# Understanding Linear Algebra

## Section3.5Subspaces

In this chapter, we have been looking at bases for $$\real^p\text{,}$$ sets of vectors that are linearly independent and span $$\real^p\text{.}$$ Frequently, however, we focus on only a subset of $$\real^p\text{.}$$ In particular, if we are given an $$m\times n$$ matrix $$A\text{,}$$ we have been interested in both the span of the columns of $$A$$ and the solution space to the homogeneous equation $$A\xvec = \zerovec\text{.}$$ In this section, we will expand the concept of basis to describe sets like these.

### Preview Activity3.5.1.

Let’s consider the following matrix $$A$$ and its reduced row echelon form.
\begin{equation*} A = \left[\begin{array}{rrrr} 2 \amp -1 \amp 2 \amp 3 \\ 1 \amp 0 \amp 0 \amp 2 \\ -2 \amp 2 \amp -4 \amp -2 \\ \end{array}\right] \sim \left[\begin{array}{rrrr} 1 \amp 0 \amp 0 \amp 2 \\ 0 \amp 1 \amp -2 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation*}
1. Are the columns of $$A$$ linearly independent? Is the span of the columns $$\real^3\text{?}$$
2. Give a parametric description of the solution space to the homogeneous equation $$A\xvec = \zerovec\text{.}$$
3. Explain how this parametric description produces two vectors $$\wvec_1$$ and $$\wvec_2$$ whose span is the solution space to the equation $$A\xvec = \zerovec\text{.}$$
4. What can you say about the linear independence of the set of vectors $$\wvec_1$$ and $$\wvec_2\text{?}$$
5. Let’s denote the columns of $$A$$ as $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ $$\vvec_3\text{,}$$ and $$\vvec_4\text{.}$$ Explain why $$\vvec_3$$ and $$\vvec_4$$ can be written as linear combinations of $$\vvec_1$$ and $$\vvec_2\text{.}$$
6. Explain why $$\vvec_1$$ and $$\vvec_2$$ are linearly independent and $$\laspan{\vvec_1,\vvec_2} = \laspan{\vvec_1, \vvec_2, \vvec_3, \vvec_4}\text{.}$$

### Subsection3.5.1Subspaces

Our goal is to develop a common framework for describing subsets like the span of the columns of a matrix and the solution space to a homogeneous equation. That leads us to the following definition.

#### Definition3.5.1.

A subspace of $$\real^p$$ is a subset of $$\real^p$$ that is the span of a set of vectors.
Since we have explored the concept of span in some detail, this definition just gives us a new word to describe something familiar. Let’s look at some examples.

#### Example3.5.2.Subspaces of $$\real^3$$.

In Activity 2.3.3 and the following discussion, we looked at subspaces in $$\real^3$$ without explicitly using that language. Let’s recall some of those examples.
• Suppose we have a single nonzero vector $$\vvec\text{.}$$ The span of $$\vvec$$ is a subspace, which we’ll write as $$S = \laspan{\vvec}\text{.}$$ As we have seen, the span of a single vector consists of all scalar multiples of that vector, and these form a line passing through the origin. • If instead we have two linearly independent vectors $$\vvec_1$$ and $$\vvec_2\text{,}$$ the subspace $$S=\laspan{\vvec_1,\vvec_2}$$ is a plane passing through the origin. • Consider the three vectors $$\evec_1\text{,}$$ $$\evec_2\text{,}$$ and $$\evec_3\text{.}$$ Since we know that every 3-dimensional vector can be written as a linear combination, we have $$S = \laspan{\evec_1, \evec_2, \evec_3} = \real^3\text{.}$$
• One more subspace worth mentioning is $$S=\laspan{\zerovec}\text{.}$$ Since any linear combination of the zero vector is itself the zero vector, this subspace consists of a single vector, $$\zerovec\text{.}$$
In fact, any subspace of $$\real^3$$ is one of these types: the origin, a line, a plane, or all of $$\real^3\text{.}$$

#### Activity3.5.2.

We will look at some sets of vectors and the subspaces they form.
1. If $$\vvec_1, \vvec_2,\ldots,\vvec_n$$ is a set of vectors in $$\real^m\text{,}$$ explain why $$\zerovec$$ can be expressed as a linear combination of these vectors. Use this fact to explain why the zero vector $$\zerovec$$ belongs to any subspace in $$\real^m\text{.}$$
2. Explain why the line on the left of Figure 3.5.3 is not a subspace of $$\real^2$$ and why the line on the right is. Figure 3.5.3. Two lines in $$\real^2\text{,}$$ one of which is a subspace and one of which is not.
3. Consider the vectors
\begin{equation*} \vvec_1=\threevec101,~~~ \vvec_2=\threevec011,~~~ \vvec_3=\threevec110, \end{equation*}
and describe the subspace $$S=\laspan{\vvec_1,\vvec_2,\vvec_3}$$ of $$\real^3\text{.}$$


4. Consider the vectors
\begin{equation*} \wvec_1=\threevec210,~~~ \wvec_2=\threevec{-1}1{-1},~~~ \wvec_3=\threevec03{-2} \end{equation*}
1. Write $$\wvec_3$$ as a linear combination of $$\wvec_1$$ and $$\wvec_2\text{.}$$
2. Explain why $$\laspan{\wvec_1,\wvec_2,\wvec_3} = \laspan{\wvec_1, \wvec_2}\text{.}$$
3. Describe the subspace $$S = \laspan{\wvec_1,\wvec_2,\wvec_3}$$ of $$\real^3\text{.}$$
5. Suppose that $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ $$\vvec_3\text{,}$$ and $$\vvec_4$$ are four vectors in $$\real^3$$ and that
\begin{equation*} \begin{bmatrix} \vvec_1 \amp \vvec_2 \amp \vvec_3 \amp \vvec_4 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 2 \amp 0 \amp -2 \\ 0 \amp 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{bmatrix}. \end{equation*}
Give a description of the subspace $$S=\laspan{\vvec_1,\vvec_2,\vvec_3,\vvec_4}$$ of $$\real^3\text{.}$$
As the activity shows, it is possible to represent some subspaces as the span of more than one set of vectors. We are particularly interested in representing a subspace as the span of a linearly independent set of vectors.

#### Definition3.5.4.

A basis for a subspace $$S$$ of $$\real^p$$ is a set of vectors in $$S$$ that are linearly independent and whose span is $$S\text{.}$$ We say that the dimension of the subspace $$S\text{,}$$ denoted $$\dim S\text{,}$$ is the number of vectors in any basis.

#### Example3.5.5.A subspace of $$\real^4$$.

Suppose we have the 4-dimensional vectors $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ and $$\vvec_3$$ that define the subspace $$S = \laspan{\vvec_1,\vvec_2,\vvec_3}$$ of $$\real^4\text{.}$$ Suppose also that
\begin{equation*} \begin{bmatrix} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{bmatrix} \sim \begin{bmatrix} 1 \amp -1 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ \end{bmatrix}. \end{equation*}
From the reduced row echelon form of the matrix, we see that $$\vvec_2 = -\vvec\text{.}$$ Therefore, any linear combination of $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ and $$\vvec_3$$ can be rewritten
\begin{equation*} c_1\vvec_1 + c_2\vvec_2 + c_3 \vvec_3 = (c_1-c_2)\vvec_1 + c_2\vvec_3 \end{equation*}
as a linear combination of $$\vvec_1$$ and $$\vvec_3\text{.}$$ This tells us that
\begin{equation*} S = \laspan{\vvec_1,\vvec_2,\vvec_3} = \laspan{\vvec_1,\vvec_3}. \end{equation*}
Furthermore, the reduced row echelon form of the matrix shows that $$\vvec_1$$ and $$\vvec_3$$ are linearly independent. Therefore, $$\{\vvec_1,\vvec_3\}$$ is a basis for $$S\text{,}$$ which means that $$S$$ is a two-dimensional subspace of $$\real^4\text{.}$$
Subspaces of $$\real^3$$ are either
• 0-dimensional, consisting of the single vector $$\zerovec\text{,}$$
• a 1-dimensional line,
• a 2-dimensional plane, or
• the 3-dimensional subspace $$\real^3\text{.}$$
There is no 4-dimensional subspace of $$\real^3$$ because there is no linearly independent set of four vectors in $$\real^3\text{.}$$
There are two important subspaces associated to any matrix, each of which springs from one of our two fundamental questions, as we will now see.

### Subsection3.5.2The column space of $$A$$

The first subspace associated to a matrix that we’ll consider is its column space.

#### Definition3.5.6.

If $$A$$ is an $$m\times n$$ matrix, we call the span of its columns the column space of $$A$$ and denote it as $$\col(A)\text{.}$$
Notice that the columns of $$A$$ are vectors in $$\real^m\text{,}$$ which means that any linear combination of the columns is also in $$\real^m\text{.}$$ Since the column space is described as the span of a set of vectors, we see that $$\col(A)$$ is a subspace of $$\real^m\text{.}$$

#### Activity3.5.3.

We will explore some column spaces in this activity.
1. Consider the matrix
\begin{equation*} A= \left[\begin{array}{rrr} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{array}\right] = \left[\begin{array}{rrr} 1 \amp 3 \amp -1 \\ -2 \amp 0 \amp -4 \\ 1 \amp 2 \amp 0 \\ \end{array}\right]. \end{equation*}
Since $$\col(A)$$ is the span of the columns, we have
\begin{equation*} \col(A) = \laspan{\vvec_1,\vvec_2,\vvec_3}. \end{equation*}
Explain why $$\vvec_3$$ can be written as a linear combination of $$\vvec_1$$ and $$\vvec_2$$ and why $$\col(A)=\laspan{\vvec_1,\vvec_2}\text{.}$$
2. Explain why the vectors $$\vvec_1$$ and $$\vvec_2$$ form a basis for $$\col(A)$$ and why $$\col(A)$$ is a 2-dimensional subspace of $$\real^3$$ and therefore a plane.
3. Now consider the matrix $$B$$ and its reduced row echelon form:
\begin{equation*} B = \left[\begin{array}{rrrr} -2 \amp -4 \amp 0 \amp 6 \\ 1 \amp 2 \amp 0 \amp -3 \\ \end{array}\right] \sim \left[\begin{array}{rrrr} 1 \amp 2 \amp 0 \amp -3 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]. \end{equation*}
Explain why $$\col(B)$$ is a 1-dimensional subspace of $$\real^2$$ and is therefore a line.
4. For a general matrix $$A\text{,}$$ what is the relationship between the dimension $$\dim~\col(A)$$ and the number of pivot positions in $$A\text{?}$$
5. How does the location of the pivot positions indicate a basis for $$\col(A)\text{?}$$
6. If $$A$$ is an invertible $$9\times9$$ matrix, what can you say about the column space $$\col(A)\text{?}$$
7. Suppose that $$A$$ is an $$8\times 10$$ matrix and that $$\col(A) = \real^8\text{.}$$ If $$\bvec$$ is an 8-dimensional vector, what can you say about the equation $$A\xvec = \bvec\text{?}$$

#### Example3.5.7.

Consider the matrix $$A$$ and its reduced row echelon form:
\begin{equation*} A = \left[\begin{array}{rrrrr} 2 \amp 0 \amp -4 \amp -6 \amp 0 \\ -4 \amp -1 \amp 7 \amp 11 \amp 2 \\ 0 \amp -1 \amp -1 \amp -1 \amp 2 \\ \end{array}\right] \sim \left[\begin{array}{rrrrr} 1 \amp 0 \amp -2 \amp -3 \amp 0 \\ 0 \amp 1 \amp 1 \amp 1 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right], \end{equation*}
and denote the columns of $$A$$ as $$\vvec_1,\vvec_2,\ldots,\vvec_5\text{.}$$
It is certainly true that $$\col(A) = \laspan{\vvec_1,\vvec_2,\ldots,\vvec_5}$$ by the definition of the column space. However, the reduced row echelon form of the matrix shows us that the vectors are not linearly independent so $$\vvec_1,\vvec_2,\ldots,\vvec_5$$ do not form a basis for $$\col(A)\text{.}$$
From the reduced row echelon form, however, we can see that
\begin{equation*} \begin{aligned} \vvec_3 \amp {}={} -2\vvec_1 + \vvec_2 \\ \vvec_4 \amp {}={} -3\vvec_1 + \vvec_2 \\ \vvec_5 \amp {}={} -2\vvec_2 \\ \end{aligned}\text{.} \end{equation*}
This means that any linear combination of $$\vvec_1,\vvec_2,\ldots,\vvec_5$$ can be written as a linear combination of just $$\vvec_1$$ and $$\vvec_2\text{.}$$ Therefore, we see that $$\col(A) = \laspan{\vvec_1,\vvec_2}\text{.}$$
Moreover, the reduced row echelon form shows that $$\vvec_1$$ and $$\vvec_2$$ are linearly independent, which implies that they form a basis for $$\col(A)\text{.}$$ This means that $$\col(A)$$ is a 2-dimensional subspace of $$\real^3\text{,}$$ which is a plane in $$\real^3\text{,}$$ having basis
In general, a column without a pivot position can be written as a linear combination of the columns that have pivot positions. This means that a basis for $$\col(A)$$ will always be given by the columns of $$A$$ having pivot positions. This leads us to the following definition and proposition.

#### Definition3.5.8.

The rank of a matrix $$A$$ is the number of pivot positions in $$A$$ and is denoted by $$\rank(A)\text{.}$$
For example, the rank of the matrix $$A$$ in Example 3.5.7 is two because there are two pivot positions. A basis for $$\col(A)$$ is given by the first two columns of $$A$$ since those columns have pivot positions.
As a note of caution, we determine the pivot positions by looking at the reduced row echelon form of $$A\text{.}$$ However, we form a basis of $$\col(A)$$ from the columns of $$A$$ rather than the columns of the reduced row echelon matrix.

### Subsection3.5.3The null space of $$A$$

The second subspace associated to a matrix is its null space.

#### Definition3.5.10.

If $$A$$ is an $$m\times n$$ matrix, we call the subset of vectors $$\xvec$$ in $$\real^n$$ satisfying $$A\xvec = \zerovec$$ the null space of $$A$$ and denote it by $$\nul(A)\text{.}$$
Remember that a subspace is a subset that can be represented as the span of a set of vectors. The column space of $$A\text{,}$$ which is simply the span of the columns of $$A\text{,}$$ fits this definition. It may not be immediately clear how the null space of $$A\text{,}$$ which is the solution space of the equation $$A\xvec = \zerovec\text{,}$$ does, but we will see that $$\nul(A)$$ is a subspace of $$\real^n\text{.}$$

#### Activity3.5.4.

We will explore some null spaces in this activity and see why $$\nul(A)$$ satisfies the definition of a subspace.
1. Consider the matrix
\begin{equation*} A=\begin{bmatrix} 1 \amp 3 \amp -1 \amp 2 \\ -2 \amp 0 \amp -4 \amp 2 \\ 1 \amp 2 \amp 0 \amp 1 \end{bmatrix} \end{equation*}
and give a parametric description of the solution space to the equation $$A\xvec = \zerovec\text{.}$$ In other words, give a parametric description of $$\nul(A)\text{.}$$


2. This parametric description shows that the vectors satisfying the equation $$A\xvec=\zerovec$$ can be written as a linear combination of a set of vectors. In other words, this description shows why $$\nul(A)$$ is the span of a set of vectors and is therefore a subspace. Identify a set of vectors whose span is $$\nul(A)\text{.}$$
3. Use this set of vectors to find a basis for $$\nul(A)$$ and state the dimension of $$\nul(A)\text{.}$$
4. The null space $$\nul(A)$$ is a subspace of $$\real^p$$ for which value of $$p\text{?}$$
5. Now consider the matrix $$B$$ whose reduced row echelon form is given by
\begin{equation*} B \sim \left[\begin{array}{rrrr} 1 \amp 2 \amp 0 \amp -3 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]. \end{equation*}
Give a parametric description of $$\nul(B)\text{.}$$
6. The parametric description gives a set of vectors that span $$\nul(B)\text{.}$$ Explain why this set of vectors is linearly independent and hence forms a basis. What is the dimension of $$\nul(B)\text{?}$$
7. For a general matrix $$A\text{,}$$ how does the number of pivot positions indicate the dimension of $$\nul(A)\text{?}$$
8. Suppose that the columns of a matrix $$A$$ are linearly independent. What can you say about $$\nul(A)\text{?}$$

#### Example3.5.11.

Consider the matrix $$A$$ along with its reduced row echelon form:
\begin{equation*} A = \left[\begin{array}{rrrrr} 2 \amp 0 \amp -4 \amp -6 \amp 0 \\ -4 \amp -1 \amp 7 \amp 11 \amp 2 \\ 0 \amp -1 \amp -1 \amp -1 \amp 2 \\ \end{array}\right] \sim \left[\begin{array}{rrrrr} 1 \amp 0 \amp -2 \amp -3 \amp 0 \\ 0 \amp 1 \amp 1 \amp 1 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation*}
To find a parametric description of the solution space to $$A\xvec=\zerovec\text{,}$$ imagine that we augment both $$A$$ and its reduced row echelon form by a column of zeroes, which leads to the equations
\begin{equation*} \begin{alignedat}{6} x_1 \amp \amp \amp {}-{} \amp 2x_3 \amp {}-{} \amp 3x_4 \amp \amp \amp {}={} \amp 0 \\ \amp \amp x_2 \amp {}+{} \amp x_3 \amp {}+{} \amp x_4 \amp {}-{} \amp 2x_5 \amp {}={} \amp 0. \\ \end{alignedat} \end{equation*}
Notice that $$x_3\text{,}$$ $$x_4\text{,}$$ and $$x_5$$ are free variables so we rewrite these equations as
\begin{equation*} \begin{aligned} x_1 \amp {}={} 2x_3 + 3x_4 \\ x_2 \amp {}={} -x_3 - x_4 + 2x_5. \\ \end{aligned} \end{equation*}
In vector form, we have
\begin{equation*} \begin{aligned} \xvec \amp {}={} \fivevec{x_1}{x_2}{x_3}{x_4}{x_5} = \fivevec{2x_3 + 3x_4}{-x_3-x_4+2x_5}{x_3}{x_4}{x_5} \\ \\ \amp {}={} x_3\fivevec{2}{-1}{1}{0}{0} +x_4\fivevec{3}{-1}{0}{1}{0} +x_5\fivevec{0}{2}{0}{0}{1}. \end{aligned} \end{equation*}
This expression says that any vector $$\xvec$$ satisfying $$A\xvec= \zerovec$$ is a linear combination of the vectors
\begin{equation*} \vvec_1 = \fivevec{2}{-1}{1}{0}{0},~~~ \vvec_2 = \fivevec{3}{-1}{0}{1}{0},~~~ \vvec_3 = \fivevec{0}{2}{0}{0}{1}. \end{equation*}
It is straightforward to check that these vectors are linearly independent, which means that $$\vvec_1\text{,}$$ $$\vvec_2\text{,}$$ and $$\vvec_3$$ form a basis for $$\nul(A)\text{,}$$ a 3-dimensional subspace of $$\real^5\text{.}$$
As illustrated in this example, the dimension of $$\nul(A)$$ is equal to the number of free variables in the equation $$A\xvec=\zerovec\text{,}$$ which equals the number of columns of $$A$$ without pivot positions or the number of columns of $$A$$ minus the number of pivot positions.
Combining Proposition 3.5.9 and Proposition 3.5.12 shows that

### Subsection3.5.4Summary

Once again, we find ourselves revisiting our two fundamental questions concerning the existence and uniqueness of solutions to linear systems. The column space $$\col(A)$$ contains all the vectors $$\bvec$$ for which the equation $$A\xvec = \bvec$$ is consistent. The null space $$\nul(A)$$ is the solution space to the equation $$A\xvec = \zerovec\text{,}$$ which reflects on the uniqueness of solutions to this and other equations.
• A subspace $$S$$ of $$\real^p$$ is a subset of $$\real^p$$ that can be represented as the span of a set of vectors. A basis of $$S$$ is a linearly independent set of vectors whose span is $$S\text{.}$$
• If $$A$$ is an $$m\times n$$ matrix, the column space $$\col(A)$$ is the span of the columns of $$A$$ and forms a subspace of $$\real^m\text{.}$$
• A basis for $$\col(A)$$ is found from the columns of $$A$$ that have pivot positions. The dimension is therefore $$\dim~\col(A) = \rank(A)\text{.}$$
• The null space $$\nul(A)$$ is the solution space to the homogeneous equation $$A\xvec = \zerovec$$ and is a subspace of $$\real^n\text{.}$$
• A basis for $$\nul(A)$$ is found through a parametric description of the solution space of $$A\xvec = \zerovec\text{,}$$ and we have that $$\dim~\nul(A) = n - \rank(A)\text{.}$$

### Exercises3.5.5Exercises

#### 1.

Suppose that $$A$$ and its reduced row echelon form are
\begin{equation*} A = \left[\begin{array}{rrrrrr} 0 \amp 2 \amp 0 \amp -4 \amp 0 \amp 6 \\ 0 \amp -4 \amp -1 \amp 7 \amp 0 \amp -16 \\ 0 \amp 6 \amp 0 \amp -12 \amp 3 \amp 15 \\ 0 \amp 4 \amp -1 \amp -9 \amp 0 \amp 8 \\ \end{array}\right] \sim \left[\begin{array}{rrrrrr} 0 \amp 1 \amp 0 \amp -2 \amp 0 \amp 3 \\ 0 \amp 0 \amp 1 \amp 1 \amp 0 \amp 4 \\ 0 \amp 0 \amp 0 \amp 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation*}
1. The null space $$\nul(A)$$ is a subspace of $$\real^p$$ for what $$p\text{?}$$ The column space $$\col(A)$$ is a subspace of $$\real^p$$ for what $$p\text{?}$$
2. What are the dimensions $$\dim~\nul(A)$$ and $$\dim~\col(A)\text{?}$$
3. Find a basis for the column space $$\col(A)\text{.}$$
4. Find a basis for the null space $$\nul(A)\text{.}$$

#### 2.

Suppose that
\begin{equation*} A = \left[\begin{array}{rrrr} 2 \amp 0 \amp -2 \amp -4 \\ -2 \amp -1 \amp 1 \amp 2 \\ 0 \amp -1 \amp -1 \amp -2 \end{array}\right]\text{.} \end{equation*}
1. Is the vector $$\threevec{0}{-1}{-1}$$ in $$\col(A)\text{?}$$
2. Is the vector $$\fourvec{2}{1}{0}{2}$$ in $$\col(A)\text{?}$$
3. Is the vector $$\threevec{2}{-2}{0}$$ in $$\nul(A)\text{?}$$
4. Is the vector $$\fourvec{1}{-1}{3}{-1}$$ in $$\nul(A)\text{?}$$
5. Is the vector $$\fourvec{1}{0}{1}{-1}$$ in $$\nul(A)\text{?}$$

#### 3.

Determine whether the following statements are true or false and provide a justification for your response. Unless otherwise stated, assume that $$A$$ is an $$m\times n$$ matrix.
1. If $$A$$ is a $$127\times 341$$ matrix, then $$\nul(A)$$ is a subspace of $$\real^{127}\text{.}$$
2. If $$\dim~\nul(A) = 0\text{,}$$ then the columns of $$A$$ are linearly independent.
3. If $$\col(A) = \real^m\text{,}$$ then $$A$$ is invertible.
4. If $$A$$ has a pivot position in every column, then $$\nul(A) = \real^n\text{.}$$
5. If $$\col(A) = \real^m$$ and $$\nul(A) = \{\zerovec\}\text{,}$$ then $$A$$ is invertible.

#### 4.

Explain why the following statements are true.
1. If $$B$$ is invertible, then $$\nul(BA) = \nul(A)\text{.}$$
2. If $$B$$ is invertible, then $$\col(AB) = \col(A)\text{.}$$
3. If $$A\sim A'\text{,}$$ then $$\nul(A) = \nul(A')\text{.}$$

#### 5.

For each of the following conditions, construct a $$3\times 3$$ matrix having the given properties.
1. $$\dim~\nul(A) = 0\text{.}$$
2. $$\dim~\nul(A) = 1\text{.}$$
3. $$\dim~\nul(A) = 2\text{.}$$
4. $$\dim~\nul(A) = 3\text{.}$$

#### 6.

Suppose that $$A$$ is a $$3\times 4$$ matrix.
1. Is it possible that $$\dim~\nul(A) = 0\text{?}$$
2. If $$\dim~\nul(A) = 1\text{,}$$ what can you say about $$\col(A)\text{?}$$
3. If $$\dim~\nul(A) = 2\text{,}$$ what can you say about $$\col(A)\text{?}$$
4. If $$\dim~\nul(A) = 3\text{,}$$ what can you say about $$\col(A)\text{?}$$
5. If $$\dim~\nul(A) = 4\text{,}$$ what can you say about $$\col(A)\text{?}$$

#### 7.

Consider the vectors
\begin{equation*} \vvec_1 = \threevec{2}{3}{-1}, \vvec_2 = \threevec{-1}{2}{4}, \wvec_1 = \fourvec{3}{-1}{1}{0}, \wvec_2 = \fourvec{-2}{4}{0}{1} \end{equation*}
and suppose that $$A$$ is a matrix such that $$\col(A)=\laspan{\vvec_1,\vvec_2}$$ and $$\nul(A) = \laspan{\wvec_1,\wvec_2}\text{.}$$
1. What are the dimensions of $$A\text{?}$$
2. Find such a matrix $$A\text{.}$$

#### 8.

Suppose that $$A$$ is an $$8\times 8$$ matrix and that $$\det A = 14\text{.}$$
1. What can you conclude about $$\nul(A)\text{?}$$
2. What can you conclude about $$\col(A)\text{?}$$

#### 9.

Suppose that $$A$$ is a matrix and there is an invertible matrix $$P$$ such that
\begin{equation*} A = P~\left[\begin{array}{rrr} 2 \amp 0 \amp 0 \\ 0 \amp -3 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{array}\right]~P^{-1}\text{.} \end{equation*}
1. What can you conclude about $$\nul(A)\text{?}$$
2. What can you conclude about $$\col(A)\text{?}$$

#### 10.

In this section, we saw that the solution space to the homogeneous equation $$A\xvec = \zerovec$$ is a subspace of $$\real^p$$ for some $$p\text{.}$$ In this exercise, we will investigate whether the solution space to another equation $$A\xvec = \bvec$$ can form a subspace.
Let’s consider the matrix
\begin{equation*} A = \left[\begin{array}{rr} 2 \amp -4 \\ -1 \amp 2 \\ \end{array}\right]\text{.} \end{equation*}
1. Find a parametric description of the solution space to the homogeneous equation $$A\xvec = \zerovec\text{.}$$
2. Graph the solution space to the homogeneous equation to the right. 3. Find a parametric description of the solution space to the equation $$A\xvec = \twovec{4}{-2}$$ and graph it above.
4. Is the solution space to the equation $$A\xvec = \twovec{4}{-2}$$ a subspace of $$\real^2\text{?}$$
5. Find a parametric description of the solution space to the equation $$A\xvec=\twovec{-8}{4}$$ and graph it above.
6. What can you say about all the solution spaces to equations of the form $$A\xvec = \bvec$$ when $$\bvec$$ is a vector in $$\col(A)\text{?}$$
7. Suppose that the solution space to the equation $$A\xvec = \bvec$$ forms a subspace. Explain why it must be true that $$\bvec = \zerovec\text{.}$$