got a row of
zeroes when we row-reduced). So one of them is redundant.
We can get the same span by leaving out the column vector
that corresponds to our free variable in the row-reduced
matrix (this is the 3rd column).
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A simple example.
Here is a homogeneous linear system:
It represents a linear transformation from ℝ3↦ℝ3.
Let’s look at the matrix of this linear transformation.
The matrix will take a vector from ℝ3 (the domain) and
transform it into a different vector from ℝ3 (the co-domain).
x1  2x3  0
x 2  x3  0
x1  x2  x3  0
1 0 2 
A  0 1  1
1 1 1 
Next we will find the Column Space of A.
By definition, Col(A) is the span of the columns of A.
We have 3 columns, but as we will see, the column space
will be a 2-dimensional subspace of ℝ3.
1 0  2  


Col(A)  span0, 1,  1 
1 1  1  
      
1 0 


Col(A)  span0, 1 
1 1 
    
This set of vectors is linearly dependent (we got a row of
zeroes when we row-reduced). So one of them is redundant.
We can get the same span by leaving out the column vector
that corresponds to our free variable in the row-reduced
matrix (this is the 3rd column).
So our pared-down set only has 2 vectors, and the span of this
set is a 2-dimensional subspace of ℝ3 (i.e. a plane).
This new set of 2 vectors is a basis for the column space of A.
A graph is on the next slide.
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(1,0,1)
(0,1,1)
The column space of A is this plane in ℝ3.
The basis vectors are shown in the diagram.
All points in Col(A) are linear combinations of these basis vectors.
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Another simple example.
The matrix of a linear transformation from ℝ5↦ℝ3 is given.
The matrix will take a vector from ℝ5 (the domain) and
transform it into a different vector from ℝ3 (the co-domain).
1 0  2 0 3 
A  0 1 2 0 1 
0 0 0 1  2
This matrix is already in RREF. Notice that columns 1,2 and 4
are pivots, and columns 3 and 5 represent free variables.
We can find Nul(A) by writing down the parametric
form of the solution.
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Another simple example.
The matrix of a linear transformation from ℝ5↦ℝ3 is given.
The matrix will take a vector from ℝ5 (the domain) and
transform it into a different vector from ℝ3 (the co-domain).
1 0  2 0 3 
A  0 1 2 0 1 
0 0 0 1  2
This matrix is already in RREF. Notice that columns 1,2 and 4
are pivots, and columns 3 and 5 represent free variables.
We can find Nul(A) by writing down the parametric
form of the solution.
There are 2 parameters (s is for x3 and t is for x5).
Thus this is a 2-dimensional subspace of ℝ5.
 x1   2 
   
x2   2
x3    1   s 
   
x 4   0 
 x5   0 
 3
  1
 
 0 t
 
2 
 1 
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Assistance Services at UCSB
Another simple example.
The matrix of a linear transformation from ℝ5↦ℝ3 is given.
The matrix will take a vector from ℝ5 (the domain) and
transform it into a different vector from ℝ3 (the co-domain).
1 0  2 0 3 
A  0 1 2 0 1 
0 0 0 1  2
This matrix is already in RREF. Notice that columns 1,2 and 4
are pivots, and columns 3 and 5 represent free variables.
We can find Nul(A) by writing down the parametric
form of the solution.
There are 2 parameters (s is for x3 and t is for x5).
Thus this is a 2-dimensional subspace of ℝ5.
 x1   2 
   
x2   2
x3    1   s 
   
x 4   0 
 x5   0 
 3
  1
 
 0 t
 
2 
 1 
For the column space, we need the span of the 5
column vectors. In this case we have a set of 5
1 0 0 
vectors from ℝ3. For a basis of Col(A) we don’t need


to include columns 3 and 5, because they
Col(A)  span0, 1, 0   R 3
corresponded to free variables in our reduced matrix.


We have 3 independent vectors in ℝ3, so the span is a
3-dimensional subspace of ℝ3 (i.e. all of ℝ3).
0 0 1 
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