Definition 5.3.1.
Let \(A\) be an \(m\times n\) matrix. The row space of \(A\text{,}\) denoted by \(\mbox{row}(A)\text{,}\) is the subspace of \(\R^n\) spanned by the rows of \(A\text{.}\)
Section 5.3 Subspaces of \(\R^n\) Associated with Matrices
Subsection 5.3.1 Row Space of a Matrix
Recall when learning Gaussian elimination, we observed that every row-echelon form of a given matrix has the same number of nonzero rows. This result suggests that there are certain characteristics associated with the rows of a matrix that are not affected by elementary row operations. We are now in the position to examine this question and to supply the proof we omitted earlier.
Exploration 5.3.1.
Consider the matrix
\begin{equation*}
A=\begin{bmatrix}-2\amp 2\amp 1\\4\amp -2\amp 1\end{bmatrix}.
\end{equation*}
Let \(\mathbf{r}_1\) and \(\mathbf{r}_2\) be the rows of \(A\text{:}\)
\begin{equation*}
\mathbf{r}_1=\begin{bmatrix}-2\amp 2\amp 1\end{bmatrix},\quad \mathbf{r}_2=\begin{bmatrix}4\amp -2\amp 1\end{bmatrix}.
\end{equation*}
Then \(\mbox{row}(A)=\mbox{span}(\mathbf{r}_1, \mathbf{r}_2)\) is a plane through \((0,0)\) containing \(\mathbf{r}_1\) and \(\mathbf{r}_2\text{.}\)
We will use elementary row operations to reduce \(A\) to \(\mbox{rref}(A)\text{,}\)namely
\begin{equation*}
\begin{bmatrix}-2\amp 2\amp 1\\4\amp -2\amp 1\end{bmatrix}\rightsquigarrow\begin{bmatrix}1\amp 0\amp 1\\0\amp 1\amp 3/2\end{bmatrix}.
\end{equation*}
Let \(\mathbf{\rho}_1\) and \(\mathbf{\rho}_2\) be the rows of \(\mbox{rref}(A)\text{:}\)
\begin{equation*}
\mathbf{\rho}_1=\begin{bmatrix}1\amp 0\amp 1\end{bmatrix},\quad \mathbf{\rho}_2=\begin{bmatrix}0\amp 1\amp 3/2\end{bmatrix}.
\end{equation*}
What do you think \(\mbox{span}(\mathbf{\rho}_1, \mathbf{\rho}_2)\) looks like? The following video will help us visualize \(\mbox{span}(\mathbf{\rho}_1, \mathbf{\rho}_2)\) and compare it to \(\mbox{span}(\mathbf{r}_1, \mathbf{r}_2)\text{.}\)
Based on what we observed in the video, we may conjecture that
\begin{equation*}
\mbox{span}(\mathbf{\rho}_1, \mathbf{\rho}_2)=\mbox{span}(\mathbf{r}_1, \mathbf{r}_2)
\end{equation*}
But why does this make sense? Vectors \(\mathbf{\rho}_1\) and \(\mathbf{\rho}_2\) were obtained from \(\mathbf{r}_1\) and \(\mathbf{r}_2\) by repeated applications of elementary row operations. At every stage of the row reduction process, the rows of the matrix are linear combinations of \(\mathbf{r}_1\) and \(\mathbf{r}_2\text{.}\) Thus, at every stage of the row reduction process, the rows of the matrix lie in the span of \(\mathbf{r}_1\) and \(\mathbf{r}_2\text{.}\) Our next video shows a step-by-step row reduction process accompanied by sketches of vectors.
The observations made in Exploration 5.3.1 makes a convincing case for the following theorem.
Theorem 5.3.2.
If matrix \(B\) was obtained from matrix \(A\) by applying an elementary row operation to \(A\) then
\begin{equation*}
\mbox{row}(B)=\mbox{row}(A).
\end{equation*}
Proof.
Let \(\mathbf{r}_1,\ldots ,\mathbf{r}_m\) be the rows of \(A\text{.}\) There are three elementary row operations. Clearly, switching the order of vectors in \(\mbox{span}(\mathbf{r}_1,\ldots ,\mathbf{r}_m)\) will not affect the span. Suppose that \(B\) was obtained from \(A\) by multiplying the \(i^{th}\) row of \(A\) by a non-zero constant \(k\text{.}\) We need to show that
\begin{equation*}
\mbox{span}(\mathbf{r}_1,\ldots ,k\mathbf{r}_i,\ldots ,\mathbf{r}_m)=\mbox{span}(\mathbf{r}_1,\ldots ,\mathbf{r}_i,\ldots ,\mathbf{r}_m)
\end{equation*}
To do this we will assume that some vector \(\mathbf{v}\) is in
\begin{equation*}
\mbox{span}(\mathbf{r}_1,\ldots ,k\mathbf{r}_i,\ldots ,\mathbf{r}_m)
\end{equation*}
and show that \(\mathbf{v}\) is in \(\mbox{span}(\mathbf{r}_1,\ldots ,\mathbf{r}_i,\ldots ,\mathbf{r}_m)\text{.}\) We will then assume that some vector \(\mathbf{w}\) is in
\begin{equation*}
\mbox{span}(\mathbf{r}_1,\ldots ,\mathbf{r}_i,\ldots ,\mathbf{r}_m)
\end{equation*}
and show that \(\mathbf{w}\) must be in \(\mbox{span}(\mathbf{r}_1,\ldots ,k\mathbf{r}_i,\ldots ,\mathbf{r}_m)\text{.}\)
Suppose that \(\mathbf{v}\) is in \(\mbox{span}(\mathbf{r}_1,\ldots ,k\mathbf{r}_i,\ldots ,\mathbf{r}_m)\text{.}\) Then
\begin{equation*}
\mathbf{v}=a_1\mathbf{r}_1+\ldots +a_i(k\mathbf{r}_i)+\ldots +a_m\mathbf{r}_m
\end{equation*}
But this implies
\begin{equation*}
\mathbf{v}=a_1\mathbf{r}_1+\ldots +(a_ik)\mathbf{r}_i+\ldots +a_m\mathbf{r}_m.
\end{equation*}
So \(\mathbf{v}\) is in \(\mbox{span}(\mathbf{r}_1,\ldots ,\mathbf{r}_i,\ldots ,\mathbf{r}_m)\text{.}\)
Now suppose \(\mathbf{w}\) is in \(\mbox{span}(\mathbf{r}_1,\ldots ,\mathbf{r}_i,\ldots ,\mathbf{r}_m)\text{,}\) then
\begin{equation*}
\mathbf{w}=b_1\mathbf{r}_1+\ldots +b_i\mathbf{r}_i+\ldots +b_m\mathbf{r}_m.
\end{equation*}
But because \(k\neq 0\text{,}\) we can do the following:
\begin{equation*}
\mathbf{w}=b_1\mathbf{r}_1+\ldots +\frac{b_i}{k}(k\mathbf{r}_i)+\ldots +b_m\mathbf{r}_m.
\end{equation*}
Therefore, \(\mathbf{w}\) is in \(\mbox{span}(\mathbf{r}_1,\ldots ,k\mathbf{r}_i,\ldots ,\mathbf{r}_m)\text{.}\)
We leave it to the reader to verify that adding a multiple of one row of \(A\) to another does not change the row space. See also Exercise 5.3.5.17.)
Corollary 5.3.3.
If matrix \(B\) was obtained from matrix \(A\) by applying a sequence of elementary row operations to \(A\) then
\begin{equation*}
\mbox{row}(B)=\mbox{row}(A).
\end{equation*}
Proof.
This follows from repeated applications of Theorem 5.3.2.
Corollary 5.3.4.
\begin{equation*}
\mbox{row}(A)=\mbox{row}(\mbox{rref}(A)).
\end{equation*}
Example 5.3.5.
Let
\begin{equation*}
A=\begin{bmatrix}2\amp -1\amp 1\amp -4\amp 1\\1\amp 0\amp 3\amp 3\amp 0\\-2\amp 1\amp -1\amp 5\amp 2\\4\amp -1\amp 7\amp 2\amp 1\end{bmatrix}.
\end{equation*}
Find two distinct bases for \(\mbox{row}(A)\text{.}\)
Answer.
By Corollary 5.3.4 a basis for \(\mbox{row}(\mbox{rref}(A))\) will also be a basis for \(\mbox{row}(A)\text{.}\) Row reduction gives us:
\begin{equation*}
\begin{bmatrix}2\amp -1\amp 1\amp -4\amp 1\\1\amp 0\amp 3\amp 3\amp 0\\-2\amp 1\amp -1\amp 5\amp 2\\4\amp -1\amp 7\amp 2\amp 1\end{bmatrix}\rightsquigarrow\begin{bmatrix}1\amp 0\amp 3\amp 0\amp -9\\0\amp 1\amp 5\amp 0\amp -31\\0\amp 0\amp 0\amp 1\amp 3\\0\amp 0\amp 0\amp 0\amp 0\end{bmatrix}=\mbox{rref}(A).
\end{equation*}
Since the zero row contributes nothing to the span, we conclude that the nonzero rows of \(\mbox{rref}(A)\) span \(\mbox{row}(\mbox{rref}(A))\text{.}\) Therefore, a basis for the row space consists of the vecotrs
\begin{equation*}
[1, 0, 3, 0, -9], \
[0, 1, 5, 0, -3], \
[0, 0, 0, 1, 3].
\end{equation*}
It follows that the nonzero rows of \(\mbox{rref}(A)\) are linearly independent.
To find a second basis for \(\mbox{row}(A)\text{,}\) observe that by Corollary 5.3.3 the row space of any row-echelon form of \(A\) will be equal to \(\mbox{row}(A)\text{.}\) Matrix \(A\) has many row-echelon forms. Here is one of them:
\begin{equation*}
B=\begin{bmatrix}1\amp 0\amp 3\amp 3\amp 0\\0\amp -1\amp -5\amp -10\amp 1\\0\amp 0\amp 0\amp 1\amp 3\\0\amp 0\amp 0\amp 0\amp 0 \end{bmatrix}.
\end{equation*}
The nonzero rows of \(B\) span \(\mbox{row}(A)\text{.}\) Once again the nonzero rows of \(B\) are linearly independent. Thus the nonzero rows of \(B\) form a basis for \(\mbox{row}(A)\text{.}\)
Our observations in Example 5.3.5 can be generalized to all matrices. Given any matrix \(A\text{,}\)
- The nonzero rows of \(\mbox{rref}(A)\) are linearly independent (Why?) and span \(\mbox{row}(A)\) (Corollary 5.3.4).
- The nonzero rows of any row-echelon form of \(A\) are linearly independent (Why?) and span \(\mbox{row}(A)\) (Corollary 5.3.3).
Therefore nonzero rows of \(\mbox{rref}(A)\) or the nonzero rows of any row-echelon form of \(A\) constitute a basis of \(\mbox{row}(A)\text{.}\) Since all bases for \(\mbox{row}(A)\) must have the same number of elements ( Theorem 5.2.13), we have just proved the following theorem.
Theorem 5.3.6.
This result was first introduced without proof when learning of Guassian elimination, where we used it to define the rank of a matrix as the number of nonzero rows in its row-echelon forms. We can now update the definition of rank as follows.
Definition 5.3.7.
For any matrix \(A\text{,}\)
\begin{equation*}
\mbox{rank}(A)=\mbox{dim}\Big(\mbox{row}(A)\Big).
\end{equation*}
Subsection 5.3.2 Column Space of a Matrix
Definition 5.3.8.
Let \(A\) be an \(m\times n\) matrix. The column space of \(A\text{,}\) denoted by \(\mbox{col}(A)\text{,}\) is the subspace of \(\R^m\) spanned by the columns of \(A\text{.}\)
Exploration 5.3.2.
Let
\begin{equation*}
B=\begin{bmatrix}2\amp -1\amp 3\amp 1\\1\amp -1\amp 2\amp 2\\1\amp 3\amp -2\amp -3\end{bmatrix}.
\end{equation*}
Our goal is to find a basis for \(\mbox{col}(B)\text{.}\) To do this we need to find a linearly independent subset of the columns of \(B\) that spans \(\mbox{col}(B)\text{.}\)
Consider the linear relation:
\begin{equation}
a_1\begin{bmatrix}2\\1\\1\end{bmatrix}+a_2\begin{bmatrix}-1\\-1\\3\end{bmatrix}+a_3\begin{bmatrix}3\\2\\-2\end{bmatrix}+a_4\begin{bmatrix}1\\2\\-3\end{bmatrix}=\mathbf{0}.\tag{5.3.1}
\end{equation}
Solving this homogeneous equation amounts to finding \(\mbox{rref}(B)\text{.}\) So,
\begin{equation*}
\begin{bmatrix}2\amp -1\amp 3\amp 1\\1\amp -1\amp 2\amp 2\\1\amp 3\amp -2\amp -3\end{bmatrix}\rightsquigarrow\begin{bmatrix}1\amp 0\amp 1\amp 0\\0\amp 1\amp -1\amp 0\\0\amp 0\amp 0\amp 1\end{bmatrix}=\mbox{rref}(B).
\end{equation*}
We now see that (5.3.1) has infinitely many solutions.
Observe that the homogeneous equation
\begin{equation}
a_1\begin{bmatrix}1\\0\\0\end{bmatrix}+a_2\begin{bmatrix}0\\1\\0\end{bmatrix}+a_3\begin{bmatrix}1\\-1\\0\end{bmatrix}+a_4\begin{bmatrix}0\\0\\1\end{bmatrix}=\mathbf{0}.\tag{5.3.2}
\end{equation}
has the same solution set as (5.3.1). In particular, \(a_1=1\text{,}\) \(a_2=-1\text{,}\) \(a_3=-1\text{,}\) \(a_4=0\) is a non-trivial solution of (5.3.1) and (5.3.2). This means that the third column of \(B\) and the third column of \(\mbox{rref}(B)\) can be expressed as the first column minus the second column of their respective matrices.
We conclude that the third column of \(B\) can be eliminated from the spanning set for \(\mbox{col}(B)\) and
\begin{align*}
\mbox{col}(B) \amp =\mbox{span}\left(\begin{bmatrix}2\\1\\1\end{bmatrix},\begin{bmatrix}-1\\-1\\3\end{bmatrix}, \begin{bmatrix}3\\2\\-2\end{bmatrix}, \begin{bmatrix}1\\2\\-3\end{bmatrix}\right) \\
\amp =\mbox{span}\left(\begin{bmatrix}2\\1\\1\end{bmatrix},\begin{bmatrix}-1\\-1\\3\end{bmatrix}, \begin{bmatrix}1\\2\\-3\end{bmatrix}\right).
\end{align*}
Having gotten rid of one of the vectors, we need to determine whether the remaining three vectors are linearly independent. To do this we need to find all solutions of
\begin{equation}
b_1\begin{bmatrix}2\\1\\1\end{bmatrix}+b_2\begin{bmatrix}-1\\-1\\3\end{bmatrix}+b_3\begin{bmatrix}1\\2\\-3\end{bmatrix}=\mathbf{0}.\tag{5.3.3}
\end{equation}
Fortunately, we do not have to start from scratch. Observe that crossing out the third column in the previous row reduction process yields the desired reduced row-echelon form.
This time the reduced row-echelon form tells us that ((5.3.3) has only the trivial solution. We conclude that the three vectors are linearly independent and
\begin{equation*}
\left\{\begin{bmatrix}2\\1\\1\end{bmatrix},\begin{bmatrix}-1\\-1\\3\end{bmatrix}, \begin{bmatrix}1\\2\\-3\end{bmatrix}\right\}
\end{equation*}
is a basis for \(\mbox{col}(B)\text{.}\)
The approach we took to find a basis for \(\mbox{col}(B)\) in Exploration 5.3.2 uses the reduced row-echelon form of \(B\text{.}\) It is true, however, that any row-echelon form of \(B\) could have been used in place of \(\mbox{rref}(B)\text{.}\) (Why?). We generalize the steps as follows:
Algorithm 5.3.9.
Given a matrix \(B\text{,}\) a basis for \(\mbox{col}(B)\) can be found as follows:
- Find \(\mbox{rref}(B)\) (or any row-echelon form \(B'\) of \(B\text{.}\))
- Identify the pivot columns of \(\mbox{rref}(B)\) (or \(B'\)).
- The columns of \(B\) corresponding to the pivot columns of \(\mbox{rref}(B)\) (or \(B'\)) form a basis for \(\mbox{col}(B)\text{.}\)
Proof.
Let \(\mathbf{b}_1,\ldots ,\mathbf{b}_n\) be the columns of \(B\text{,}\) and let \(\mathbf{b}'_1,\ldots ,\mathbf{b}'_n\) be the columns of \(\mbox{rref}(B)\) (or \(B'\)). Observe that the equations
\begin{equation}
a_1\mathbf{b}_1+\ldots +a_n\mathbf{b}_n=\mathbf{0}\tag{5.3.4}
\end{equation}
\begin{equation}
a_1\mathbf{b}'_1+\ldots +a_n\mathbf{b}'_n=\mathbf{0}\tag{5.3.5}
\end{equation}
have the same solution set. This means that any non-trivial relation among the columns of \(\mbox{rref}(B)\) (or \(B'\)) translates into a non-trivial relation among the columns of \(B\text{.}\) Likewise, any collection of linearly independent columns of \(\mbox{rref}(B)\) (or \(B'\)) corresponds to linearly independent columns of \(B\text{.}\)
Now, the pivot columns of \(\mbox{rref}(B)\) (or \(B'\)) are linearly independent. Therefore the corresponding columns of \(B\) are linearly independent whereas non-pivot columns can be expressed as linear combinations of the pivot columns, therefore they contribute nothing to the span and can be removed from the spanning set.
The proof of Algorithm 5.3.9 shows that the number of basis elements for the column space of a matrix is equal to the number of pivot columns. But the number of pivot columns is the same as the number of pivots in a row-echelon form, which is equal to the number of nonzero rows and the rank of the matrix. This gives us the following important result.
Theorem 5.3.10.
Let \(A\) be a matrix.
\begin{equation*}
\mbox{dim}\Big(\mbox{row}(A)\Big)=\mbox{rank}(A)=\mbox{dim}\Big(\mbox{col}(A)\Big).
\end{equation*}
Example 5.3.11.
Subsection 5.3.3 The Null Space
Definition 5.3.12.
Let \(A\) be an \(m\times n\) matrix. The null space of \(A\text{,}\) denoted by \(\mbox{null}(A)\text{,}\) is the set of all vectors \(\mathbf{x}\) in \(\R^n\) such that \(A\mathbf{x}=\mathbf{0}\text{.}\)
Before digging further, let us examine the notion through an example.
Example 5.3.13.
Find \(\mbox{null}(A)\) if
\begin{equation*}
A=\begin{bmatrix}3\amp -1\\-6\amp 2\end{bmatrix}.
\end{equation*}
Answer.
We need to solve the equation \(A\mathbf{x}=\mathbf{0}\text{.}\) Row reduction gives us
\begin{equation*}
\begin{bmatrix}3\amp -1\\-6\amp 2\end{bmatrix}\rightsquigarrow\begin{bmatrix}1\amp -1/3\\0\amp 0\end{bmatrix}=\mbox{rref}(A).
\end{equation*}
We conclude that \(\mathbf{x}=\begin{bmatrix}1/3\\1\end{bmatrix}t\text{.}\) Thus \(\mbox{null}(A)\) consists of all vectors of the form
\begin{equation*}
t \begin{bmatrix}1/3\\1\end{bmatrix}.
\end{equation*}
We might write
\begin{equation*}
\mbox{null}(A)=\left\{\begin{bmatrix}1/3\\1\end{bmatrix}t\right\}
\end{equation*}
or
\begin{equation*}
\mbox{null}(A)=\mbox{span}\left(\begin{bmatrix}1/3\\1\end{bmatrix}\right).
\end{equation*}
The approach in Example 5.3.13 allows us to make an important observation. Note that every scalar multiple of \([1/3,1]\) is contained in \(\mbox{null}(A)\text{.}\) This means that \(\mbox{null}(A)\) is closed under vector addition and scalar multiplication.
Recall that this property makes \(\mbox{null}(A)\) a subspace of \(\R^n\text{.}\) This result was first presented as Exercise 5.1.4.8. We now formalize it as a theorem.
Theorem 5.3.14.
Let \(A\) be an \(m\times n\) matrix. Then \(\mbox{null}(A)\) is a subspace of \(\R^n\text{.}\)
Proof.
To deduce that \(\mbox{null}(A)\) is closed under vector addition and scalar multiplication we will show that a linear combination of any two elements of \(\mbox{null}(A)\) is contained in \(\mbox{null}(A)\text{.}\)
Suppose \(\mathbf{x}_1\) and \(\mathbf{x}_2\) are in \(\mbox{null}(A)\text{.}\) Then \(A\mathbf{x}_1=\mathbf{0}\) and \(A\mathbf{x}_2=\mathbf{0}\text{.}\) But then
\begin{equation*}
A(a_1\mathbf{x}_1+a_2\mathbf{x}_2)=a_1A\mathbf{x}_1+a_2A\mathbf{x}_2=\mathbf{0}
\end{equation*}
We conclude that \(a_1\mathbf{x}_1+a_2\mathbf{x}_2\) is also in \(\mbox{null}(A)\text{.}\)
Example 5.3.15.
It is not a coincidence that the steps we used in Example 5.3.15 produced linearly independent vectors, and it is worth while to try to understand why this procedure will always produce linearly independent vectors. Take a closer look at the elements of the null space:
\begin{equation*}
\mathbf{x}=\begin{bmatrix}9t-3s\\31t-5s\\{\color{red}\fbox{}}\\-3t\\{\color{blue}\fbox{}}\end{bmatrix}=\begin{bmatrix}-3\\-5\\{\color{red}\fbox{}}\\0\\{\color{blue}\fbox{}}\end{bmatrix}s+\begin{bmatrix}9\\31\\{\color{red}\fbox{}}\\-3\\{\color{blue}\fbox{}}\end{bmatrix}t
\end{equation*}
The parameter \(s\) in the third component of \(\mathbf{x}\) produces a \(1\) in the third component of the first vector and a \(0\) in the third component of the second vector, while parameter \(t\) in the fifth component of \(\mathbf{x}\) produces a \(1\) in the fifth component of the second vector and a \(0\) in the fifth component of the first vector. This makes it clear that the two vectors are linearly independent.
This pattern will hold for any number of parameters, each parameter producing a \(1\) in exactly one vector and \(0's\) in the corresponding components of the other vectors.
\begin{equation*}
\begin{bmatrix}\vdots \\t_1\\\vdots\\t_2\\\vdots\\t_3\\\vdots\\t_n\\\vdots \end{bmatrix}=\begin{bmatrix}\vdots \\1\\\vdots\\0\\\vdots\\0\\\vdots\\0\\\vdots \end{bmatrix}t_1+\ldots +\begin{bmatrix}\vdots \\0\\\vdots\\1\\\vdots\\0\\\vdots\\0\\\vdots \end{bmatrix}t_2+\ldots+\begin{bmatrix}\vdots \\0\\\vdots\\0\\\vdots\\1\\\vdots\\0\\\vdots \end{bmatrix}t_3+\ldots+\begin{bmatrix}\vdots \\0\\\vdots\\0\\\vdots\\0\\\vdots\\1\\\vdots \end{bmatrix}t_n
\end{equation*}
Therefore, vectors obtained in this way will always be linearly independent.
Subsection 5.3.4 Rank and Nullity Theorem
Definition 5.3.16.
Let \(A\) be a matrix. The dimension of the null space of \(A\) is called the nullity of \(A\text{.}\)
\begin{equation*}
\mbox{dim}\Big(\mbox{null}(A)\Big)=\mbox{nullity}(A).
\end{equation*}
We know that the dimension of the row space and the dimension of the column space of a matrix are the same and are equal to the rank of the matrix (or the number of nonzero rows in any row-echelon form of the matrix).
As we observed in Example 5.3.15, the dimension of the null space of a matrix is equal to the number of free variables in the solution vector of the homogeneous system associated with the matrix. Since the number of pivots and the number of free variables add up to the number of columns in a matrix (Theorem 2.3.11) we have the following significant result.
Theorem 5.3.17.
Let \(A\) be an \(m\times n\) matrix. Then
\begin{equation*}
\mbox{rank}(A)+\mbox{nullity}(A)=n
\end{equation*}
We will see the geometric implications of this theorem when we study linear transformations.
Exercises 5.3.5 Exercises
Exercise Group.
In the following four problems, the matrix \(A\) given below is studied.
\begin{equation}
A=\begin{bmatrix}2\amp 0\amp 2\amp 4\\1\amp 3\amp -2\amp -1\\-1\amp -2\amp 1\amp 0\end{bmatrix}.\tag{5.3.6}
\end{equation}
1.
Let Find \(\mbox{rref}(A)\text{.}\)
Answer.
\begin{equation*}
\mbox{rref}(A)=\begin{bmatrix}1\amp 0\amp 1\amp 2\\0\amp 1\amp -1\amp -1\\0\amp 0\amp 0\amp 0\end{bmatrix}.
\end{equation*}
2.
Compute \(\text{rank}(A)\text{,}\) \(\text{dim}(\text{row}(A))\) and \(\text{dim}(\text{col}(A)) \)
Answer.
\begin{equation*}
\mbox{rank}(A)=\mbox{dim}(\mbox{row}(A))=\mbox{dim}(\mbox{col}(A))=2.
\end{equation*}
3.
Use \(\mbox{rref}(A)\) and the procedure outlined in Example 5.3.5 to find a basis for \(\mbox{row}(A)\text{.}\)
Answer.
A basis for \(\mbox{row}(A)\) is
\begin{equation*}
\left\{\begin{bmatrix}1\amp 0\amp 1 \amp 2\end{bmatrix},\begin{bmatrix}0 \amp 1\amp -1 \amp -1\end{bmatrix} \right\}.
\end{equation*}
4.
Use Algorithm 5.3.9 to find a basis for \(\mbox{col}(A)\text{.}\)
Answer.
A basis for \(\mbox{col}(A) \) is
\begin{equation*}
\left\{ \begin{bmatrix}2\\1\\-1\end{bmatrix}, \begin{bmatrix}0\\3\\-2\end{bmatrix}\right\}.
\end{equation*}
Exercise Group.
In the following four problems, we matrix in question is
\begin{equation}
B=\begin{bmatrix}1\amp 2\amp 3\\-1\amp 1\amp 3\\2\amp 0\amp -2\\1\amp -2\amp -5\\0\amp 1\amp 2\end{bmatrix}.\tag{5.3.7}
\end{equation}
5.
Find \(\mbox{rref}(B)\text{.}\)
Answer.
\begin{equation*}
\mbox{rref}(B)=\begin{bmatrix}1\amp 0\amp -1\\0\amp 1\amp 2\\0\amp 0\amp 0\\0\amp 0\amp 0\\0\amp 0\amp 0\end{bmatrix}.
\end{equation*}
6.
Find \(\mbox{rank}(B)=\mbox{dim}(\mbox{row}(B))=\mbox{dim}(\mbox{col}(B)).\)
Answer.
The answer is \(\mbox{rank}(B)=\mbox{dim}(\mbox{row}(B))=\mbox{dim}(\mbox{col}(B)) = 2 \text{.}\)
7.
Use \(\mbox{rref}(B)\) and the procedure outlined in Example 5.3.5 to find a basis for \(\mbox{row}(B)\text{.}\)
Answer.
A basis for \(\mbox{row}(B)\) is
\begin{equation*}
\left\{\begin{bmatrix}1\amp 0\amp -1\end{bmatrix},\begin{bmatrix}0 \amp 1\amp 2\end{bmatrix} \right\}.
\end{equation*}
8.
Use Algorithm 5.3.9 to find a basis for \(\mbox{col}(B)\text{.}\)
Answer.
A basis for \(\mbox{col}(B)\) is
\begin{equation*}
\left\{ \begin{bmatrix}1\\-1\\2\\1\\0\end{bmatrix}, \begin{bmatrix}2\\1\\0\\-2\\1\end{bmatrix}\right\}.
\end{equation*}
9.
Prove that \(\mbox{rank}(A)=\mbox{rank}(A^T)\)
10.
Find a basis for \(V\text{,}\) where
\begin{equation*}
V=\mbox{span}\left( \begin{bmatrix}1\\0\\2\end{bmatrix}, \begin{bmatrix}-1\\2\\-1\end{bmatrix}, \begin{bmatrix}1\\2\\3\end{bmatrix}, \begin{bmatrix}3\\-1\\0\end{bmatrix}, \begin{bmatrix}3\\1\\1\end{bmatrix}\right).
\end{equation*}
Hint.
Find a basis for the column space of a matrix whose columns are the given vectors.
Exercise Group.
11.
Find a basis for \(\mbox{null}(A)\text{,}\) demonstrate that the Rank-Nullity Theorem (see Theorem 5.3.17) holds for \(A\) and explain how you can quickly tell that the vectors you selected for your basis are linearly independent.
Answer.
Basis for \(\mbox{null}(A) \)
\begin{equation*}
\left\{ \begin{bmatrix}-1\\1\\1\\0\end{bmatrix}, \begin{bmatrix}-2\\1\\0\\1\end{bmatrix}\right\}.
\end{equation*}
12.
Find a basis for \(\mbox{null}(B)\) and demonstrate that the Rank-Nullity Theorem (see Theorem 5.3.17) holds for \(B\text{.}\)
Answer.
Basis for \(\mbox{null}(B) \text{:}\)
\begin{equation*}
\left\{ \begin{bmatrix}1\\-2\\1\end{bmatrix}\right\}
\end{equation*}
Exercise Group.
Suppose matrix \(M\) is such that
\begin{equation*}
\mbox{rref}(M)=\begin{bmatrix}1\amp 0\amp 2\amp 0\amp 3\amp 1\\0\amp 1\amp -1\amp 0\amp 1\amp -2\\0\amp 0\amp 0\amp 1\amp -2\amp 1\\0\amp 0\amp 0\amp 0\amp 0\amp 0\end{bmatrix}
\end{equation*}
13.
Follow the process used in Example 5.3.15 to find a basis for \(\mbox{null}(M)\text{.}\) Explain why the basis elements obtained in this way are linearly independent.
Answer.
Basis of \(\mbox{null}(M):\quad\left\{\begin{bmatrix}-2\\1\\1\\0\\0\\0\end{bmatrix}, \begin{bmatrix}-3\\-1\\0\\2\\1\\0\end{bmatrix}, \begin{bmatrix}-1\\2\\0\\-1\\0\\1\end{bmatrix} \right\}\)
14.
Let \(\mathbf{v}_1,\ldots,\mathbf{v}_6\) denote the columns of \(M\text{.}\) Express \(\mathbf{v}_3\) as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\)
Answer.
\begin{equation*}
\mathbf{v}_3=2\mathbf{v}_1+-1\mathbf{v}_2
\end{equation*}
15.
- \(\mbox{dim}(\mbox{col}(A))=5\)
- \(\mbox{dim}(\mbox{row}(A))=3\)
- \(\mbox{dim}(\mbox{null}(A))=1\)
- \(\mbox{dim}(\mbox{null}(A))=2\)
- \(\mbox{dim}(\mbox{null}(A))=3\)
Suppose \(A\) is a \(3\times 5\) matrix. Which of the following statements could be true?
16.
- \(\mbox{dim}(\mbox{col}(A))=3\)
- \(\mbox{dim}(\mbox{row}(A))=3\)
- \(\mbox{dim} (\mbox{row}(A))=7\)
- \(\mbox{dim}(\mbox{null}(A))=0 \)
- \(\mbox{dim}(\mbox{null}(A))=4\)
Suppose \(A\) is a \(7\times 3\) matrix. Which of the following statements could be true?
17.
Complete the proof of Theorem 5.3.2 by showing that adding a scalar multiple of one row of a matrix to another row does not change the row space.
