Skip to main content

Section 4.6 Subspaces associated to matrices

In Section 3.4 we introduced the notion of a subspace of \(\mathbb{R}^n\text{.}\) In this section we will describe three important subspaces associated to any matrix, and also see how understanding these subspaces can shed light on some earlier topics from the course.

Subsection 4.6.1 The row space

Definition 4.6.1.

Let \(A\) be an \(m \times n\) matrix. The row space of \(A\text{,}\) denoted \(\row(A)\text{,}\) is defined to be the span of the rows of \(A\text{.}\)

Note that \(\row(A)\) is a subspace of \(\mathbb{R}^n\text{,}\) by Theorem 3.4.6.

It follows from the definition of the reduced row echelon form that the non-zero rows of \(\RREF(A)\) form a basis for \(\row(\RREF(A))\text{,}\) so all we need to show is that \(\row(\RREF(A)) = \row(A)\text{.}\) To do that, it suffices to show that if we perform one row operation on \(A\) then we do not change the span of the rows. The verification that this is true is left as an exercise.

Find a basis for \(\row(A)\text{,}\) where \(A = \begin{bmatrix}1 \amp 3 \amp -2 \amp 4 \\ 4 \amp -4 \amp 3 \amp 2 \\ 3 \amp -7 \amp 5 \amp -2 \\ 1 \amp -1 \amp 2 \amp -2\end{bmatrix}\text{.}\)

Solution.

We row reduce:

\begin{equation*} A = \begin{bmatrix}1 \amp 3 \amp -2 \amp 4 \\ 4 \amp -4 \amp 3 \amp 2 \\ 3 \amp -7 \amp 5 \amp -2 \\ 1 \amp -1 \amp 2 \amp -2\end{bmatrix} \to \begin{bmatrix}1 \amp 0 \amp 0 \amp 3/2 \\ 0 \amp 1 \amp 0 \amp -1/2 \\ 0 \amp 0 \amp 1 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0\end{bmatrix}\text{.} \end{equation*}

By Theorem 4.6.2 a basis for \(\row(A)\) is \(\left\{\begin{bmatrix}1 \\ 0 \\ 0 \\ 3/2\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 0 \\ -1/2\end{bmatrix}, \begin{bmatrix}0 \\ 0\\ 1 \\-2\end{bmatrix}\right\}\text{.}\)

Note 4.6.4.

Warning: The rows of \(A\) corresponding to the non-zero rows of \(\RREF(A)\) are not necessarily a basis for \(\row(A)\text{.}\) Indeed, in the example above, the first three rows of \(A\) itself are \(not\) a basis for \(\row(A)\text{.}\)

By Theorem 4.6.2, \(\dim(\row(A))\) is equal to the number of non-zero rows of \(\RREF(A)\text{.}\) On the other hand, \(\rank(A)\) is also equal to the number of non-zero rows of \(\RREF(A)\text{,}\) by Lemma 3.3.3.

Subsection 4.6.2 The column space

Definition 4.6.6.

Let \(A\) be an \(m \times n\) matrix. The column space of \(A\text{,}\) denoted \(\col(A)\text{,}\) is defined to be the span of the columns of \(A\text{.}\)

As with the row space, Theorem 3.4.6 tells us that \(\col(A)\) is a subspace (this time it is a subspace of \(\mathbb{R}^m\)).

One way that we could look for a basis for \(\col(A)\) is to notice that \(\col(A) = \row(A^t)\text{,}\) so we could apply Theorem 4.6.2 to \(A^t\) to give us a basis for \(\col(A)\text{.}\) However, doing this requires row reducing \(A^t\text{.}\) It is useful to have a method for finding a basis for \(\col(A)\) that can be calculated from \(\RREF(A)\text{,}\) rather than \(\RREF(A^t)\text{.}\)

We won't give a complete proof of this fact, but the key idea is that when we do row operations we do not change the linear dependency relations between the columns. Thus, for instance, if in \(\RREF(A)\) the third column is equal to the sum of the first two then also in \(A\) the third column is the sum of the first two, and vice versa. From the definition of the reduced row echelon form we see that the pivot columns of \(\RREF(A)\) form a basis for \(\col(\RREF(A))\text{,}\) and therefore the corresponding columns of \(A\) form a basis for \(\col(A)\text{.}\)

Find a basis for \(\col(A)\text{,}\) where \(A = \begin{bmatrix} 2 \amp -1 \amp 5 \amp 3 \\ 1 \amp 1 \amp -2 \amp 1 \\ 4 \amp 0 \amp 4 \amp -1 \\ -1 \amp 1 \amp -4 \amp 0\end{bmatrix}\text{.}\)

Solution.

Start by row reducing:

\begin{equation*} A = \begin{bmatrix} 2 \amp -1 \amp 5 \amp 3 \\ 1 \amp 1 \amp -2 \amp 1 \\ 4 \amp 0 \amp 4 \amp -1 \\ -1 \amp 1 \amp -4 \amp 0\end{bmatrix} \to \begin{bmatrix} 1 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp -3 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0\end{bmatrix}\text{.} \end{equation*}

We see that columns \(1\text{,}\) \(2\text{,}\) and \(4\) are the pivot columns. Therefore, by Theorem 4.6.7, a basis for \(\col(A)\) is \(\left\{\begin{bmatrix}2\\1\\4\\-1\end{bmatrix}, \begin{bmatrix}-1\\1\\0\\1\end{bmatrix}, \begin{bmatrix}3\\1\\-1\\0\end{bmatrix}\right\}\text{.}\)

Note 4.6.9.

Warning: It usually happens that \(\col(A) \neq \col(\RREF(A))\text{,}\) so the pivot columns of \(\RREF(A)\) are not necessarily a basis for \(\col(A)\text{.}\)

Notice that the process for the row and column spaces are different, even though both use the reduced row echelon form. When finding a basis for the row space we use the rows of \(\RREF(A)\text{.}\) When finding a basis for the column space we use the columns of \(A\text{,}\) and use \(\RREF(A)\) to determine which columns to take.

From Theorem 4.6.7 we have that \(\dim(\col(A))\) is the number of pivot columns of \(A\text{,}\) which is by definition \(\rank(A)\text{.}\) Using Theorem 4.6.5 we therefore have

\begin{equation*} \dim(\col(A)) = \rank(A) = \dim(\row(A))\text{.} \end{equation*}

Note 4.6.11.

The conclusion of the above theorem is surprising! For instance, if \(A\) is a \(17 \times 3\) matrix then \(\col(A)\) is a subspace of \(\mathbb{R}^{17}\) and \(\row(A)\) is a subspace of \(\mathbb{R}^{3}\text{.}\) A priori it is surprising that these two subspaces should have anything to do with each other - they certainly do not contain the same vectors! Yet Theorem 4.6.10 tells us that they do have the same dimension. In particular, even though \(\col(A)\) is a subspace of \(\mathbb{R}^{17}\text{,}\) we know that \(\dim(\col(A)) = \dim(\row(A)) \leq 3\text{.}\)

\begin{equation*} \rank(A) = \dim(\col(A)) = \dim(\row(A^t)) = \rank(A^t)\text{.} \end{equation*}

The column space of a matrix has a nice geometric interpretation, though we won't spend much time on this point.

Suppose that \(T : \mathbb{R}^n \to \mathbb{R}^m\) is a linear transformation. The image of \(T\) is the collection of all of the vectors in \(\mathbb{R}^m\) that actually appear as outputs of \(T\text{;}\) that is, it is the set of all vectors of the form \(T(\vec{v})\) where \(\vec{v}\) is a vector in \(\mathbb{R}^n\text{.}\) Using the fact that \(T(\vec{v}) = [T]\vec{v}\) (Theorem 4.1.13) and the fact that multiplying a matrix by a vector gives a linear combination of the columns (Definition 4.1.12), one can show that the image of \(T\) is exactly the same as the column space of \([T]\text{.}\)

Subsection 4.6.3 The null space

Definition 4.6.13.

Let \(A\) be an \(m \times n\) matrix. The null space of \(A\text{,}\) denoted \(\NullSp(A)\text{,}\) is the collection of all vectors \(\vec{v}\) in \(\mathbb{R}^n\) such that \(A\vec{v} = \vec{0}\text{.}\)

We check the conditions of Definition 3.4.1.

First, since \(A\vec{0} = \vec{0}\) the vector \(\vec{0}\) is in \(\NullSp(A)\text{,}\) and hence \(\NullSp(A)\) is not empty.

Second, suppose that \(\vec{v}\) and \(\vec{w}\) are in \(\NullSp(A)\text{.}\) Then \(A\vec{v} = \vec{0}\) and \(A\vec{w} = \vec{0}\text{,}\) so

\begin{equation*} A(\vec{v} + \vec{w}) = A\vec{v}+A\vec{w} = \vec{0}+\vec{0} = \vec{0}\text{.} \end{equation*}

This shows that \(\vec{v}+\vec{w}\) is in \(\NullSp(A)\text{.}\)

Finally, suppose that \(\vec{v}\) is in \(\NullSp(A)\) and \(c\) is a scalar. Then \(A\vec{v} = \vec{0}\text{,}\) so

\begin{equation*} A(c\vec{v}) = cA\vec{v} = c\vec{0} = \vec{0}\text{.} \end{equation*}

This shows that \(c\vec{v}\) is in \(\NullSp(A)\text{.}\)

Definition 4.6.15.

Let \(A\) be an \(m \times n\) matrix. The nullity of \(A\) is \(\nullity(A) = \dim(\NullSp(A))\text{.}\)

We've seen both the null space and the nullity before, but they were in disguise. Notice that, by Theorem 4.1.19, we have that \(\vec{v}\) is in \(\NullSp(A)\) if and only if \(\vec{v}\) is a solution to the homogeneous system \([A|\vec{0}]\text{.}\) Thus to find a basis for \(\NullSp(A)\) we just take \([A|\vec{0}]\) to reduced row echelon form and extract a collection of linearly independent solutions, as we have done before. Moreover, as we have seen before, the number of free variables of the system will be the dimension of the space of solutions (see Section 3.3). These observations give us a reformulation of Theorem 3.3.4:

Let \(A = \begin{bmatrix}3 \amp 2 \amp -1 \amp 0 \\ 1 \amp 2 \amp -1 \amp 3\end{bmatrix}\text{.}\) Then \(\RREF(A) = \begin{bmatrix}1 \amp 0 \amp 0 \amp -3/2 \\ 0 \amp 1 \amp -1/2 \amp 9/4\end{bmatrix}\text{,}\) so we have that \(\begin{bmatrix}x\\y\\z\\w\end{bmatrix}\) is in \(\NullSp(A)\) if and only if \(x=\frac{3}{2}w\) and \(y = \frac{1}{2}z - \frac{9}{4}w\text{.}\) That is,

\begin{equation*} \begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}\frac{3}{2}w \\ \frac{1}{2}z - \frac{9}{4}w \\ z \\ w\end{bmatrix} = z\begin{bmatrix}0\\1/2\\1\\0\end{bmatrix} + w\begin{bmatrix}3/2 \\ -9/4 \\ 0 \\1\end{bmatrix}\text{.} \end{equation*}

Therefore \(\left\{\begin{bmatrix}0 \\ 1/2 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}3/2 \\ -9/4 \\ 0 \\ 1\end{bmatrix}\right\}\) is a basis for \(\NullSp(A)\text{,}\) and \(\nullity(A) = 2\text{.}\)

In Corollary 3.3.6 we proved that any system of linear equations has either no solution, exactly one solution, or infinitely many solutions. The material we have developed in this chapter allows us to give a different proof, which also gives some information about what the solutions to a system look like.

First, suppose that \(\vec{w}\) is a solution to \([A|\vec{b}]\text{.}\) Then by Theorem 4.1.19 we have \(A\vec{w} = \vec{b}\text{,}\) and since \(\vec{v}\) is also a solution we have \(A\vec{v} = \vec{b}\text{.}\) If we let \(\vec{z} = \vec{w}-\vec{v}\) and subtract the above equations, we get:

\begin{equation*} \vec{0} = \vec{b}-\vec{b} = A\vec{w} - A\vec{v} = A(\vec{w}-\vec{v}) = A\vec{z}\text{.} \end{equation*}

Thus \(\vec{z}\) is in \(\NullSp(A)\text{,}\) and \(\vec{v}+\vec{z} = \vec{v} + (\vec{w}-\vec{v}) = \vec{w}\text{.}\)

For the other direction, suppose that \(\vec{z}\) is in \(\NullSp(A)\text{.}\) By Theorem 4.1.19 we have \(A\vec{v} = \vec{b}\text{,}\) and so:

\begin{equation*} A(\vec{v}+\vec{z}) = A\vec{v} + A\vec{z} = \vec{b}+\vec{0} = \vec{b}\text{,} \end{equation*}

and so (again by Theorem 4.1.19) we have that \(\vec{v}+\vec{z}\) is a solution to \([A|\vec{b}]\text{.}\)

Suppose that a system \([A|\vec{b}]\) has more than one solution, say \(\vec{v_1}\) and \(\vec{v_2}\) are two different solutions. By Theorem 4.6.18 there is a vector \(\vec{z}\) in \(\NullSp(A)\) such that \(\vec{v_1} = \vec{v_2}+\vec{z}\text{.}\) For any scalar \(a\) the vector \(a\vec{z}\) is also in \(\NullSp(A)\text{,}\) because \(\NullSp(A)\) is a subspace. Therefore, by the other direction of Theorem 4.6.18, each of the vectors \(\vec{v_1} + a\vec{z}\) is a solution to \([A|\vec{b}]\text{.}\) Finally, since \(\vec{v_1} \neq \vec{v_2}\) we have \(\vec{z} \neq \vec{0}\text{,}\) so if \(a \neq b\) then \(\vec{v_1} + a\vec{z} \neq \vec{v_1} + b\vec{z}\text{.}\) Thus the vectors of the form \(\vec{v_1}+a\vec{z}\) are infinitely many different solutions to the system \([A|\vec{b}]\text{.}\)

Subsection 4.6.4 The Fundamental Theorem, again

In this section we introduced a lot of terminology related to ideas that already appeared in the Fundamental Theorem, so we take this opportunity to re-state that theorem. Each of the new items on this version of the theorem is just a rephrasing of something that was already on the list in Theorem 4.5.14.

The new entries in this version of the theorem are (5), (10), (11), and (13). (5) is just a rephrasing of (4), and (10) is a rephrasing of (9). (11) is equivalent to (2) because \(\row(A) = \mathbb{R}^n\) if and only if \(\dim(\row(A)) = n\) if and only if \(\rank(A) = n\) (by Theorem 4.6.5). Finally, (13) is equivalent to (12) by the Rank-Nullity Theorem.

Exercises 4.6.5 Exercises

1.

Determine the rank and nullity and find a basis of the column space, row space, and null space of each of the following matrices.

  1. \begin{equation*} A = \begin{bmatrix} 1 \amp 3 \amp 2 \\ 3 \amp 9 \amp 6 \\ 1 \amp 3 \amp 2 \end{bmatrix} \end{equation*}
    Solution.

    We row-reduce \(A\text{:}\)
    \begin{align*} \begin{bmatrix} 1 \amp 3 \amp 2 \\ 3 \amp 9 \amp 6 \\ 1 \amp 3 \amp 2 \end{bmatrix} \overset{-3R_{1}+R_{2}}{\underset{-R_{1}+R_{3}}{\longrightarrow}} \begin{bmatrix} 1 \amp 3 \amp 2 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{bmatrix}. \end{align*}
    We have found a row echelon form of \(A\) (actually, we have found its RREF). It has one non-zero row, so that
    \begin{gather*} \rank (A) = 1 \text{ and } \nullity (A) = 3-1 = 2. \end{gather*}
    Moreover, using Hint 4.6.5.1.3 for the row space and Hint 4.6.5.1.4 for the column space, we find the following bases:
    \begin{align*} \text{basis of } \row (A):\quad \amp \left\{\begin{bmatrix}1 \\ 3 \\ 2\end{bmatrix}\right\} \\ \text{basis of } \col (A):\quad \amp \left\{\begin{bmatrix}1 \\ 3 \\ 1\end{bmatrix}\right\}. \end{align*}
    Lastly, for the null space, we use Hint 4.6.5.1.1: From the RREF of \(A\text{,}\) we are looking for the solution set of
    \begin{equation*} \matr{ccc|c}{ 1 \amp 3 \amp 2 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 }. \end{equation*}
    We have that \(x_{2}\) and \(x_{3}\) are free variables, and \(x_{1}=-3x_{2}-2x_{3}\text{,}\) so that
    \begin{align*} \text{basis of } \NullSp (B):\quad \amp \left\{ \begin{bmatrix} -3 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1\end{bmatrix} \right\}. \end{align*}
    Answer.

    \begin{align*} \text{basis of } \row (A):\quad \amp \left\{\begin{bmatrix}1 \\ 3 \\ 2\end{bmatrix}\right\} ,\quad\rank (A) = 1\\ \text{basis of } \col (A):\quad \amp \left\{\begin{bmatrix}1 \\ 3 \\ 1\end{bmatrix}\right\}\\ \text{basis of } \NullSp (A):\quad \amp \left\{ \begin{bmatrix} -3 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1\end{bmatrix} \right\},\quad\nullity (A) = 2 \end{align*}

  2. \begin{equation*} B = \begin{bmatrix} 1 \amp 3 \amp 0 \amp 2 \\ 3 \amp 9 \amp 1 \amp 7 \\ 1 \amp 3 \amp 1 \amp 3 \end{bmatrix} \end{equation*}
    Solution.

    We row-reduce \(B\text{:}\)
    \begin{align*} \begin{bmatrix} 1 \amp 3 \amp 0 \amp 2 \\ 3 \amp 9 \amp 1 \amp 7 \\ 1 \amp 3 \amp 1 \amp 3 \end{bmatrix} \overset{-3R_{1}+R_{2}}{\underset{-R_{1}+R_{3}}{\longrightarrow}} \amp\begin{bmatrix} 1 \amp 3 \amp 0 \amp 2 \\ 0 \amp 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 1 \amp 1 \end{bmatrix}\\ \overset{-R_{2}+R_{3}}{\longrightarrow} \amp\begin{bmatrix} 1 \amp 3 \amp 0 \amp 2 \\ 0 \amp 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}. \end{align*}
    We have found a row echelon form of \(B\) (actually, we have found its RREF).It has two non-zero rows, so that
    \begin{gather*} \rank (B) = 2 \text{ and } \nullity (B) = 4-2 = 2. \end{gather*}
    Moreover, using Hint 4.6.5.1.3 for the row space and Hint 4.6.5.1.4 for the column space, we find the following bases:
    \begin{align*} \text{basis of } \row (B):\quad \amp \left\{\begin{bmatrix}1 \\ 3 \\ 0 \amp 2\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1 \\ 1\end{bmatrix}\right\} \\ \text{basis of } \col (B):\quad \amp \left\{ \begin{bmatrix}1 \\ 3 \\ 1\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix} \right\}. \end{align*}
    Lastly, for the null space, we use Hint 4.6.5.1.1: From the RREF of \(B\text{,}\) we are looking for the solution set of
    \begin{equation*} \matr{cccc|c}{ 1 \amp 3 \amp 0 \amp 2 \amp 0 \\ 0 \amp 0 \amp 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 } \end{equation*}
    We have that \(x_{2}\) and \(x_{4}\) are free variables, and \(x_{1}=-3x_{2}-2x_{4}\text{,}\) \(x_{3}=-x_{4}\text{,}\) so that
    \begin{align*} \text{basis of } \NullSp (B):\quad \amp \left\{ \begin{bmatrix} -3 \\ 1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ -1 \\ 1\end{bmatrix} \right\}. \end{align*}
    Answer.

    \begin{align*} \text{basis of } \row (B):\quad \amp \left\{\begin{bmatrix}1 \\ 3 \\ 0 \\ 2\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1 \\ 1\end{bmatrix}\right\}, \quad \rank (B) = 2\\ \text{basis of } \col (B):\quad \amp \left\{ \begin{bmatrix}1 \\ 3 \\ 1\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix} \right\}\\ \text{basis of } \NullSp (B):\quad \amp \left\{ \begin{bmatrix} -3 \\ 1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ -1 \\ 1\end{bmatrix} \right\} , \quad \nullity (B) = 2. \end{align*}

  3. \begin{equation*} C = \begin{bmatrix} 1 \amp 0 \amp 3 \\ 3 \amp 1 \amp 10 \\ 1 \amp 1 \amp 4 \\ 1 \amp -1 \amp 2 \end{bmatrix} \end{equation*}
    Solution.

    We row-reduce \(C\text{:}\)
    \begin{align*} \begin{bmatrix} 1 \amp 0 \amp 3 \\ 3 \amp 1 \amp 10 \\ 1 \amp 1 \amp 4 \\ 1 \amp -1 \amp 2 \end{bmatrix} \overset{-3R_{1}+R_{2}}{\underset{-R_{1}+R_{3}}{\longrightarrow}} \amp\begin{bmatrix} 1 \amp 0 \amp 3 \\ 0 \amp 1 \amp 1 \\ 0 \amp 1 \amp 1 \\ 1 \amp -1 \amp 2 \end{bmatrix}\\ \overset{-R_{1}+R_{4}}{\longrightarrow} \amp\begin{bmatrix} 1 \amp 0 \amp 3 \\ 0 \amp 1 \amp 1 \\ 0 \amp 1 \amp 1 \\ 0 \amp -1 \amp -1 \end{bmatrix}\\ \overset{R_{2}+R_{4}}{\underset{-R_{2}+R_{3}{\longrightarrow}} \amp\begin{bmatrix} 1 \amp 0 \amp 3 \\ 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{bmatrix}. \end{align*}
    We have found a row echelon form of \(C\) (actually, we have found its RREF). It has two non-zero rows, so that
    \begin{gather*} \rank (C) = 2 \text{ and } \nullity (C) = 3-2 = 1. \end{gather*}
    Moreover, using Hint 4.6.5.1.3 for the row space and Hint 4.6.5.1.4 for the column space, we find the following bases:
    \begin{align*} \text{basis of } \row (C):\quad \amp \left\{ \begin{bmatrix}1 \\ 0 \\ 3\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}\right\}\\ \text{basis of } \col (C):\quad \amp \left\{ \begin{bmatrix}1 \\ 3 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1 \\ -1\end{bmatrix} \right\}. \end{align*}
    Lastly, for the null space, we use Hint 4.6.5.1.1: From the RREF of \(C\text{,}\) we are looking for the solution set of
    \begin{equation*} \matr{ccc|c}{ 1 \amp 0 \amp 3 \amp 0 \\ 0 \amp 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 } \end{equation*}
    We have that \(x_{3}\) is a free variable, and \(x_{1}=-3x_{3}\text{,}\) \(x_{2}=-x_{3}\text{,}\) so that
    \begin{align*} \text{basis of } \NullSp (C):\quad \amp \left\{ \begin{bmatrix} -3 \\ -1 \\ 1\end{bmatrix} \right\}. \end{align*}
    Answer.

    \begin{align*} \text{basis of } \row (C):\quad \amp \left\{ \begin{bmatrix}1 \\ 0 \\ 3\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}\right\}, \quad \rank (C) = 2\\ \text{basis of } \col (C):\quad \amp \left\{ \begin{bmatrix}1 \\ 3 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1 \\ -1\end{bmatrix} \right\}.\\ \text{basis of } \NullSp (C):\quad \amp \left\{ \begin{bmatrix} -3 \\ -1 \\ 1\end{bmatrix} \right\}, \quad \nullity (C) = 1 \end{align*}

  4. \begin{equation*} D = \begin{bmatrix} 0 \amp 0 \amp -1 \amp 0 \amp 1 \\ 1 \amp 2 \amp 3 \amp -2 \amp -18\\ 1 \amp 2 \amp 2 \amp 1 \amp 11 \\ -1 \amp -2 \amp -2 \amp 1 \amp 11 \end{bmatrix} \end{equation*}
    Solution.

    We row-reduce \(D\text{:}\)
    \begin{align*} \begin{bmatrix} 0 \amp 0 \amp -1 \amp 0 \amp 1 \\ 1 \amp 2 \amp 3 \amp -2 \amp -18\\ 1 \amp 2 \amp 2 \amp 1 \amp 11 \\ -1 \amp -2 \amp -2 \amp 1 \amp 11 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 \amp 2 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \end{align*}
    The reduced row echelon form of \(D\) has four non-zero rows, so that
    \begin{gather*} \rank (D) = 4 \text{ and } \nullity (D) = 5-4 = 1. \end{gather*}
    Moreover, using Hint 4.6.5.1.3 for the row space and Hint 4.6.5.1.4 for the column space, we find the following bases:
    \begin{align*} \text{basis of } \row (D):\quad \amp \left\{ \begin{bmatrix}1 \\ 2 \\ 0 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1 \\ 0 \\ 0\end{bmatrix}, \right.\\ \amp \left. \begin{bmatrix}0 \\ 0 \\ 0 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{bmatrix} \right\}\\ \text{basis of } \col (D):\quad \amp \left\{ \begin{bmatrix}0 \\ 1 \\ 1 \\ -1\end{bmatrix}, \begin{bmatrix}-1 \\ 3 \\ 2 \\ -2\end{bmatrix}, \begin{bmatrix}0 \\ -2 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}1 \\ -18 \\ 11 \\ 11\end{bmatrix} \right\}. \end{align*}
    Lastly, for the null space, we use Hint 4.6.5.1.1: From the RREF of \(D\text{,}\) we are looking for the solution set of
    \begin{equation*} \matr{ccccc|c}{ 1 \amp 2 \amp 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 1 \amp 0 } \end{equation*}
    We have that \(x_{2}\) is the only free variable, and
    \begin{align*} \text{basis of } \NullSp (D):\quad \amp \left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix} \right\}. \end{align*}
    Answer.

    \begin{align*} \text{basis of } \row (D):\quad \amp left\{ \begin{bmatrix}1 \\ 2 \\ 0 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1 \\ 0 \\ 0\end{bmatrix}, \right.\\ \amp \left. \begin{bmatrix}0 \\ 0 \\ 0 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{bmatrix} \right\},\\ \amp \rank (D) = 4\\ \text{basis of } \col (D):\quad \amp \left\{ \begin{bmatrix}0 \\ 1 \\ 1 \\ -1\end{bmatrix}, \begin{bmatrix}-1 \\ 3 \\ 2 \\ -2\end{bmatrix}, \begin{bmatrix}0 \\ -2 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}1 \\ -18 \\ 11 \\ 11\end{bmatrix} \right\}\\ \text{basis of } \NullSp (D):\quad \amp \left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0\end{bmatrix} \right\}, \quad \nullity (D) = 1. \end{align*}

  5. \begin{equation*} E = \begin{bmatrix} 1 \amp 0 \amp 3 \amp 0 \\ 3 \amp 1 \amp 10 \amp 0 \\ -1 \amp 1 \amp -2 \amp 1\\ 1 \amp -1 \amp 2 \amp -2 \end{bmatrix} \end{equation*}
    Solution.

    We row-reduce \(E\text{:}\)
    \begin{align*} \begin{bmatrix} 1 \amp 0 \amp 3 \amp 0 \\ 3 \amp 1 \amp 10 \amp 0 \\ -1 \amp 1 \amp -2 \amp 1\\ 1 \amp -1 \amp 2 \amp -2 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 \amp 0 \amp 3 \amp 0 \\ 0 \amp 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \end{bmatrix} \end{align*}
    The reduced row echelon form of \(E\) has three non-zero rows, so that
    \begin{gather*} \rank (D) = 3 \text{ and } \nullity (D) = 4-3 = 1. \end{gather*}
    Moreover, using Hint 4.6.5.1.3 for the row space and Hint 4.6.5.1.4 for the column space, we find the following bases:
    \begin{align*} \text{basis of } \row (E):\quad \amp \left\{ \begin{bmatrix}1 \\ 0 \\ 3 \\ 0 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 0 \\ 1\end{bmatrix}\right\}\\ \text{basis of } \col (E):\quad \amp \left\{ \begin{bmatrix}1 \\ 3 \\ -1 \\ 1\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1 \\ -1\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1 \\ -2\end{bmatrix} \right\}. \end{align*}
    Lastly, for the null space, we use Hint 4.6.5.1.1: From the RREF of \(E\text{,}\) we are looking for the solution set of
    \begin{equation*} \matr{cccc|c}{ 1 \amp 0 \amp 3 \amp 0\amp 0 \\ 0 \amp 1 \amp 1 \amp 0\amp 0 \\ 0 \amp 0 \amp 0 \amp 1\amp 0 \\ 0 \amp 0 \amp 0 \amp 0\amp 0 } \end{equation*}
    We have that \(x_{3}\) is the only free variable, and \(x_{1}=-3x_{3}\) and \(x_{2}=-x_{3}\text{.}\) Thus,
    \begin{align*} \text{basis of } \NullSp (E):\quad \amp \left\{ \begin{bmatrix} -3 \\ -1 \\ 1 \\ 0 \end{bmatrix} \right\}. \end{align*}
    Answer.

    \begin{align*} \text{basis of } \row (E):\quad \amp \left\{ \begin{bmatrix}1 \\ 0 \\ 3 \\ 0 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 0 \\ 1\end{bmatrix}\right\},\\ \amp \rank (E) = 3\\ \text{basis of } \col (E):\quad \amp \left\{ \begin{bmatrix}1 \\ 3 \\ -1 \\ 1\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1 \\ -1\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1 \\ -2\end{bmatrix} \right\}\\ \text{basis of } \NullSp (E):\quad \amp \left\{ \begin{bmatrix} -3 \\ -1 \\ 1 \\ 0 \end{bmatrix} \right\}, \quad \nullity (E) = 1 \end{align*}

Hint 1.
The null space of \(A\) is the solution space of the homogeneous system \(A\mathbf{x}=\mathbf{0}\text{.}\) Thus, finding a basis of the null space is the same as finding a set of basic solutions. From the reduced echelon form, we can easily find the general solution of \(A\mathbf{x}=\mathbf{0}\text{,}\) using parameters corresponding to the non-pivot columns of \(\RREF(A)\text{.}\)
Hint 2.
Let \(A\) be a matrix and consider any echelon form of \(A\text{.}\) Then, the number of pivot entries of \(A\) does not depend on the echelon form we choose, and is called the rank of \(A\). We denote it by \(\rank(A)\text{.}\) The dimension of the null space is called the nullity of \(A\text{,}\) and is written \(\nullity(A)\text{.}\)
Hint 3.
Let \(A\) and \(B\) be row equivalent matrices. Then \(\row(A) =\row(B)\) and \(\NullSp(A) =\NullSp(B)\text{.}\) In particular, the non-zero rows of a row echelon form of \(A\) form a basis of the row space, \(\row(A)\text{.}\)
Hint 4.
Casting-out algorithm:

  1. Input: a list of \(k\) vectors \(u_1,...,u_k\in \mathbb{R}^n\text{.}\)

  2. Output: the set of indices \(j\) such that \(u_j\) is redundant.

  3. Algorithm: Write the vectors \(u_1,\ldots,u_k\) as the columns of an \(n\times k\) matrix, and reduce to echelon form. Every non-pivot column, if any, corresponds to a redundant vector.

2.

Find a basis and calculate the dimension of the following subspaces of \(\mathbb{R}^4 \text{.}\)

  1. \begin{equation*} \text{span}\{ (1,-1,2,0), (2,3,0,3), (1,9,-6,6)\} \end{equation*}
    Solution.

    We will write the vectors as rows of a matrix and then use Hint 4.6.5.1.3 to determine a basis of its row space. We row-reduce:
    \begin{align*} \matr{cccc}{ 1 \amp -1 \amp 2 \amp 0 \\ 2 \amp 3 \amp 0 \amp 3 \\ 1 \amp 9 \amp -6 \amp 6 } \overset{-2R_{1}+R_{2}}{\underset{-R_{1}+R_{3}}{\longrightarrow}} \amp\matr{cccc}{ 1 \amp -1 \amp 2 \amp 0 \\ 0 \amp 5 \amp -4 \amp 3 \\ 0 \amp 10 \amp -8 \amp 6 }\\ \overset{-2R_{2}+R_{3}}{\longrightarrow} \amp\matr{cccc}{ 1 \amp -1 \amp 2 \amp 0 \\ 0 \amp 5 \amp -4 \amp 3 \\ 0 \amp 0 \amp 0 \amp 0 }. \end{align*}
    We see that the rank of the matrix and thus the dimension of the subspace is \(2\text{.}\) Moreover, a basis for the space in question is given by
    \begin{equation*} \left\{ (1 , -1 , 2 , 0), (0 , 5 , -4 , 3)\right\}. \end{equation*}
    Answer.

    The dimension is \(2\) and a basis is given by
    \begin{equation*} \left\{ (1 , -1 , 2 , 0), (0 , 5 , -4 , 3)\right\}. \end{equation*}

  2. \begin{equation*} \text{span}\{(2,1,0,-1), (-1,1,1,1), (2,7,4,1)\} \end{equation*}
    Solution.

    We will write the vectors as rows of a matrix and then use Hint 4.6.5.1.3 to determine a basis of its row space. We row-reduce:
    \begin{align*} \matr{cccc}{ 2 \amp 1 \amp 0 \amp -1 \\ -1 \amp 1 \amp 1 \amp 1 \\ 2 \amp 7 \amp 4 \amp 1 } \overset{-R_{1}+R_{2}}{\underset{R_{1}+2R_{3}}{\longrightarrow}} \amp\matr{cccc}{ 2 \amp 1 \amp 0 \amp -1 \\ 0 \amp 3 \amp 2 \amp 1 \\ 0 \amp 6 \amp 4 \amp 2 }\\ \overset{-2R_{2}+R_{3}}{\longrightarrow} \amp\matr{cccc}{ 2 \amp 1 \amp 0 \amp -1 \\ 0 \amp 3 \amp 2 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 }. \end{align*}
    We see that the rank of the matrix and thus the dimension of the subspace is \(2\text{.}\) Moreover, a basis for the space in question is given by
    \begin{equation*} \left\{ (2 , 1 , 0 , -1), (0 , 3 , 2 , 1)\right\}. \end{equation*}
    Answer.

    The dimension is \(2\) and a basis is given by
    \begin{equation*} \left\{ (2 , 1 , 0 , -1), (0 , 3 , 2 , 1)\right\}. \end{equation*}

  3. \begin{equation*} \text{span}\{(-1,2,1,0), (2,0,3,-1), (4,4,11,-3), (3, -2, 2, -1)\} \end{equation*}
    Solution.

    We will write the vectors as rows of a matrix and then use Hint 4.6.5.1.3 to determine a basis of its row space. We row-reduce:
    \begin{align*} \matr{cccc}{ -1 \amp 2 \amp 1 \amp 0 \\ 2 \amp 0 \amp 3 \amp -1 \\ 4 \amp 4 \amp 11 \amp -3 \\ 3 \amp -2 \amp 2 \amp -1 } \overset{2R_{1}+R_{2}, 4R_{1}+R_{3}}{\underset{3R_{1}+R_{4}}{\longrightarrow}} \amp\matr{ccc}{ -1 \amp 2 \amp 1 \amp 0 \\ 0 \amp 4 \amp 5 \amp -1 \\ 0 \amp 12 \amp 15 \amp -3 \\ 0 \amp 4 \amp 5 \amp -1 }\\ \overset{-R_{2}+R_{4}}{\underset{-2R_{2}+R_{3}}{\longrightarrow}} \amp\matr{cccc}{ -1 \amp 2 \amp 1 \amp 0 \\ 0 \amp 4 \amp 5 \amp -1 \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 }. \end{align*}
    We see that the rank of the matrix and thus the dimension of the subspace is \(2\text{.}\) Moreover, a basis for the space in question is given by
    \begin{equation*} \left\{ (-1 , 2 , 1 , 0), (0 , 4 , 5 , -1)\right\}. \end{equation*}
    Answer.

    The dimension is \(2\) and a basis is given by
    \begin{equation*} \left\{ (-1 , 2 , 1 , 0), (0 , 4 , 5 , -1)\right\}. \end{equation*}

  4. \begin{equation*} \text{span}\{(-2,0,3,1), (1,2,-1,0), (-2,8,5,3), (-1, 2, 2, 1)\} \end{equation*}
    Solution.

    We will write the vectors as rows of a matrix and then use Hint 4.6.5.1.3 to determine a basis of its row space. We row-reduce:
    \begin{align*} \matr{cccc}{ -2 \amp 0 \amp 3 \amp 1 \\ 1 \amp 2 \amp -1 \amp 0 \\ -2 \amp 8 \amp 5 \amp 3 \\ -1 \amp 2 \amp 2 \amp 1 } \overset{R_{1}+2R_{2}, -R_{1}+R_{3}}{\underset{-R_{1}+2R_{4}}{\longrightarrow}} \amp\matr{cccc}{ -2 \amp 0 \amp 3 \amp 1 \\ 0 \amp 4 \amp 1 \amp 1 \\ 0 \amp 8 \amp 2 \amp 2 \\ 0 \amp 4 \amp 1 \amp 1 }\\ \overset{-R_{2}+R_{4}}{\underset{-2R_{2}+R_{3}}{\longrightarrow}} \amp\matr{cccc}{ -2 \amp 0 \amp 3 \amp 1 \\ 0 \amp 4 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 }. \end{align*}
    We see that the rank of the matrix and thus the dimension of the subspace is \(2\text{.}\) Moreover, a basis for the space in question is given by
    \begin{equation*} \left\{ (-2 , 0 , 3 , 1), (0 , 4 , 1 , 1)\right\}. \end{equation*}
    Answer.

    The dimension is \(2\) and a basis is given by
    \begin{equation*} \left\{ (-2 , 0 , 3 , 1), (0 , 4 , 1 , 1)\right\}. \end{equation*}

Hint.
You can write the vectors as rows of a matrix. Then use Hint 4.6.5.1.3 to determine a basis of its row space.

3.

  1. Can a \(3 \times 4 \) matrix have independent columns? Independent rows? Explain. Solution.

    Recall that the dimension of the row space always coincides with the dimension of the column space. Since the matrix only has three rows, its column space can have at most dimension three. Since four vectors in a three dimensional space are always linearly dependent, the answer is: No, it cannot have independent columns. For the other statement, consider
    \begin{equation*} \matr{cccc}{ 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 } \end{equation*}
    This matrix has three rows and four columns (i.e. it is \(3 \times 4 \)) and its rows are linearly independent. So the answer is yes, it can have independent rows.
    Answer.

    It cannot have independent columns, it can have independent rows.

  2. If \(A \) is \(4 \times 3 \) and \(\rank (A) \) =2, can \(A \) have independent columns? Independent rows? Explain. Hint.

    It might help to look back into Hint 4.6.5.1.3 and Hint 4.6.5.1.2.
    Solution.
    By Hint 4.6.5.1.3, we know that the number of non-zero rows in a row-echelon form of \(A\) is the number of basis vectors for the row space of \(A\text{,}\) i.e. the dimension of \(\row (A)\text{.}\) Recall the definition of \(\rank (A)\) from Hint 4.6.5.1.2: since \(\rank(A)\) is the number of pivot entries in a row-echelon form of \(A\text{,}\) we see that \(\dim \row (A)= \rank(A)\text{,}\) which was assumed to be \(2\text{.}\) Since the matrix has \(4\) rows but its row space is only of dimension \(2\text{,}\) it cannot have linearly independent rows. Similarly, we know that \(\dim\row(A)=\dim\col(A)\text{,}\) so the column space of \(A\) has dimension \(2\text{,}\) which is strictly smaller than the number of columns, \(3\text{.}\) Thus, it cannot have linearly independent columns.
    Answer.

    No and no.

  3. If \(A \) is an \(m \times n \) matrix and \(\rank(A) = m \text{,}\) show that \(m \le n \text{.}\) Solution.

    As we have argued in Solution 4.6.5.3.b.1, \(\rank (A) = \dim\row(A)\text{,}\) and this number also coincides with \(\dim\col(A)\text{.}\) Thus, \(\dim\col(A) = m\) by assumption. This means, in particular, that among the \(n\) many columns of \(A\text{,}\) there must exist \(m\) many linearly independent ones. Thus, \(m \le n \text{.}\)

  4. Can a nonsquare matrix have its rows independent and its columns independent? Explain. Solution.

    Suppose \(A\) is of size \(m\times n\text{.}\) If all rows of \(A\) are linearly independent, then the row space dimension the number of rows in \(A\text{,}\) i.e. \(\dim\row(A)=m\text{.}\) Similarly, if all columns of \(A\) are linearly independent, then the column space dimension the number of columns in \(A\text{,}\) i.e. \(\dim\col(A)=n\text{.}\) Since we always have \(\dim\row(A) = \dim\col(A)\text{,}\) this implies \(m=n\text{,}\) i.e. \(A\) is square. Thus, the answer is no.
    Answer.

    No.

  5. Can the nullspace of a \(3 \times 6 \) matrix have dimension 2? Explain. Hint.

    It might help to look back into Hint 4.6.5.1.1.
    Solution.

    No. By Theorem 4.6.16, we have for an \(3\times 6\) matrix \(A\text{:}\)
    \begin{equation*} \nullity(A) = \dim\NullSp(A) = 6 - \rank(A). \end{equation*}
    Thus, if \(\nullity(A)=2\text{,}\) then the rank of the matrix would be \(4\text{.}\) Since \(\rank(A) = \dim\row(A)\text{,}\) this would imply that \(A\) must have \(4\) independent rows; this is impossible since \(A\) only has \(3\) rows.
    Answer.

    No.

  6. Suppose that \(A \) is \(5 \times 4 \) and \(\NullSp(A) = \SpanS(\vec{x}) \) for some column vector \(\vec{x}\ne \mathbf{0} \text{.}\) Can \(\dim(\col(A))=2? \) Solution.

    No. The given information tells us that \(\dim(\NullSp(A)) = 1\text{,}\) i.e., \(\nullity(A) = 1\text{.}\) By Theorem 4.6.10 we have \(\rank(A) = \dim(\col(A))\text{,}\) and so by Theorem 4.6.16 we have
    \begin{equation*} 1 + \dim(\col(A)) = \nullity(A) + \rank(A) = 4\text{,} \end{equation*}
    meaning that the only possibility is \(\dim(\col(A)) = 3\text{.}\)
    Answer.

    No.

Hint.
See Theorem 4.6.16 and Theorem 4.6.10.