5.4E: Rank of a Matrix Exercises

Exercise \(\PageIndex{1}\)

In each case find bases for the row and column spaces of \(A\) and determine the rank of \(A\).

1. \(\left[ \begin{array}{rrrr} 2 & -4 & 6 & 8 \\ 2 & -1 & 3 & 2 \\ 4 & -5 & 9 & 10\\ 0 & -1 & 1 & 2 \end{array} \right]\)
2. \(\left[ \begin{array}{rrr} 2 & -1 & 1 \\ -2 & 1 & 1 \\ 4 & -2 & 3 \\ -6 & 3 & 0 \end{array} \right]\)
3. \(\left[ \begin{array}{rrrrr} 1 & -1 & 5 & -2 & 2 \\ 2 & -2 & -2 & 5 & 1 \\ 0 & 0 & -12 & 9 & -3 \\ -1 & 1 & 7 & -7 & 1 \end{array} \right]\)
4. \(\left[ \begin{array}{rrrr} 1 & 2 & -1 & 3 \\ -3 & -6 & 3 & -2 \end{array} \right]\)

Answer

2. \(\left\{ \left[ \begin{array}{r} 2\\-1\\1\end{array} \right], \left[ \begin{array}{r}0\\0\\1\end{array} \right]\right\}; \left\{\left[ \begin{array}{r}2\\-2\\4\\-6\end{array} \right], \left[ \begin{array}{r}1\\1\\3\\0\end{array} \right]\right\}\)
3. \(\left\{\left[ \begin{array}{r}1\\2\\-1\\3\end{array} \right], \left[ \begin{array}{r}0\\0\\0\\1\end{array} \right]\right\};\left\{\left[ \begin{array}{r}1\\-3\end{array} \right], \left[ \begin{array}{r}3\\-2\end{array} \right]\right\}\)

Exercise \(\PageIndex{2}\)

In each case find a basis of the subspace \(U\).

1. \(U = \text{span} \{(1, -1, 0, 3), (2, 1, 5, 1), (4, -2, 5, 7)\}\)
2. \(U = \text{span} \{(1, -1, 2, 5, 1), (3, 1, 4, 2, 7), (1, 1, 0, 0, 0), (5, 1, 6, 7, 8)\}\)
3. \(U = \text{span} \left\{\left[ \begin{array}{r}1\\1\\0\\0\end{array}\right], \left[ \begin{array}{r}0\\0\\1\\1\end{array}\right], \left[ \begin{array}{r}1\\0\\1\\0\end{array} \right]\left[ \begin{array}{r}0\\1\\0\\1\end{array} \right]\right\}\)
4. \(U = \text{span}\left\{\left[ \begin{array}{r}1\\5\\-6 \end{array} \right], \left[ \begin{array}{r}2\\6\\-8\end{array} \right], \left[ \begin{array}{r}3\\7\\-10\end{array} \right], \left[ \begin{array}{r}4\\8\\12\end{array} \right]\right\}\)

Answer

2. \(\left\{\left[ \begin{array}{r}1\\1\\0\\0\\0\end{array} \right], \left[ \begin{array}{r}0\\-2\\2\\5\\1\end{array} \right], \left[ \begin{array}{r}0\\0\\2\\-3\\6\end{array} \right]\right\}\)
3. \(\left\{\left[ \begin{array}{r}1\\5\\-6\end{array} \right], \left[ \begin{array}{r}0\\1\\-1\end{array} \right]\left[ \begin{array}{r}0\\0\\1\end{array} \right]\right\}\)

Exercise \(\PageIndex{3}\)

1. Can a \(3 \times 4\) matrix have independent columns? \newline Independent rows? Explain.
2. If \(A\) is \(4 \times 3\) and \(\text{rank} A = 2\), can \(A\) have independent columns? Independent rows? Explain.
3. If \(A\) is an \(m \times n\) matrix and \(\text{rank} A = m\), show that \(m \leq n\).
4. Can a nonsquare matrix have its rows independent and its columns independent? Explain.
5. Can the null space of a \(3 \times 6\) matrix have dimension \(2\)? Explain.
6. Suppose that \(A\) is \(5 \times 4\) and \(\text{null}(A) = \mathbb{R}\textbf{x}\) for some column \(\textbf{x} \neq \textbf{0}\). Can \(\text{dim}(\text{im} A) = 2\)?

Answer

2. No; no
3. No
4. Otherwise, if \(A\) is \(m \times n\), we have \(m = \text{dim}(\text{row} A) = \text{rank} A = \text{dim}(\text{col} A) = n\)

Exercise \(\PageIndex{4}\)

If \(A\) is \(m \times n\) show that

\[\text{col}(A) = \{A\textbf{x} \mid \textbf{x} \mbox{ in }\mathbb{R}^n\}\]

Answer

Let \(A =\left[ \begin{array}{ccc} \textbf{c}_{1} & \dots & \textbf{c}_{n} \end{array} \right]\). Then \(\text{col} A = \text{span}\{\textbf{c}_{1}, \dots, \textbf{c}_{n}\} = \{x_{1}\textbf{c}_{1} + \dots + x_{n}\textbf{c}_{n} \mid x_{i} \mbox{ in }\mathbb{R}\} = \{A\textbf{x} \mid \textbf{x} \mbox{ in }\mathbb{R}^n\)

Exercise \(\PageIndex{5}\)

If \(A\) is \(m \times n\) and \(B\) is \(n \times m\), show that \(AB = 0\) if and only if \(\text{col} B \subseteq \text{null} A\).

Exercise \(\PageIndex{6}\)

Show that the rank does not change when an elementary row or column operation is performed on a matrix.

Exercise \(\PageIndex{7}\)

In each case find a basis of the null space of \(A\). Then compute rank \(A\) and verify (1) of Theorem 5.4.2.

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

Answer

2. The basis is \(\left\{ \left[ \begin{array}{r} 6\\ 0\\ -4\\ 1\\ 0 \end{array} \right], \left[ \begin{array}{r} 5\\ 0\\ -3\\ 0\\ 1 \end{array} \right] \right\}\) so the dimension is 2. Have \(\text{rank} A = 3\) and \(n - 3 = 2\).

Exercise \(\PageIndex{8}\)

Let \(A = \textbf{c}\textbf{r}\) where \(\textbf{c} \neq \textbf{0}\) is a column in \(\mathbb{R}^m\) and \(\textbf{r} \neq \textbf{0}\) is a row in \(\mathbb{R}^n\).

1. Show that \(\text{col} A = \text{span}\{\textbf{c}\}\) and \newline \(\text{row} A = \text{span}\{\textbf{r}\}\).
2. Find \(\text{dim}(\text{null} A\)).
3. Show that \(\text{null} A = \text{null} \textbf{r}\).

Answer

1. \(n - 1\)

Exercise \(\PageIndex{9}\)

Let \(A\) be \(m \times n\) with columns \(\textbf{c}_{1}, \textbf{c}_{2}, \dots, \textbf{c}_{n}\).

1. If \(\{\textbf{c}_{1}, \dots, \textbf{c}_{n}\}\) is independent, show \(\text{null} A = \{\textbf{0}\}\).
2. If \(\text{null} A = \{\textbf{0}\}\), show that \(\{\textbf{c}_{1}, \dots, \textbf{c}_{n}\}\) is independent.

Answer

2. If \(r_{1}\textbf{c}_{1} + \dots + r_{n}\textbf{c}_{n} = \textbf{0}\), let \(\textbf{x} = \left[ r_{1}, \dots, r_{n} \right]^{T}\). Then \(C\textbf{x} = r_{1}\textbf{c}_{1} + \dots + r_{n}\textbf{c}_{n} = \textbf{0}\), so \(\textbf{x}\) is in \(\text{null} A = 0\). Hence each \(r_{i} = 0\).

Exercise \(\PageIndex{10}\)

Let \(A\) be an \(n \times n\) matrix.

1. Show that \(A^{2} = 0\) if and only if \(\text{col} A \subseteq \text{null} A\).
2. Conclude that if \(A^{2} = 0\), then \(\text{rank} A \leq \frac{n}{2}\).
3. Find a matrix \(A\) for which \(\text{col} A = \text{null} A\).

Answer

2. Write \(r = \text{rank} A\). Then \textbf{(a)} gives \(r = \text{dim}(\text{col} A \leq \text{dim}(\text{null} A) = n - r\).

Exercise \(\PageIndex{11}\)

Let \(B\) be \(m \times n\) and let \(AB\) be \(k \times n\). If \(\text{rank} B = \text{rank}(AB)\), show that \(\text{null} B = \text{null}(AB)\). [Hint: Theorem 5.4.1.]

Exercise \(\PageIndex{12}\)

Give a careful argument why \(\text{rank}(A^{T}) = \text{rank} A\).

Answer

We have \(\text{rank}(A) = \text{dim}[\text{col}(A)]\) and \(\text{rank}(A^T) = \text{dim}[\text{row}(A^T)]\). Let \(\{\textbf{c}_1, \textbf{c}_2, \dots, \textbf{c}_k\}\) be a basis of \(\text{col}(A)\); it suffices to show that \(\{\textbf{c}_1^T, \textbf{c}_2^T, \dots, \textbf{c}_k^T\}\) is a basis of \(\text{row}(A^T)\). But if \(t_1\textbf{c}_1^T + t_2\textbf{c}_2^T + \dots + t_k\textbf{c}_k^T = \textbf{0}\), \(t_j\) in \(\mathbb{R}\), then (taking transposes) \(t_1\textbf{c}_1 + t_2\textbf{c}_2 + \dots + t_k\textbf{c}_k = \textbf{0}\) so each \(t_j = 0\). Hence \(\{\textbf{c}_1^T, \textbf{c}_2^T, \dots, \textbf{c}_k^T\}\) is independent. Given \(\textbf{v}\) in \(\text{row}(A^T)\) then \(\textbf{v}^T\) is in \(\text{col}(A)\); say \(\textbf{v}^T = s_1\textbf{c}_1 + s_2\textbf{c}_2 + \dots + s_k\textbf{c}_k, s_j \) in \(\mathbb{R}\):
Hence \(\textbf{v} = s_1\textbf{c}_1^T + s_2\textbf{c}_2^T + \dots + s_k\textbf{c}_k^T,\) so \(\{\textbf{c}_1^T, \textbf{c}_2^T, \dots, \textbf{c}_k^T \}\) spans \(\text{row}(A^T)\), as required.

Exercise \(\PageIndex{13}\)

Let \(A\) be an \(m \times n\) matrix with columns \(\textbf{c}_{1}, \textbf{c}_{2}, \dots, \textbf{c}_{n}\). If \(\text{rank} A = n\), show that \(\{A^{T}\textbf{c}_{1}, A^{T}\textbf{c}_{2}, \dots, A^{T}\textbf{c}_{n}\}\) is a basis of \(\mathbb{R}^n\).

Exercise \(\PageIndex{14}\)

If \(A\) is \(m \times n\) and \(\textbf{b}\) is \(m \times 1\), show that \(\textbf{b}\) lies in the column space of \(A\) if and only if \newline \(\text{rank} [A \ \textbf{b}] = \text{rank} A\).

Exercise \(\PageIndex{15}\)

1. Show that \(A\textbf{x} = \textbf{b}\) has a solution if and only if \(\text{rank} A\) = \(\text{rank}[A \ \textbf{b}]\). [\textit{Hint}: Exercises \ref{ex:5_4_12} and \ref{ex:5_4_14}.]
2. If \(A\textbf{x} = \textbf{b}\) has no solution, show that \newline \(\text{rank}[A \ \textbf{b}] = 1 + \text{rank} A\).

Answer

2. Let \(\{\textbf{u}_{1}, \dots, \textbf{u}_{r}\}\) be a basis of \(\text{col}(A)\). Then \(\textbf{b}\) is \textit{not} in \(\text{col}(A)\), so \(\{\textbf{u}_{1}, \dots, \textbf{u}_{r}, \textbf{b}\}\) is linearly independent. Show that \(\text{col}[A \ \textbf{b}] = \text{span}\{\textbf{u}_{1}, \dots, \textbf{u}_{r}, \textbf{b}\}\).

Exercise \(\PageIndex{16}\)

Let \(X\) be a \(k \times m\) matrix. If \(I\) is the \(m \times m\) identity matrix, show that \(I + X^{T}X\) is invertible.

[\textit{Hint}: \(I + X^{T}X = A^{T}A\) where \(A = \left[ \begin{array}{c}I\\X\end{array} \right]\) in block form.]

Exercise \(\PageIndex{17}\)

If \(A\) is \(m \times n\) of rank \(r\), show that \(A\) can be factored as \(A = PQ\) where \(P\) is \(m \times r\) with \(r\) independent columns, and \(Q\) is \(r \times n\) with \(r\) independent rows. [\textit{Hint}: Let

\(UAV = \left[ \begin{array}{rr} I_r & 0 \\ 0 & 0 \end{array} \right]\)

by Theorem 2.5.3, and write

\(U^{-1} = \left[ \begin{array}{rr} U_1 & U_2 \\ U_3 & U_4 \end{array} \right]\)

and

\(V^{-1} = \left[ \begin{array}{rr} V_1 & V_2 \\ V_3 & V_4 \end{array} \right]\)

in block form, where \(U_{1}\) and \(V_{1}\) are \(r \times r]).]

Exercise \(\PageIndex{18}\)

1. Show that if \(A\) and \(B\) have independent columns, so does \(AB\).
2. Show that if \(A\) and \(B\) have independent rows, so does \(AB\).

Exercise \(\PageIndex{19}\)

A matrix obtained from \(A\) by deleting rows and columns is called a \textbf{submatrix}\index{matrix!submatrix}\index{submatrix} of \(A\). If \(A\) has an invertible \(k \times k\) submatrix, show that \(\text{rank} A \geq k\). [\textit{Hint}: Show that row and column operations carry \[A \rightarrow \left[ \begin{array}{rr} I_k & P \\ 0 & Q \end{array} \right] \nonumber\] in block form.

Remark: It can be shown that \(\text{rank} A\) is the largest integer \(r\) such that \(A\) has an invertible \(r \times r\) submatrix.