Vector Spaces

Vector Spaces#

Let \(\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3\) and \(\alpha, \beta \in \mathbb{R}\) then

A1: \(\vec{u} + (\vec{v} + \vec{w}) = (\vec{u} + \vec{v}) + \vec{w} \checkmark\)
A2: \(\vec{u} + \vec{v} = \vec{v} + \vec{u} \checkmark\)
A3: \(\vec{u} + \vec{0} = \vec{u} \checkmark\)
A4: \(\vec{u} + (-\vec{u}) = \vec{0} \checkmark\)
M1: \(\alpha(\beta \vec{u}) = (\alpha \beta) \vec{u} \checkmark\)
M2: \(1 \vec{u} = \vec{u} \checkmark\)
M3: \(\alpha(\vec{u} + \vec{v}) = \alpha\vec{u} + \alpha \vec{v} \checkmark\)
M4: \((\alpha + \beta) \vec{u} = \alpha \vec{u} + \beta \vec{u} \checkmark\)

All of the axioms of vector spaces hold for \(\mathbb{R}^3\).

(a) \(U\) is non-empty since \(\vec{0} \in U\). Let \(\vec{u} = (u_1, u_2, 0)^\mathsf{T}, \vec{v} = (v_1, v_2, 0)^\mathsf{T} \in U\) and \(\alpha \in \mathbb{R}\) then

\[\begin{split} \vec{u} + \alpha \vec{v} = \begin{pmatrix} u_1 \\ u_2 \\ 0 \end{pmatrix} + \alpha \begin{pmatrix} v_1 \\ v_2 \\ 0 \end{pmatrix} = \begin{pmatrix} u_1 + \alpha v_1 \\ u_2 + \alpha v_2 \\ 0 \end{pmatrix} \in U, \end{split}\]

therefore \(U\) is a subspace.

(b) \(V\) is non-empty since \((1,2,0)^\mathsf{T} \in V\). However \(\alpha (1, 2, 0)^\mathsf{T} = (\alpha , 2\alpha , 0)^\mathsf{T} \notin V\) for \(\alpha \in \mathbb{R}\) so \(V\) is not a subspace.

(c) \(W\) is non-empty since \(\vec{0} \in W\). Let \(\vec{u} = (0, u_2, 0)^\mathsf{T}, \vec{v} = (0, v_2, 0)^\mathsf{T} \in U\) and \(\alpha \in \mathbb{R}\) then

\[\begin{split} \vec{u} + \alpha \vec{v} = \begin{pmatrix} 0 \\ u_2 \\ 0 \end{pmatrix} + \alpha \begin{pmatrix} 0 \\ v_2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ u_2 + \alpha v_2 \\ 0 \end{pmatrix} \in W, \end{split}\]

therefore \(W\) is a subspace. Note that \(W \subseteq U\) so since we showed \(U\) is a subspace then \(W\) must also be a subspace.

(d) \(X\) is not a subspace since if \(\vec{u} = (1, 1, 0)^\mathsf{T}, \vec{v} = (-1, 1, 0)^\mathsf{T} \in X\) then \(\vec{u} + \vec{v} = (0, 2, 0)^\mathsf{T} \notin X\).

Solution to Exercise 5.3

(a) \(A\) is non-empty since \(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \in A\). Let \(U = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \in A\) then

\[\begin{split} 2U = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \notin A, \end{split}\]

so \(A\) is not a subspace.

(b) \(B\) is non-empty since \(0_{2\times 2} \in B\). Let \(U = \begin{pmatrix} u_{11} & 0 \\ u_{11} & u_{11} \end{pmatrix}, V = \begin{pmatrix} v_{11} & 0 \\ v_{11} & v_{11} \end{pmatrix} \in B\) and \(\alpha \in \mathbb{R}\) then

\[\begin{split} \begin{align*} U + \alpha V = \begin{pmatrix} u_{11} & 0 \\ u_{11} & u_{11} \end{pmatrix} + \alpha \begin{pmatrix} v_{11} & 0 \\ v_{11} & v_{11}\end{pmatrix} = \begin{pmatrix} u_{11} + \alpha v_{11} & 0 \\ u_{11} + \alpha v_{11} & u_{11} + \alpha v_{11} \end{pmatrix} \in B, \end{align*} \end{split}\]

so \(B\) is a subspace.

(c) \(C\) is not a subspace since \(U = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \in C\) but \(2U = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \notin C\).

Solution to Exercise 5.4

We need to show that the vectors in the set are linearly independent.

\[\begin{split} \begin{align*} & \left( \begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 2 & 5 & 1 & 0 \\ 0 & 7 & 3 & 0 \end{array} \right) \begin{matrix} \\ R_2 - 2R_1 \\ \phantom{x} \end{matrix} & \longrightarrow & \left( \begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 5 & 3 & 0 \\ 0 & 7 & 3 & 0 \end{array} \right) \begin{matrix} \\ \frac{1}{5}R_2 \\ \phantom{x} \end{matrix} \\ \\ \longrightarrow & \left( \begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 1 & 3/5 & 0 \\ 0 & 7 & 3 & 0 \end{array} \right) \begin{matrix} \\ \\ R_3 - 7R_2 \end{matrix} & \longrightarrow & \left( \begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 1 & 3/5 & 0 \\ 0 & 0 & -6/5 & 0 \end{array} \right) \begin{matrix} \\ \\ -\frac{5}{6}R_3 \end{matrix} \\ \\ \longrightarrow & \left( \begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 1 & 3/5 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right) \begin{matrix} R_1 + R_3 \\ R_2 - \frac{3}{5}R_3 \\ \phantom{x} \end{matrix} & \longrightarrow & \left( \begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right) \end{align*} \end{split}\]

So this set of vectors is a basis for \(\mathbb{R}^3\), calculating the inverse of the coefficient matrix

\[\begin{split} \begin{align*} \det \begin{pmatrix} 1 & 0 & -1 \\ 2 & 5 & 1 \\ 0 & 7 & 3 \end{pmatrix} &= 8 - 14 = -6, \\ \operatorname{adj} \begin{pmatrix} 1 & 0 & -1 \\ 2 & 5 & 1 \\ 0 & 7 & 3 \end{pmatrix} &= \begin{pmatrix} 8 & -6 & 14 \\ -7 & 3 & -7 \\ 5 & -3 & 5 \end{pmatrix}^\mathsf{T} = \begin{pmatrix} 8 & -7 & 5 \\ -6 & 3 & -3 \\ 14 & -7 & 5 \end{pmatrix}, \\ \therefore \begin{pmatrix} 1 & 0 & -1 \\ 2 & 5 & 1 \\ 0 & 7 & 3 \end{pmatrix}^{-1} &= \frac{1}{6} \begin{pmatrix} -8 & 7 & -5 \\ 6 & -3 & 3 \\ -14 & 7 & -5 \end{pmatrix}. \end{align*} \end{split}\]

Let \(U = \left\{ \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 5 \\ 7 \end{pmatrix}, \begin{pmatrix} -1 \\ 1 \\ 3 \end{pmatrix} \right\}\) then

\[\begin{split} \begin{align*} \left[ \begin{pmatrix} 0 \\ 13 \\ 17 \end{pmatrix} \right]_U &= \frac{1}{6} \begin{pmatrix} -8 & 7 & -5 \\ 6 & -3 & 3 \\ -14 & 7 & -5 \end{pmatrix} \begin{pmatrix} 0 \\ 13 \\ 17 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \\ \left[ \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \right]_U &= \frac{1}{6} \begin{pmatrix} -8 & 7 & -5 \\ 6 & -3 & 3 \\ -14 & 7 & -5 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}. \end{align*} \end{split}\]

Solution to Exercise 5.5

We need to find two vectors in \(\mathbb{R}^4\) that are linearly independent to \((1, 1, 2, 4)^\mathsf{T}\) and \((2, -1, -5, 2)^\mathsf{T}\) and one another. Let’s choose \((1, 0, 0, 0)^\mathsf{T}\) and \((0, 1, 0, 0)^\mathsf{T}\) and check for linear dependence

\[\begin{split} \begin{align*} & \left( \begin{array}{cccc|c} 1 & 0 & 1 & 2 & 0 \\ 0 & 1 & 1 & -1 & 0 \\ 0 & 0 & 2 & -5 & 0 \\ 0 & 0 & 4 & 2 & 0 \end{array} \right) \begin{matrix} \\ \\ \frac{1}{2}R_3 \\ \phantom{x} \end{matrix} & \longrightarrow & \left( \begin{array}{cccc|c} 1 & 0 & 1 & 2 & 0 \\ 0 & 1 & 1 & -1 & 0 \\ 0 & 0 & 1 & -5/2 & 0 \\ 0 & 0 & 4 & 2 & 0 \end{array} \right) \begin{matrix} R_1 - R_3 \\ R_2 - R_3 \\ \\ R_4 - 4R_1 \end{matrix} \\ \\ \longrightarrow & \left( \begin{array}{cccc|c} 1 & 0 & 0 & 9/2 & 0 \\ 0 & 1 & 0 & 3/2 & 0 \\ 0 & 0 & 1 & -5/2 & 0 \\ 0 & 0 & 0 & 12 & 0 \end{array} \right) \begin{matrix} \\ \\ \\ \frac{1}{12}R_4 \end{matrix} & \longrightarrow & \left( \begin{array}{cccc|c} 1 & 0 & 0 & 9/2 & 0 \\ 0 & 1 & 0 & 3/2 & 0 \\ 0 & 0 & 1 & -5/2 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array} \right) \begin{matrix} R_1 - \frac{9}{2}R_4 \\ R_2 - \frac{3}{2}R_4 \\ R_3 + \frac{5}{2}R_4 \\ \phantom{x} \end{matrix} \\ \\ \longrightarrow & \left( \begin{array}{cccc|c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array} \right) \end{align*} \end{split}\]

Therefore \(\left\{ \begin{pmatrix} 1 \\ 1 \\ 2 \\ 4 \end{pmatrix}, \begin{pmatrix} 2 \\ -1 \\ -5 \\ 2 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} \right\}\) is a basis for \(\mathbb{R}^4\). Note that we could have used any two vectors in \(\mathbb{R}^4\) that form a linearly independent set of vectors.

Solution to Exercise 5.6

We need to find which of the vectors \(\vec{u}\), \(\vec{v}\), \(\vec{w}\), \(\vec{x}\) and \(\vec{y}\) are linearly dependent (and therefore remove them to from the basis).

\[\begin{split} \begin{align*} & \left( \begin{array}{ccccc|c} 1 & 1 & 2 & 1 & 1 & 0 \\ 2 & -1 & 1 & -1 & -1 & 0 \\ 3 & 2 & 5 & 0 & 0 & 0 \\ 4 & 0 & 3 & 3 & 4 & 0 \end{array} \right) \begin{matrix} \\ R_2 - 2R_1 \\ R_3 - 3R_1 \\ R_4 - 4R_1 \end{matrix} & \longrightarrow & \left( \begin{array}{ccccc|c} 1 & 1 & 2 & 1 & 1 & 0 \\ 0 & -3 & -3 & -3 & -3 & 0 \\ 0 & -1 & -1 & -3 & -3 & 0 \\ 0 & -4 & -5 & -1 & 0 & 0 \end{array} \right) \begin{matrix} \\ -\frac{1}{3}R_2 \\ \phantom{x} \\ \phantom{x} \end{matrix} \\ \\ \longrightarrow & \left( \begin{array}{ccccc|c} 1 & 1 & 2 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0 \\ 0 & -1 & -1 & -3 & -3 & 0 \\ 0 & -4 & -5 & -1 & 0 & 0 \end{array} \right) \begin{matrix} R_1 - R_2 \\ \\ R_3 + R_2 \\ R_4 + 4R_2 \end{matrix} & \longrightarrow & \left( \begin{array}{ccccc|c} 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & -2 & -2 & 0 \\ 0 & 0 & -1 & 3 & 4 & 0 \end{array} \right) \begin{matrix} \\ \\ R_3 \leftrightarrow R_4 \\ \phantom{x} \end{matrix} \\ \\ \longrightarrow & \left( \begin{array}{ccccc|c} 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & -1 & 3 & 4 & 0 \\ 0 & 0 & 0 & -2 & -2 & 0 \end{array} \right) \begin{matrix} \\ \\ -R_3 \\ \phantom{x} \end{matrix} & \longrightarrow & \left( \begin{array}{ccccc|c} 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & -3 & -4 & 0 \\ 0 & 0 & 0 & -2 & -2 & 0 \end{array} \right) \begin{matrix} R_1 - R_3 \\ R_2 - R_3 \\ \phantom{x} \\ \phantom{x} \end{matrix} \\ \\ \longrightarrow & \left( \begin{array}{ccccc|c} 1 & 0 & 0 & 3 & 4 & 0 \\ 0 & 1 & 0 & 4 & 5 & 0 \\ 0 & 0 & 1 & -3 & -4 & 0 \\ 0 & 0 & 0 & -2 & -2 & 0 \end{array} \right) \begin{matrix} \\ \\ \\ -\frac{1}{2}R_4 \end{matrix} & \longrightarrow & \left( \begin{array}{ccccc|c} 1 & 0 & 0 & 3 & 4 & 0 \\ 0 & 1 & 0 & 4 & 5 & 0 \\ 0 & 0 & 1 & -3 & -4 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 \end{array} \right) \begin{matrix} R_1 - 3 R_4 \\ R_2 - 4R_4 \\ R_3 + 3R_4 \\ \phantom{x} \end{matrix} \\ \\ \longrightarrow & \left( \begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 \end{array} \right) \end{align*} \end{split}\]

The fifth column does not have a pivot element so \(\vec{y}\) is linearly dependent on the other vectors, therefore a basis for \(W\) is \(\{ \vec{u}, \vec{v}, \vec{w}, \vec{x}\}\) and \(\dim(W) = 4\).