Exercises for 1
solutions
2
Which of the following are subspaces of \(\mathbf{P}_{3}\)? Support your answer.
- \(U = \{f(x) \mid f(x) \in\|{P}_{3}, f(2) = 1\}\)
- \(U = \{xg(x) \mid g(x) \in\|{P}_{2}\}\)
- \(U = \{xg(x) \mid g(x) \in\|{P}_{3}\}\)
- \(U = \{xg(x) + (1 - x)h(x) \mid g(x) \mbox{ and } h(x) \in\|{P}_{2}\}\)
- \(U =\) The set of all polynomials in \(\mathbf{P}_{3}\) with constant term \(0\)
- \(U = \{f(x) \mid f(x) \in\|{P}_{3}, \text{deg} f(x) = 3\}\)
- Yes
- Yes
- No; not closed under addition or scalar multiplication, and \(0\) is not in the set.
Which of the following are subspaces of \(\mathbf{M}_{22}\)? Support your answer.
- \(U = \left\{ \left[ \begin{array}{rr} a & b \\ 0 & c \end{array} \right] \middle|\ a, b, \mbox{ and } c \mbox{ in } \mathbb{R} \right\}\)
- \(U = \left\{ \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] \middle|\ a + b = c + d;\ a, b, c, d \mbox{ in } \mathbb{R} \right\}\)
- \(U = \{A \mid A \in\|{M}_{22}, A = A^{T}\}\)
- \(U = \{A \mid A \in\|{M}_{22}, AB = 0\}\), \(B\) a fixed \(2 \times 2\) matrix
- \(U = \{A \mid A \in\|{M}_{22}, A^{2} = A\}\)
- \(U = \{A \mid A \in\|{M}_{22}, A \mbox{ is not invertible}\}\)
- \(U = \{A \mid A \in\|{M}_{22}, BAC = CAB\}\), \(B\) and \(C\) fixed \(2 \times 2\) matrices
- Yes.
- Yes.
- No; not closed under addition.
Which of the following are subspaces of \(\mathbf{F}[0, 1]\)? Support your answer.
- \(U = \{f \mid f(0) = 0\}\)
- \(U = \{f \mid f(0) = 1\}\)
- \(U = \{f \mid f(0) = f(1)\}\)
- \(U = \{f \mid f(x) \geq 0 \mbox{ for all } x \mbox{ in } [0, 1]\}\)
- \(U = \{f \mid f(x) = f(y) \mbox{ for all } x \mbox{ and } y \mbox{ in } [0, 1]\}\)
- \(U = \{f \mid f(x + y) = f(x) + f(y) \mbox{ for all } \\ x \mbox{ and } y \mbox{ in } [0, 1]\}\)
- \(U = \{f \mid f \mbox{ is integrable and } \int_{0}^{1} f(x)dx = 0\}\)
- No; not closed under addition.
- No; not closed under scalar multiplication.
- Yes.
Let \(A\) be an \(m \times n\) matrix. For which columns \(\mathbf{b}\) in \(\mathbb{R}^m\) is \(U = \{\mathbf{x} \mid \mathbf{x} \in \mathbb{R}^n, A\mathbf{x} = \mathbf{b}\}\) a subspace of \(\mathbb{R}^n\)? Support your answer.
Let \(\mathbf{x}\) be a vector in \(\mathbb{R}^n\) (written as a column), and define \(U = \{A\mathbf{x} \mid A \in\|{M}_{mn}\}\).
- Show that \(U\) is a subspace of \(\mathbb{R}^m\).
- Show that \(U = \mathbb{R}^m\) if \(\mathbf{x} \neq \mathbf{0}\).
- If entry \(k\) of \(\mathbf{x}\) is \(x_{k} \neq 0\), and if \(\mathbf{y}\) is in \(\mathbb{R}^n\), then \(\mathbf{y} = A\mathbf{x}\) where the column of \(A\) is \(x_{k}^{-1}\mathbf{y}\), and the other columns are zero.
Write each of the following as a linear combination of \(x + 1\), \(x^{2} + x\), and \(x^{2} + 2\).
\(x^{2} + 3x + 2\) \(2x^{2} - 3x + 1\) \(x^{2} + 1\) \(x\)
- \(-3(x + 1) + 0(x^{2} + x) + 2(x^{2} + 2)\)
- \(\frac{2}{3}(x + 1) + \frac{1}{3}(x^{2} + x) - \frac{1}{3}(x^{2} + 2)\)
Determine whether \(\mathbf{v}\) lies in \(span \;\{\mathbf{u}, \mathbf{w}\}\) in each case.
- \(\mathbf{v} = 3x^{2} - 2x - 1\); \(\mathbf{u} = x^{2} + 1\), \(\mathbf{w} = x + 2\)
- \(\mathbf{v} = x\); \(\mathbf{u} = x^{2} + 1\), \(\mathbf{w} = x + 2\)
- \(\mathbf{v} = \left[ \begin{array}{rr} 1 & 3 \\ -1 & 1 \end{array} \right]\); \(\mathbf{u} = \left[ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right]\), \(\mathbf{w} = \left[ \begin{array}{rr} 2 & 1 \\ 1 & 0 \end{array} \right]\)
- \(\mathbf{v} = \left[ \begin{array}{rr} 1 & -4 \\ 5 & 3 \end{array} \right]\); \(\mathbf{u} = \left[ \begin{array}{rr} 1 & -1 \\ 2 & 1 \end{array} \right]\), \(\mathbf{w} = \left[ \begin{array}{rr} 2 & 1 \\ 1 & 0 \end{array} \right]\)
- No.
- Yes; \(\mathbf{v} = 3\mathbf{u} - \mathbf{w}\).
Which of the following functions lie in \(span \;\{\cos^{2}x, \sin^{2} x\}\)? (Work in \(\mathbf{F}[0, \pi]\).)
\(\cos 2x\) \(1\) \(x^{2}\) \(1 + x^{2}\)
- Yes; \(1 = \cos^{2} x + \sin^{2} x\)
- No. If \(1 + x^{2} = a \cos^{2} x + b \sin^{2} x\), then taking \(x = 0\) and \(x = \pi\) gives \(a = 1\) and \(a = 1 + \pi^{2}\).
-
Show that \(\mathbb{R}^3\) is spanned by
\(\{(1, 0, 1), (1, 1, 0), (0, 1, 1)\}\).
- Show that \(\mathbf{P}_{2}\) is spanned by \(\{1 + 2x^{2}, 3x, 1 + x\}\).
-
Show that \(\mathbf{M}_{22}\) is spanned by
\(\left\{ \left[ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right] , \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right] , \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right] , \left[ \begin{array}{rrr} 1 & 1 \\ 0 & 1 \end{array} \right] \right\}\).
- Because \(\mathbf{P}_{2} = span \;\{1, x, x^{2}\}\), it suffices to show that \(\{1, x, x^{2}\} \subseteq\) \(span \;\{1 + 2x^{2}, 3x, 1 + x\}\). But \(x = \frac{1}{3}(3x); 1 = (1 + x) - x\) and \(x^{2} = \frac{1}{2}[(1 + 2x^{2}) - 1]\).
If \(X\) and \(Y\) are two sets of vectors in a vector space \(V\), and if \(X \subseteq Y\), show that \(span \; X \subseteq span \; Y\).
Let \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) denote vectors in a vector space \(V\). Show that:
- \(span \;\{\mathbf{u}, \mathbf{v}, \mathbf{w}\} = span \;\{\mathbf{u} + \mathbf{v}, \mathbf{u} + \mathbf{w}, \mathbf{v} + \mathbf{w}\}\)
- \(span \;\{\mathbf{u}, \mathbf{v}, \mathbf{w}\} = span \;\{\mathbf{u} - \mathbf{v}, \mathbf{u} + \mathbf{w}, \mathbf{w}\}\)
- \(\mathbf{u} = (\mathbf{u} + \mathbf{w}) - \mathbf{w}\), \(\mathbf{v} = -(\mathbf{u} - \mathbf{v}) + (\mathbf{u} + \mathbf{w}) - \mathbf{w}\), and \(\mathbf{w} = \mathbf{w}\)
Show that
\[span \;\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}, \mathbf{0}\} = span \;\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}\} \nonumber \]
holds for any set of vectors \(\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}\}\).
If \(X\) and \(Y\) are nonempty subsets of a vector space \(V\) such that \(span \; X = span \; Y = V\), must there be a vector common to both \(X\) and \(Y\)? Justify your answer.
Is it possible that \(\{(1, 2, 0), (1, 1, 1)\}\) can span the subspace \(U = \{(a, b, 0) \mid a \mbox{ and } b \mbox{ in } \mathbb{R}\}\)?
No.
Describe \(span \;\{\mathbf{0}\}\).
Let \(\mathbf{v}\) denote any vector in a vector space \(V\). Show that \(span \;\{\mathbf{v}\} = span \;\{a\mathbf{v}\}\) for any \(a \neq 0\).
Determine all subspaces of \(\mathbb{R}\mathbf{v}\) where \(\mathbf{v} \neq \mathbf{0}\) in some vector space \(V\).
- Yes.
Suppose \(V = span \;\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}\}\). If \(\mathbf{u} = a_{1}\mathbf{v}_{1} + a_{2}\mathbf{v}_{2} + \dots+ a_{n}\mathbf{v}_{n}\) where the \(a_{i}\) are in \(\mathbb{R}\) and \(a_{1} \neq 0\), show that \(V = span \;\{\mathbf{u}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}\}\).
\(\mathbf{v}_1 = \frac{1}{a_1}\mathbf{u} - \frac{a_2}{a_1}\mathbf{v}_2 - \dots - \frac{a_n}{a_1}\mathbf{v}_n\), so \(V \subseteq span \;\{\mathbf{u}, \mathbf{v}_2, \dots, \mathbf{v}_n \}\)
If \(\mathbf{M}_{nn} = span \;\{A_1, A_2, \dots, A_k\}\), show that \(\mathbf{M}_{nn} = span \;\{A_1^T, A_2^T, \dots, A_k^T\}\).
If \(\mathbf{P}_{n} = span \;\{p_{1}(x), p_{2}(x), \dots, p_{k}(x)\}\) and \(a\) is in \(\mathbb{R}\), show that \(p_{i}(a) \neq 0\) for some \(i\).
Let \(U\) be a subspace of a vector space \(V\).
- If \(a\mathbf{u}\) is in \(U\) where \(a \neq 0\), show that \(\mathbf{u}\) is in \(U\).
- If \(\mathbf{u}\) and \(\mathbf{u} + \mathbf{v}\) are in \(U\), show that \(\mathbf{v}\) is in \(U\).
- \(\mathbf{v} = (\mathbf{u} + \mathbf{v}) - \mathbf{u}\) is in \(U\).
Let \(U\) be a nonempty subset of a vector space \(V\). Show that \(U\) is a subspace of \(V\) if and only if \(\mathbf{u}_{1} + a\mathbf{u}_{2}\) lies in \(U\) for all \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) in \(U\) and all \(a\) in \(\mathbb{R}\).
Given the condition and \(\mathbf{u} \in U\), \(\mathbf{0} = \mathbf{u} + (-1)\mathbf{u} \in U\). The converse holds by the subspace test.
Let \(U = \{p(x) \mbox{ in }\|{P} \mid p(3) = 0\}\) be the set in Example [exa:018124]. Use the factor theorem (see Section [sec:6_5]) to show that \(U\) consists of multiples of \(x - 3\); that is, show that \(U = \{(x - 3)q(x) \mid q(x) \in\|{P}\}\). Use this to show that \(U\) is a subspace of \(\mathbf{P}\).
Let \(A_{1}, A_{2}, \dots, A_{m}\) denote \(n \times n\) matrices. If \(\mathbf{0} \neq \mathbf{y} \in \mathbb{R}^n\) and \(A_{1}\mathbf{y} = A_{2}\mathbf{y} = \dots = A_{m}\mathbf{y} = \mathbf{0}\), show that \(\{A_{1}, A_{2}, \dots, A_{m}\}\) cannot \(span \;\|{M}_{nn}\).
Let \(\{\mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{n}\}\) and \(\{\mathbf{u}_{1}, \mathbf{u}_{2}, \dots, \mathbf{u}_{n}\}\) be sets of vectors in a vector space, and let
\[X = \left[ \begin{array}{c} \mathbf{v}_1 \\ \vdots \\ \mathbf{v}_n \end{array} \right] \ Y = \left[ \begin{array}{c} \mathbf{u}_1 \\ \vdots \\ \mathbf{u}_n \end{array} \right] \nonumber \]
as in Exercise [ex:6_1_18].
- Show that \(span \;\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\} \subseteq span \;\{\mathbf{u}_{1}, \dots, \mathbf{u}_{n}\}\) if and only if \(AY = X\) for some \(n \times n\) matrix \(A\).
- If \(X = AY\) where \(A\) is invertible, show that \(span \;\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\} = span \;\{\mathbf{u}_{1}, \dots, \mathbf{u}_{n}\}\).
If \(U\) and \(W\) are subspaces of a vector space \(V\), let \(U \cup W = \{\mathbf{v} \mid \mathbf{v}\) is in \(U\) or \(\mathbf{v}\) is in \(W\}\). Show that \(U \cup W\) is a subspace if and only if \(U \subseteq W\) or \(W \subseteq U\).
Show that \(\mathbf{P}\) cannot be spanned by a finite set of polynomials.