4  Basis and Dimension

Basis

A set of vectors v_{1}, \dots, v_{k} is a basis of a subspace \mathcal{W} of a vector space \mathcal{V} over a field \mathbb{F}, if

  • \mathcal{W} = \text{span} (v_{1}, \dots, v_{k}), and

  • v_{1}, \dots, v_{k} are linearly independent.

Properties of basis

  • If \mathcal{B} = \{ b_{1}, \dots, b_{n} \} is a basis of the subspace \mathcal{W}, \mathcal{B} is a minimal spanning set for \mathcal{W}. That is, there is no vector in \mathcal{B} that can be removed such that \mathcal{B} is still a spanning set.

    We will prove by contradiction. Assume \mathcal{B} is a basis of the subspace \mathcal{W} but is not a minimal spanning set, which means that the vectors in \mathcal{B} are linear independent and \mathcal{B} contains at least one vector b such that \hat{\mathcal{B}} = \mathcal{B} \setminus \{ b \} is still a spanning set.

    Therefore, there exists a set of coefficients \alpha_{1}, \dots, \alpha_{n - 1} such that b is a linear combination of the vectors in \hat{\mathcal{B}}

    b = \sum_{b_{i} \in \hat{\mathcal{B}}} \alpha_{i} b_{i}.

    However, this means that the vectors in \mathcal{B} are linear dependent, which violates our assumption.

  • If \mathcal{B} = \{ b_{1}, \dots, b_{n} \} is a basis of the subspace \mathcal{W}, \mathcal{B} is a maximal linearly independent subset of \mathcal{W}. That is, there is no vector in \mathcal{W} that can be added to \mathcal{B} such that \mathcal{B} is still a linearly independent set.

    We will prove by contradiction. Assume \mathcal{B} is a basis of the subspace \mathcal{W} but is not a maximal linearly independent set, which means that \mathcal{B} is a spanning set of \mathcal{W} and there exists a vector b \in \mathcal{W} that can be added to \mathcal{B} such that \hat{\mathcal{B}} = \mathcal{B} \cup \{ b \} is still a linearly independent set.

    However, since \mathcal{B} is a spanning set of \mathcal{W}, there exists a set of coefficients \alpha_{1}, \dots, \alpha_{n} such that b is a linear combination of the vectors in \mathcal{B}

    b = \sum_{b_{i} \in \mathcal{B}} \alpha_{i} b_{i}.

    which means that the vectors in \hat{\mathcal{B}} are linear dependent, which violates our assumption.

  • Existence of basis

    A subspace may NOT have a basis e.g. \mathcal{W} = \{ 0 \} has no linearly independent vector, but a subspace must have a basis if it has a finite spanning set. That is, every finite spanning set of a subspace contains a basis.

    Suppose the subspace \mathcal{W} has a finite spanning set of

    \mathcal{A}_{0} = \{ v_{i}, \dots, v_{n} \}.

    If the spanning set is linearly independent, the set \{ v_{i}, \dots, v_{n} \} is a basis of \mathcal{W}.

    If the spanning set is not linearly independent, then there exists j (1 \leq j \leq n) such that

    v_{j} = \sum_{i=1, i \neq j}^{n} \alpha_{i} v_{i},

    and the set

    \mathcal{A}_{1} = \{ v_{1}, \dots, v_{n} \} \setminus \{ v_{j} \}

    is still a spanning set.

    We can continue removing such v_{j} if the resulting set \mathcal{A}_{i} is not linearly independent.

    Since the resulting set \mathcal{A}_{i} is always a spanning set of \mathcal{W}, we will eventually get to the step i where the \mathcal{A}_{i} is linearly dependent and \mathcal{A}_{i} will be the basis for the subspace \mathcal{W}.

  • Cardinality of basis

    The numbers of elements of all bases of a given subspace are the same.

    Suppose \mathcal{A} is a basis of a subspace \mathcal{S} and \mathcal{B} is another basis of \mathcal{S}. Thus, \mathcal{A} and \mathcal{B} are both linearly independent and spanning sets.

    Since \mathcal{A} is a spanning set and \mathcal{B} is linearly independent,

    \lvert \mathcal{A} \rvert \geq \lvert \mathcal{B} \rvert.

    Since \mathcal{B} is a spanning set and \mathcal{A} is linearly independent,

    \lvert \mathcal{A} \rvert \leq \lvert \mathcal{B} \rvert.

    Thus,

    \lvert \mathcal{A} \rvert = \lvert \mathcal{B} \rvert.

  • Extension of a basis

    If the subspace \mathcal{U} is a subset of a finite dimensional subspace \mathcal{V} and \{ b_{1}, \dots, b_{k} \} is a basis of \mathcal{U}, then there exists an extension of \{ b_{1}, \dots, b_{k} \}

    \{ b_{1}, \dots, b_{k}, b_{k + 1}, \dots, b_{n} \}

    that is a basis for \mathcal{V}

    TODO

Dimension

The dimension of a subspace is the common cardinality of its all bases.

Properties of dimension

  • If a subspace \mathcal{U} is a subset of a finite dimensional subspace \mathcal{V}, then

    \text{dim} (\mathcal{U}) \leq \text{dim} (\mathcal{V}).

    Suppose there is a basis

    \mathcal{A} = \{ b_{1}, \dots, b_{n} \}

    for \mathcal{U}.

    Since \mathcal{U} \subseteq \mathcal{V}, there are 2 cases.

    In the case where \mathcal{U} = \mathcal{V}, \mathcal{A} is also a basis of \mathcal{V} and thus,

    \text{dim} (\mathcal{U}) = \text{dim} (\mathcal{V}).

    In the case where \mathcal{U} \subset \mathcal{V}, there is at least 1 vector x \in \mathcal{V} such that

    x \notin \text{span} (\mathcal{A}).

  • If a subspace \mathcal{U} is a subset of a finite dimensional subspace \mathcal{V} and \text{dim} (\mathcal{U}) = \text{dim} (\mathcal{V}), then

    \mathcal{U} = \mathcal{V}.

    Since \mathcal{U} \subset \mathcal{V} and according to the property of extension of the basis, there exists a basis b_{\mathcal{U}} for \mathcal{U} and b_{\mathcal{V}} for \mathcal{V} such that

    \mathcal{B}_{\mathcal{U}} \subset \mathcal{B}_{\mathcal{V}}.

    However, since \text{dim} (\mathcal{U}) = \text{dim} (\mathcal{V}), it must follows that

    \mathcal{B}_{\mathcal{U}} = \mathcal{B}_{\mathcal{V}}.

    Since \mathcal{U} and \mathcal{V} shares the same basis,

    \mathcal{U} = \mathcal{V}.

  • Given \mathcal{U} and \mathcal{V} are subspaces, then

    \text{dim} (\mathcal{U} + \mathcal{V}) = \text{dim} (\mathcal{U}) + \text{dim} (\mathcal{V}) - \text{dim} (\mathcal{U} \cap \mathcal{V}).

    Suppose \text{dim} (\mathcal{U} \cap \mathcal{V}) = k, \text{dim} (\mathcal{U}) = m, and \text{dim} (\mathcal{V}) = n.

    Assume there is a basis

    \mathcal{B} = \{ b_{1}, \dots, b_{k} \}

    for \mathcal{U} \cap \mathcal{V}.

    Since \mathcal{U} \cap \mathcal{V} \subseteq \mathcal{U} and \mathcal{U} \cap \mathcal{V} \subseteq \mathcal{V}, there exists a basis \mathcal{A} for \mathcal{U} and \mathcal{C} for \mathcal{V}

    \mathcal{A} = \{ b_{1}, \dots, b_{k}, a_{k + 1}, \dots, a_{m} \}

    \mathcal{C} = \{ b_{1}, \dots, b_{k}, c_{k + 1}, \dots, c_{n} \}

    as extensions of \mathcal{B}, according to extension of basis property.

    Consider the set

    \mathcal{A} \cup \mathcal{C} = \{ b_{1}, \dots, b_{k}, a_{k + 1}, \dots, a_{m}, c_{k + 1}, \dots, c_{n} \}

    we will prove that it is a basis of \mathcal{U} + \mathcal{V}.

    First, consider all x \in \mathcal{U} + \mathcal{V}, which can be represented using the basis of \mathcal{U} and the basis of \mathcal{V}

    \begin{aligned} x & = \sum_{i=1}^{k} \alpha_{i} b_{i} + \sum_{i=k+1}^{m} \alpha_{i} a_{i} + \sum_{i=1}^{k} \beta_{i} b_{i} + \sum_{i=k+1}^{m} \beta_{i} c_{i} \\ & = \sum_{i=1}^{k} (\alpha_{i} + \beta_{i}) b_{i} + \sum_{i=k+1}^{m} \alpha_{i} a_{i} + \beta_{i} c_{i}. \end{aligned}

    Thus, any vector x \in \mathcal{U} + \mathcal{V} is a linear combination of vectors in \mathcal{A} \cup \mathcal{C} and we have

    \mathcal{U} + \mathcal{V} \subseteq \text{span} \left( \mathcal{A} \cup \mathcal{C} \right).

    Conversely, all x \in \text{span} \left( \mathcal{A} \cup \mathcal{C} \right) can be written as the linear combinations of vectors in \mathcal{A} and \mathcal{C}, and thus

    \text{span} \left( \mathcal{A} \cup \mathcal{C} \right) \subseteq \mathcal{U} + \mathcal{V}.

    Thus,

    \mathcal{U} + \mathcal{V} = \text{span} \left( \mathcal{A} \cup \mathcal{C} \right).

    Next, we will prove the vectors in \mathcal{A} \cup \mathcal{C} are linearly independent by contradiction.

    Consider x \in \mathcal{U} \cap \mathcal{V}, which can be represented by the basis of \mathcal{U} \cap \mathcal{V}, \mathcal{U} and, \mathcal{V} at the same time

    \begin{aligned} x & = \sum_{i=1}^{k} \alpha_{i} b_{i} + \sum_{i=k+1}^{m} \alpha_{i} a_{i} \\ & = \sum_{i=1}^{k} \beta_{i} b_{i} + \sum_{i=k+1}^{n} \beta_{i} c_{i} \\ & = \sum_{i=1}^{k} \epsilon_{i} b_{i}. \end{aligned}

    Suppose x = 0, then because

Types of the subspaces

We can classify the types of the subspaces based on their dimensions. For example, in \mathbb{R}^{n} there are n types of subspaces

  • 0 dimension subspace: \{ 0 \}.

  • 1 dimension subspaces.

    …

  • n dimension subspace: \mathbb{R}^{n} itself.