5.2E: Independence and Dimension Exercises

Exercise \(\PageIndex{1}\)

Which of the following subsets are independent? Support your answer.

1. \(\{(1, -1, 0)\), \((3, 2, -1)\), \((3, 5, -2)\}\) in \(\mathbb{R}^3\)
2. \(\{(1, 1, 1)\), \((1, -1, 1)\), \((0, 0, 1)\}\) in \(\mathbb{R}^3\)
3. \(\{(1, -1, 1, -1)\), \((2, 0, 1, 0)\), \((0, -2, 1, -2)\}\) in \(\mathbb{R}^4\)
4. \(\{(1, 1, 0, 0)\), \((1, 0, 1, 0)\), \((0, 0, 1, 1)\), \((0, 1, 0, 1)\}\) in \(\mathbb{R}^4\)

Answer

2. Yes. If \(r \left[ \begin{array}{r} 1 \\ 1 \\ 1 \end{array} \right] + s \left[ \begin{array}{r} 1 \\ 1 \\ 1 \end{array} \right] + t \left[ \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right]\), then \(r + s = 0\), \(r - s = 0\), and \(r + s + t = 0\). These equations give \(r = s = t = 0\).
3. No. Indeed: \(\left[ \begin{array}{r} 1 \\ 1 \\ 0 \\ 0 \end{array} \right] - \left[ \begin{array}{r} 1 \\ 0 \\ 1 \\ 0 \end{array} \right] + \left[ \begin{array}{r} 0 \\ 0 \\ 1 \\ 1 \end{array} \right] - \left[ \begin{array}{r} 0 \\ 1 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \end{array} \right]\).

Exercise \(\PageIndex{2}\)

Let \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{w}\}\) be an independent set in \(\mathbb{R}^n\). Which of the following sets is independent? Support your answer.

1. \(\{\mathbf{x} - \mathbf{y}, \mathbf{y} - \mathbf{z}, \mathbf{z} - \mathbf{x}\}\)
2. \(\{\mathbf{x} + \mathbf{y}, \mathbf{y} + \mathbf{z}, \mathbf{z} + \mathbf{x}\}\)
3. \(\{\mathbf{x} - \mathbf{y}, \mathbf{y} - \mathbf{z}, \mathbf{z} - \mathbf{w}, \mathbf{w} - \mathbf{x}\}\)
4. \(\{\mathbf{x} + \mathbf{y}, \mathbf{y} + \mathbf{z}, \mathbf{z} + \mathbf{w}, \mathbf{w} + \mathbf{x}\}\)

Answer

2. Yes. If \(r(\mathbf{x} + \mathbf{y}) + s(\mathbf{y} + \mathbf{z}) + t(\mathbf{z} + \mathbf{x}) = \mathbf{0}\), then \((r + t)\mathbf{x} + (r + s)\mathbf{y} + (s + t)\mathbf{z} = \mathbf{0}\). Since \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}\) is independent, this implies that \(r + t = 0\), \(r + s = 0\), and \(s + t = 0\). The only solution is \(r = s = t = 0\).
3. No. In fact, \((\mathbf{x} + \mathbf{y}) - (\mathbf{y} + \mathbf{z}) + (\mathbf{z} + \mathbf{w}) - (\mathbf{w} + \mathbf{x}) = \mathbf{0}\).

Exercise \(\PageIndex{3}\)

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

1. \(span \;\{(1, -1, 2, 0), (2, 3, 0, 3), (1, 9, -6, 6)\}\)
2. \(span \;\{(2, 1, 0, -1), (-1, 1, 1, 1), (2, 7, 4, 1)\}\)
3. \(span \;\{(-1, 2, 1, 0), (2, 0, 3, -1), (4, 4, 11, -3), \\ (3, -2, 2, -1)\}\)
4. \(span \;\{(-2, 0, 3, 1), (1, 2, -1, 0), (-2, 8, 5, 3), \\ (-1, 2, 2, 1)\}\)

Answer

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

Exercise \(\PageIndex{4}\)

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

1. \(U = \left \{ \left[ \begin{array}{c} a \\ a + b \\ a - b \\ b \end{array} \right] \, \middle| \, a \mbox{ and } b \mbox{ in } \mathbb{R} \right \}\)
2. \(U = \left \{ \left[ \begin{array}{c} a + b\\ a - b \\ b \\ a \end{array} \right] \, \middle| \, a \mbox{ and } b \mbox{ in } \mathbb{R} \right \}\)
3. \(U = \left \{ \left[ \begin{array}{c} a \\ b \\ c + a \\ c \end{array} \right] \, \middle| \, a, b, \mbox{ and } c \mbox{ in } \mathbb{R} \right \}\)
4. \(U = \left \{ \left[ \begin{array}{c} a - b\\ b + c \\ a \\ b + c \end{array} \right] \, \middle| \, a, b, \mbox{ and } c \mbox{ in } \mathbb{R} \right \}\)
5. \(U = \left \{ \left[ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right] \, \middle| \, a + b - c + d = 0 \mbox{ in } \mathbb{R} \right \}\)
6. \(U = \left \{ \left[ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right] \, \middle| \, a + b = c + d \mbox{ in } \mathbb{R} \right \}\)

Answer

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

Exercise \(\PageIndex{5}\)

Suppose that \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{w}\}\) is a basis of \(\mathbb{R}^4\). Show that:

1. \(\{\mathbf{x} + a\mathbf{w}, \mathbf{y}, \mathbf{z}, \mathbf{w}\}\) is also a basis of \(\mathbb{R}^4\) for any choice of the scalar \(a\).
2. \(\{\mathbf{x} + \mathbf{w}, \mathbf{y} + \mathbf{w}, \mathbf{z} + \mathbf{w}, \mathbf{w}\}\) is also a basis of \(\mathbb{R}^4\).
3. \(\{\mathbf{x}, \mathbf{x} + \mathbf{y}, \mathbf{x} + \mathbf{y} + \mathbf{z}, \mathbf{x} + \mathbf{y} + \mathbf{z} + \mathbf{w}\}\) is also a basis of \(\mathbb{R}^4\).

Answer

2. If \(r(\mathbf{x} + \mathbf{w}) + s(\mathbf{y} + \mathbf{w}) + t(\mathbf{z} + \mathbf{w}) + u(\mathbf{w}) = \mathbf{0}\), then \(r\mathbf{x} + s\mathbf{y} + t\mathbf{z} + (r + s + t + u)\mathbf{w} = \mathbf{0}\), so \(r = 0\), \(s = 0\), \(t = 0\), and \(r + s + t + u = 0\). The only solution is \(r = s = t = u = 0\), so the set is independent. Since \(dim \; \mathbb{R}^4 = 4\), the set is a basis by Theorem 5.2.7.

Exercise \(\PageIndex{6}\)

Use Theorem 5.2.3 to determine if the following sets of vectors are a basis of the indicated space.

1. \(\{(3, -1), (2, 2)\}\) in \(\mathbb{R}^2\)
2. \(\{(1, 1, -1), (1, -1, 1), (0, 0, 1)\}\) in \(\mathbb{R}^3\)
3. \(\{(-1, 1, -1), (1, -1, 2), (0, 0, 1)\}\) in \(\mathbb{R}^3\)
4. \(\{(5, 2, -1), (1, 0, 1), (3, -1, 0)\}\) in \(\mathbb{R}^3\)
5. \(\{(2, 1, -1, 3), (1, 1, 0, 2), (0, 1, 0, -3), \\(-1, 2, 3, 1)\}\) in \(\mathbb{R}^4\)
6. \(\{(1, 0, -2, 5), (4, 4, -3, 2), (0, 1, 0, -3), \\(1, 3, 3, -10)\}\) in \(\mathbb{R}^4\)

Answer

2. Yes
3. Yes
4. No.

Exercise \(\PageIndex{7}\)

In each case show that the statement is true or give an example showing that it is false.

1. If \(\{\mathbf{x}, \mathbf{y}\}\) is independent, then \(\{\mathbf{x}, \mathbf{y}, \mathbf{x} + \mathbf{y}\}\) is independent.
2. If \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}\) is independent, then \(\{\mathbf{y}, \mathbf{z}\}\) is independent.
3. If \(\{\mathbf{y}, \mathbf{z}\}\) is dependent, then \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}\) is dependent for any \(\mathbf{x}\).
4. If all of \(\mathbf{x}_{1}, \mathbf{x}_{2}, \dots, \mathbf{x}_{k}\) are nonzero, then \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \dots, \mathbf{x}_{k}\}\) is independent.
5. If one of \(\mathbf{x}_{1}, \mathbf{x}_{2}, \dots, \mathbf{x}_{k}\) is zero, then \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \dots, \mathbf{x}_{k}\}\) is dependent.
6. If \(a\mathbf{x} + b\mathbf{y} + c\mathbf{z} = \mathbf{0}\), then \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}\) is independent.
7. If \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}\) is independent, then \(a\mathbf{x} + b\mathbf{y} + c\mathbf{z} = \mathbf{0}\) for some \(a\), \(b\), and \(c\) in \(\mathbb{R}\).
8. If \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \dots, \mathbf{x}_{k}\}\) is dependent, then \(t_{1}\mathbf{x}_{1} + t_{2}\mathbf{x}_{2} + \dots + t_{k}\mathbf{x}_{k} = \mathbf{0}\) for some numbers \(t_{i}\) in \(\mathbb{R}\) not all zero.
9. If \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \dots, \mathbf{x}_{k}\}\) is independent, then \(t_{1}\mathbf{x}_{1} + t_{2}\mathbf{x}_{2} + \dots + t_{k}\mathbf{x}_{k} = \mathbf{0}\) for some \(t_{i}\) in \(\mathbb{R}\).
10. Every non-empty subset of a linearly independent set is again linearly independent.
11. Every set containing a spanning set is again a spanning set.

Answer

2. T. If \(r\mathbf{y} + s\mathbf{z} = \mathbf{0}\), then \(0\mathbf{x} + r\mathbf{y} + s\mathbf{z} = \mathbf{0}\) so \(r = s = 0\) because \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}\) is independent.
3. F. If \(\mathbf{x} \neq \mathbf{0}\), take \(k = 2\), \(\mathbf{x}_{1} = \mathbf{x}\) and \(\mathbf{x}_{2} = -\mathbf{x}\).
4. F. If \(\mathbf{y} = -\mathbf{x}\) and \(\mathbf{z} = \mathbf{0}\), then \(1\mathbf{x} + 1\mathbf{y} + 1\mathbf{z} = \mathbf{0}\).
5. T. This is a nontrivial, vanishing linear combination, so the \(\mathbf{x}_{i}\) cannot be independent.

Exercise \(\PageIndex{8}\)

If \(A\) is an \(n \times n\) matrix, show that \(\det A = 0\) if and only if some column of \(A\) is a linear combination of the other columns.

Exercise \(\PageIndex{9}\)

Let \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}\) be a linearly independent set in \(\mathbb{R}^4\). Show that \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{e}_{k}\}\) is a basis of \(\mathbb{R}^4\) for some \(\mathbf{e}_{k}\) in the standard basis \(\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}, \mathbf{e}_{4}\}\).

Exercise \(\PageIndex{10}\)

If \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}, \mathbf{x}_{4}, \mathbf{x}_{5}, \mathbf{x}_{6}\}\) is an independent set of vectors, show that the subset \(\{\mathbf{x}_{2}, \mathbf{x}_{3}, \mathbf{x}_{5}\}\) is also independent.

Answer

If \(r\mathbf{x}_{2} + s\mathbf{x}_{3} + t\mathbf{x}_{5} = \mathbf{0}\) then \(0\mathbf{x}_{1} + r\mathbf{x}_{2} + s\mathbf{x}_{3} + 0\mathbf{x}_{4} + t\mathbf{x}_{5} + 0\mathbf{x}_{6} = \mathbf{0}\) so \(r = s = t = 0\).

Exercise \(\PageIndex{11}\)

Let \(A\) be any \(m \times n\) matrix, and let \(\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}, \dots, \mathbf{b}_{k}\) be columns in \(\mathbb{R}^m\) such that the system \(A\mathbf{x} = \mathbf{b}_{i}\) has a solution \(\mathbf{x}_{i}\) for each \(i\). If \(\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}, \dots, \mathbf{b}_{k}\}\) is independent in \(\mathbb{R}^m\), show that \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}, \dots, \mathbf{x}_{k}\}\) is independent in \(\mathbb{R}^n\).

Exercise \(\PageIndex{12}\)

If \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}, \dots, \mathbf{x}_{k}\}\) is independent, show \(\{\mathbf{x}_{1}, \mathbf{x}_{1} + \mathbf{x}_{2}, \mathbf{x}_{1} + \mathbf{x}_{2} + \mathbf{x}_{3}, \dots, \mathbf{x}_{1} + \mathbf{x}_{2} + \dots + \mathbf{x}_{k}\}\) is also independent.

Exercise \(\PageIndex{13}\)

If \(t_{1}\mathbf{x}_{1} + t_{2}(\mathbf{x}_{1} + \mathbf{x}_{2}) + \dots + t_{k}(\mathbf{x}_{1} + \mathbf{x}_{2} + \dots + \mathbf{x}_{k}) = \mathbf{0}\), then \((t_{1} + t_{2} + \dots + t_{k})\mathbf{x}_{1} + (t_{2} + \dots + t_{k})\mathbf{x}_{2} + \dots + (t_{k-1} + t_{k})\mathbf{x}_{k-1} + (t_{k})\mathbf{x}_{k} = \mathbf{0}\). Hence all these coefficients are zero, so we obtain successively \(t_{k} = 0, t_{k-1} = 0, \dots, t_{2} = 0, t_{1} = 0\).

Exercise \(\PageIndex{14}\)

If \(\{\mathbf{y}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}, \dots, \mathbf{x}_{k}\}\) is independent, show that \(\{\mathbf{y} + \mathbf{x}_{1}, \mathbf{y} + \mathbf{x}_{2}, \mathbf{y} + \mathbf{x}_{3}, \dots, \mathbf{y} + \mathbf{x}_{k}\}\) is also independent.

Answer

If \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \dots, \mathbf{x}_{k}\}\) is independent in \(\mathbb{R}^n\), and if \(\mathbf{y}\) is not in \(span \;\{\mathbf{x}_{1}, \mathbf{x}_{2}, \dots, \mathbf{x}_{k}\}\), show that \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \dots, \mathbf{x}_{k}, \mathbf{y}\}\) is independent.

Exercise \(\PageIndex{15}\)

If \(A\) and \(B\) are matrices and the columns of \(AB\) are independent, show that the columns of \(B\) are independent.

Exercise \(\PageIndex{16}\)

Suppose that \(\{\mathbf{x}, \mathbf{y}\}\) is a basis of \(\mathbb{R}^2\), and let \(A = \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right]\).

1. If \(A\) is invertible, show that \(\{a\mathbf{x} + b\mathbf{y}, c\mathbf{x} + d\mathbf{y}\}\) is a basis of \(\mathbb{R}^2\).
2. If \(\{a\mathbf{x} + b\mathbf{y}, c\mathbf{x} + d\mathbf{y}\}\) is a basis of \(\mathbb{R}^2\), show that \(A\) is invertible.

Answer

2. We show \(A^{T}\) is invertible (then \(A\) is invertible). Let \(A^{T}\mathbf{x} = \mathbf{0}\) where \(\mathbf{x} = [s\ t]^{T}\). This means \(as + ct = 0\) and \(bs + dt = 0\), so \(s(a\mathbf{x} + b\mathbf{y}) + t(c\mathbf{x} + d\mathbf{y}) = (sa + tc)\mathbf{x} + (sb + td)\mathbf{y} = \mathbf{0}\). Hence \(s = t = 0\) by hypothesis.

Exercise \(\PageIndex{17}\)

Let \(A\) denote an \(m \times n\) matrix.

1. Show that \(\text{null} A = \text{null}(UA)\) for every invertible \(m \times m\) matrix \(U\).
2. Show that \(dim \;(\text{null} A) = dim \;(\text{null}(AV))\) for every invertible \(n \times n\) matrix \(V\). [Hint: If \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \dots, \mathbf{x}_{k}\}\) is a basis of \(\text{null} A\), show that \(\{V^{-1}\mathbf{x}_{1}, V^{-1}\mathbf{x}_{2}, \dots, V^{-1}\mathbf{x}_{k}\}\) is a basis of \(\text{null}(AV)\).]

Answer

Each \(V^{-1}\mathbf{x}_{i}\) is in \(\text{null}(AV)\) because \(AV(V^{-1}\mathbf{x}_{i}) = A\mathbf{x}_{i} = \mathbf{0}\). The set \(\{V^{-1}\mathbf{x}_{1}, \dots, V^{-1}\mathbf{x}_{k}\}\) is independent as \(V^{-1}\) is invertible. If \(\mathbf{y}\) is in \(\text{null}(AV)\), then \(V\mathbf{y}\) is in \(\text{null}(A)\) so let \(V\mathbf{y} = t_{1}\mathbf{x}_{1} + \dots + t_{k}\mathbf{x}_{k}\) where each \(t_{k}\) is in \(\mathbb{R}\). Thus \(\mathbf{y} = t_{1}V^{-1}\mathbf{x}_{1} + \dots + t_{k}V^{-1}\mathbf{x}_{k}\) is in \(span \;\{V^{-1}\mathbf{x}_{1}, \dots, V^{-1}\mathbf{x}_{k}\}\).

Exercise \(\PageIndex{18}\)

Let \(A\) denote an \(m \times n\) matrix.

1. Show that \(im \; A = im \;(AV)\) for every invertible \(n \times n\) matrix \(V\).
2. Show that \(dim \;(im \; A) = dim \;(im \;(UA))\) for every invertible \(m \times m\) matrix \(U\). [Hint: If \(\{\mathbf{y}_{1}, \mathbf{y}_{2}, \dots, \mathbf{y}_{k}\}\) is a basis of \(im \;(UA)\), show that \(\{U^{-1}\mathbf{y}_{1}, U^{-1}\mathbf{y}_{2}, \dots, U^{-1}\mathbf{y}_{k}\}\) is a basis of \(im \; A\).]

Exercise \(\PageIndex{19}\)

Let \(U\) and \(W\) denote subspaces of \(\mathbb{R}^n\), and assume that \(U \subseteq W\). If \(dim \; U = n - 1\), show that either \(W = U\) or \(W = \mathbb{R}^n\).

Exercise \(\PageIndex{20}\)

Let \(U\) and \(W\) denote subspaces of \(\mathbb{R}^n\), and assume that \(U \subseteq W\). If \(dim \; W = 1\), show that either \(U = \{\mathbf{0}\}\) or \(U = W\).

Answer

We have \(\{\mathbf{0}\} \subseteq U \subseteq W\) where \(dim \;\{\mathbf{0}\} = 0\) and \(dim \; W = 1\). Hence \(dim \; U = 0\) or \(dim \; U = 1\) by Theorem 5.2.8, that is \(U = 0\) or \(U = W\), again by Theorem 5.2.8.