1. Vector Spaces Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

2. The sum of u and v, denoted by u+v , is in V. u+v=v+u.
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.
The sum of u and v, denoted by u+v , is in V.
u+v=v+u.
(u+v)+w=u+(v+w) .
There is a zero vector 0 in V such that u+(-u)=0.
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
a subspace of the familiar vector space.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

3. 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}.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

4. 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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

5. 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).
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

6. 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).
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:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

7. 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).
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:
t is the parameter of our explicit solution. Our solution space represents a line in ℝ3.
On the next slide is a graph of the situation.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

8. Domain (ℝ3) Co-Domain (ℝ3) (0,0,0)
This line is Nul(A)
(0,0,0)
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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

9. 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).
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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

10. 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).
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.
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).
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

11. 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).
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.
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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

12. The column space of A is this plane in ℝ3.
(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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

13. 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).
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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

14. 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).
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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

15. 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).
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.
For the column space, we need the span of the 5 column vectors. In this case we have a set of 5 vectors from ℝ3. For a basis of Col(A) we don’t need to include columns 3 and 5, because they 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).
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB