4.9.E: Exercise for Section 4.7

Exercise \(\PageIndex{1}\)

Given the matrix \[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 0 & 1 & -1 & 2 \end{bmatrix} \nonumber\]

1. Find a basis for the row space of \( A \).
2. Find a basis for the column space of \( A \).
3. Find a basis for the null space of \( A \).

Answer

Given the matrix \[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 0 & 1 & -1 & 2 \end{bmatrix} \nonumber\]

1. Basis for the row space: The row space is spanned by the linearly independent rows of \( A \) after row reduction: \[ \text{Row space basis: } \left\{ \begin{bmatrix} 1 & 2 & 3 & 4 \end{bmatrix}, \begin{bmatrix} 0 & 1 & -1 & 2 \end{bmatrix} \right\}. \nonumber\]
2. Basis for the column space: The column space is spanned by the pivot columns of \( A \): \[ \text{Column space basis: } \left\{ \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 4 \\ 1 \end{bmatrix} \right\}. \nonumber\]
3. Basis for the null space: Solve \( A \mathbf{x} = 0 \) to find the null space basis: \[ \text{Null space basis: } \left\{ \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -4 \\ -2 \\ 0 \\ 1 \end{bmatrix} \right\}. \nonumber\]

Exercise \(\PageIndex{2}\)

Let \[ B = \begin{bmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \\ 3 & 6 & -3 \end{bmatrix} \nonumber\]

1. Compute the rank of \( B \).
2. Compute the nullity of \( B \).
3. Verify the Rank-Nullity Theorem for this matrix.

Answer

Let \[ B = \begin{bmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \\ 3 & 6 & -3 \end{bmatrix} \nonumber\]

1. Rank of \( B \): The rank is the number of linearly independent rows: \[ \text{Rank}(B) = 1. \nonumber\]
2. Nullity of \( B \): The nullity is computed as: \[ \text{Nullity}(B) = 3 - 1 = 2. \nonumber\]
3. Verification of Rank-Nullity Theorem: \[ \text{Rank}(B) + \text{Nullity}(B) = 1 + 2 = 3, \nonumber\] which matches the number of columns, so the theorem holds.

Exercise \(\PageIndex{3}\)

Given the matrix \[ D = \begin{bmatrix} 2 & 1 & -3 & 4 \\ 0 & 0 & 0 & 0 \\ 1 & -1 & 2 & -2 \end{bmatrix} \nonumber\]

1. Compute the rank of \( D \).
2. Compute the nullity of \( D \).
3. Find a basis for the null space of \( D \).
4. Does the Rank-Nullity Theorem hold for \( D \)?

Answer

Given the matrix \[ D = \begin{bmatrix} 2 & 1 & -3 & 4 \\ 0 & 0 & 0 & 0 \\ 1 & -1 & 2 & -2 \end{bmatrix} \nonumber\]

\textbf{Rank of \( D \):} \\ The rank is the number of linearly independent rows: \[ \text{Rank}(D) = 2. \nonumber\]

\textbf{Nullity of \( D \):} \\ \[ \text{Nullity}(D) = 4 - 2 = 2. \nonumber\]

\textbf{Basis for the null space:} \\ Solve \( D \mathbf{x} = 0 \): \[ \text{Null space basis: } \left\{ \begin{bmatrix} 1 \\ -5 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ -4 \\ 0 \\ 1 \end{bmatrix} \right\}. \nonumber\]

\textbf{Verification of Rank-Nullity Theorem:} \\ \[ \text{Rank}(D) + \text{Nullity}(D) = 2 + 2 = 4, \nonumber\] which matches the number of columns, so the theorem holds.

Exercise \(\PageIndex{4}\)

Consider the matrix \[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \end{bmatrix}. \]

1. Find a basis for the row space of \( A \).
2. Find a basis for the column space of \( A \).
3. Find a basis for the null space of \( A \).
4. Compute rank and nullity of \( A \), and verify the Rank-Nullity Theorem.

Exercise \(\PageIndex{5}\)

Let \[ B = \begin{bmatrix} 1 & -1 & 3 & 2 & 4 \\ 2 & 0 & 5 & 6 & 8 \\ 3 & -1 & 7 & 8 & 12 \end{bmatrix} \nonumber \]

1. Compute the rank of \( B \).
2. Compute the nullity of \( B \).

Answer

1. Row reduce \( B \): \[ \begin{bmatrix} 1 & -1 & 3 & 2 & 4 \\ 2 & 0 & 5 & 6 & 8 \\ 3 & -1 & 7 & 8 & 12 \end{bmatrix} \to \begin{bmatrix} 1 & -1 & 3 & 2 & 4 \\ 0 & 2 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \nonumber \] \[ \text{Rank}(B) = 2 \nonumber \]
2. Using the Rank-Nullity Theorem: \[ \text{Nullity}(B) = 5 - 2 = 3 \nonumber \]

Exercise \(\PageIndex{6}\)

Consider the matrix \[ D = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 1 & 2 \end{bmatrix}. \nonumber\] Find a basis for the null space of \( D \)

Exercise \(\PageIndex{7}\)

Show that if \(A\) is an \(m\times n\) matrix, then \(\text{ker}(A)\) is a subspace of \(\mathbb{R}^n\).

Answer

If \(\vec{x}\), \(\vec{y}\in\text{ker}(A)\), then \(A(\vec{x})=\vec 0 \) and \(A(\vec{y})=\vec 0 \). Now let's look at \(a\vec{x}+b\vec{y}\) for some \(a, b \in \mathbb{R}\). Then,

\[A(a\vec{x}+b\vec{y})=aA\vec{x}+bA\vec{y}=a\vec{0}+b\vec{0}=\vec{0}\nonumber\] and so \(\text{ker}(A)\) is closed under linear combinations. Hence it is a subspace.

Exercise \(\PageIndex{8}\)

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrrrrr}1&3&0&-2&0&3 \\ 3&9&1&-7&0&8 \\ 1&3&1&-3&1&-1 \\ 1&3&-1&-1&-2&10\end{array}\right]\nonumber\]

Answer

The rank of the given matrix is \(3\). A basis for the row space is: \[ \left\{ \begin{bmatrix} 1 & 3 & 0 & -2 & 0 & 3 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 1 & -1 & 1 & -3 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & -5 \end{bmatrix} \right\} \nonumber\] A basis for the column space is: \[ \left\{ \begin{bmatrix} 1 \\ 3 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 3 \\ 9 \\ 3 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \\ -1 \end{bmatrix} \right\} \nonumber\]

Exercise \(\PageIndex{9}\)

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrrrrr}1&3&0&-2&7&3 \\ 3&9&1&-7&23&8 \\ 1&3&1&-3&9&2 \\ 1&3&-1&-1&5&4\end{array}\right]\nonumber\]

Exercise \(\PageIndex{10}\)

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrrrrr}1&0&3&0&7&0 \\ 3&1&10&0&23&0 \\ 1&1&4&1&7&0 \\ 1&-1&2&-2&9&1\end{array}\right]\nonumber\]

Exercise \(\PageIndex{11}\)

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrr}1&0&3 \\ 3&1&10 \\ 1&1&4 \\ 1&-1&2\end{array}\right]\nonumber\]

Exercise \(\PageIndex{12}\)

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrrrr}0&0&-1&0&1 \\ 1&2&3&-2&-18 \\ 1&2&2&-1&-11 \\ -1&-2&-2&1&11\end{array}\right]\nonumber\]

Exercise \(\PageIndex{13}\)

Find the rank of the following matrix. Also find a basis for the row and column spaces. \[\left[\begin{array}{rrrr}1&0&3&0 \\ 3&1&10&0 \\ -1&1&-2&1 \\ 1&-1&2&-2\end{array}\right]\nonumber\]

Exercise \(\PageIndex{14}\)

Find \(\text{ker}(A)\) and \(\text{im}(A)\) for the following matrices.

1. \(A=\left[\begin{array}{rr}2&3 \\ 4&6\end{array}\right]\)
2. \(A=\left[\begin{array}{rrr}1&0&-1 \\ -1&1&3 \\ 3&2&1\end{array}\right]\)
3. \(A=\left[\begin{array}{rrr}2&4&0 \\ 3&6&-2 \\ 1&2&-2\end{array}\right]\)
4. \(A=\left[\begin{array}{rrrr}2&-1&3&5 \\ 2&0&1&2 \\ 6&4&-5&-6 \\ 0&2&-4&-6\end{array}\right]\)

Exercise \(\PageIndex{15}\)

Suppose \(A\) is an \(m\times n\) matrix and \(B\) is an \(n\times p\) matrix. Show that \[\text{dim}(\text{ker}(AB))\leq\text{dim}(\text{ker}(A))+\text{dim}(\text{ker}(B)).\nonumber\] Consider the subspace, \(B(\mathbb{R}^p )∩\text{ker}(A)\) and suppose a basis for this subspace is \(\{\vec{w}_1,\cdots ,\vec{w}_k\}\). Now suppose \(\{\vec{u}_1,\cdots ,\vec{u}_r\}\) is a basis for \(\text{ker}(B)\). Let \(\{\vec{z}_1,\cdots ,\vec{z}_k\}\) be such that \(B\vec{z}_1 =\vec{w}_i\) and argue that \[\text{ker}(AB)⊆ span\{\vec{u}_1,\cdots ,\vec{u}_r,\vec{z}_1,\cdots ,\vec{z}_k\}.\nonumber\]

Answer

Here is how you do this. Suppose \(AB\vec{x} =\vec{0}\). Then \(B\vec{x} ∈ \text{ker}(A) ∩ B(\mathbb{R}^p)\) and so \(B\vec{x} =\sum\limits_{i=1}^k B\vec{z}_i\) showing that \[\vec{x}-\sum\limits_{i=1}^k\vec{z}_i\in\text{ker}(B)\nonumber\] Consider \(B(\mathbb{R}^p )∩\text{ker}(A)\) and let a basis be \(\{\vec{w}_1,\cdots ,\vec{w}_k\}\). Then each \(\vec{w}_i\) is of the form \(B\vec{z}_i =\vec{w}_i\). Therefore, \(\{\vec{z}_1,\cdots ,\vec{z}_k\}\) is linearly independent and \(AB\vec{z}_i = 0\). Now let \(\{\vec{u}_1,\cdots ,\vec{u}_r\}\) be a basis for \(\text{ker}(B)\). If \(AB\vec{x} =\vec{0}\), then \(B\vec{x} ∈ \text{ker}(A)∩B(\mathbb{R}^p)\) and so \(B\vec{x} =\sum\limits_{i=1}^k c_iB\vec{z}_1\) which implies \[\vec{x}-\sum\limits_{i=1}^k c_i\vec{z}_i\in\text{ker}(B)\nonumber\] and so it is of the form \[\vec{x}-\sum\limits_{i=1}^kc_i\vec{z}_i=\sum\limits_{j=1}^r d_j\vec{u}_j\nonumber\] It follows that if \(AB\vec{x} =\vec{0}\) so that \(\vec{x} ∈ \text{ker}(AB)\), then \[\vec{x}\in span (\vec{z}_1,\cdots ,\vec{z}_k,\vec{u}_1, \cdots ,\vec{u}_r ).\nonumber\] Therefore, \[\begin{aligned}\text{dim}(\text{ker}(AB))&\leq k+r=\text{dim}(B(\mathbb{R}^p)∩\text{ker}(A))+\text{dim}(\text{ker}(B)) \\ &\leq\text{dim}(\text{ker}(A))+\text{dim}(\text{ker}(B))\end{aligned}\]

Exercise \(\PageIndex{16}\)

Suppose \(A\) is an m×nm \times nm×n matrix. Prove that every vector in \(\mathbb{R}^n\) can be uniquely written as a sum of a vector in the row space of \(A\) and a vector in the null space of \(A\).

Exercise \(\PageIndex{17}\)

Let \(A\) and \(B\) be two \(m \times n\) matrices such that BBB is obtained from \(A\) by elementary row operations. Prove that \(A\) and \(B\) have the same row space.