Bases#

Definition: Basis

A basis of a vector space \((V,F,+,\cdot)\) is a linearly independent spanning set for it.

Example

\[ \begin{aligned} \mathbf{e}_1 &= (1, 0, \dotsc, 0) \\ \mathbf{e}_2 &= (0, 1, 0, \dotsc, 0) \\ &\vdots \\ \mathbf{e}_n &= (0, \dotsc, 0, 1) \end{aligned} \]

in \(F^n\) are a basis for \(F^n\).

Example

The polynomials \(1, x, x^2, \dotsc\) are a basis for the space of all polynomials over \(F\).

Equivalent Definition

A set of vectors is a basis for a vector space \((V,F,+,\cdot)\) if and only if it is a maximal linearly independent set.

Proof

TODO

Equivalent Definition

A set of vectors is a basis for a vector space \((V,F,+,\cdot)\) if and only if it is a minimal spanning set.

Proof

TODO

Theorem: Existence of a Basis

Every vector space has at least one basis.

Proof

TODO

Theorem: Existence of a Finite Basis

A vector space has a finite basis if and only if it is finitely generated.

Proof

TODO

Dimension#

Theorem: Number of Basis Elements

All bases of a given vector space \((V, F, +, \cdot)\) have the same cardinality.

Proof

TODO

Definition: Dimension of a Vector

If these bases are finite, then we call \((V, F, +, \cdot)\) finite-dimensional. In this case, the cardinality of the bases is known as the dimension of \((V, F, +, \cdot)\).

Notation

\[ \dim(V) \]

If these bases are infinite, then we call \((V, F, +, \cdot)\) infinite-dimensional.

Theorem: Dimension Equality

Two finite dimensional vector spaces have the same dimension if and only if there is a vector space isomorphism between them.

Proof

TODO

Theorem: Basis Representation

Let \((V,F,+,\cdot)\) a vector space.

If \(B\) is a basis of \((V,F,+,\cdot)\), then each vector of \((V,F,+,\cdot)\) can be uniquely expressed as a linear combination of elements from \(B\).

Note: Uniqueness

Here, "uniquely" means that no two vectors are expressed as a linear combination which has the exact same coefficients in front of the exact same basis vectors.

Proof

Let \(\mathbf{v}_1, \dotsc, \mathbf{v}_n \in B\) be pairwise different and let \(\mathbf{v} \in V\). Furthermore, suppose that \(\mathbf{v}\) has the representations \(\sum_{i=1}^n \lambda_i \mathbf{v}_i\) and \(\sum_{i=1}^n \mu_i \mathbf{v}_i\), i.e.

\[ \mathbf{v} = \sum_{i=1}^n \lambda_i \mathbf{v}_i = \sum_{i=1}^n \mu_i \mathbf{v}_i. \]

We therefore have the following:

\[ \sum_{i = 1}^n (\lambda_i - \mu_i)\mathbf{v}_i = \mathbf{0} \]

Since \(B\) is a basis, we know that \(\mathbf{v}_1, \dotsc, \mathbf{v}_n\) are linearly independent. Therefore, we have \(\lambda_i - \mu_i = 0\) for all \(i \in \{1, \dotsc, n\}\), i.e. \(\lambda_i = mu_i\).

Theorem: Basis Criterion

Let \((V,F,+,\cdot)\) be a finite dimensional vector space.

Every linearly independent subset whose cardinality is equal to the dimension \(\dim(V)\) is a basis for \((V,F,+,\cdot)\).

Proof

TODO

Theorem: Basis Extension

Let \(V\) be a finite dimensional vector space with \(\dim V = n\).

If \(\mathbf{v}_1, \dotsc, \mathbf{v}_k\) are linearly independent, then there exist \(\mathbf{v}_{k+1}, \dotsc, \mathbf{v}_n \in V\) such that \(\mathbf{v}_1, \dotsc, \mathbf{v}_n\) is a basis of \(V\). Moreover, if \(B\) is already a basis of \(V\), then \(\mathbf{v}_{k+1}, \dotsc, \mathbf{v}_n\) can be chosen such that \(\mathbf{v}_{k+1}, \dotsc, \mathbf{v}_n \in B\).

Proof

Let \(\mathbf{w}_1, \dotsc, \mathbf{w}_n \in V\) be arbitrary.

TODO

Theorem: Steinitz Exchange Lemma

Let \(B = \{\mathbf{v}_1,\cdots,\mathbf{v}_n\}\) be a basis of a finitely generated vector space \((V, F, +,\cdot)\).

In each set \(\{\mathbf{u}_1,\cdots,\mathbf{u}_m\}\subset V\) of \(m\) linearly independent vectors there are \(n-m\) vectors in \(B\) (without loss of generality \(\mathbf{v}_{m+1}, \cdots, \mathbf{v}_n\)) such that

\[ \{\mathbf{u}_1,\cdots,\mathbf{u}_m,\mathbf{v}_{m+1},\cdots,\mathbf{v}_n\} \]

is also a basis of \((V,F,+,\cdot)\).

Tip

This means that for every set \(\{\mathbf{u}_1,\cdots,\mathbf{u}_m\}\subset V\) of \(m\) linearly independent vectors, we can find \(m\) vectors in \(B\) which we can replace with \(\mathbf{u}_1,\cdots,\mathbf{u}_m\) and still obtain a basis of \((V,K,+,\cdot)\).

Proof

TODO

Ordered Bases#

Definition: Ordered Basis

Let \((V,F,+,\cdot)\) be an finite-dimensional vector space with \(\dim(V) = n\);

An ordered basis of \((V,F,+,\cdot)\) is an \(n\)-tuple \((\mathbf{b}_1, \cdots, \mathbf{b}_n)\) such that \(\{\mathbf{b}_1, \cdots, \mathbf{b}_n\}\) is a basis of \((V,F,+,\cdot)\).

Definition: Coordinate Vector

Let \(B = (\mathbf{b}_1, \cdots, \mathbf{b}_n)\) be an ordered basis of an \(n\)-dimensional vector space \((V,F,+,\cdot)\).

If \(\mathbf{v} \in V\) has the basis representation \(\mathbf{v} = v_1\mathbf{b}_1 + \cdots + v_n \mathbf{b}_n\), then the coordinate vector of \(\mathbf{v}\) with respect to the basis \(B\) is the column vector

\[ [\mathbf{v}]_{B} \overset{\text{def}}{=} \begin{bmatrix} v_1 \\ \vdots \\ v_n \end{bmatrix} \in F^{n}. \]

Definition: Coordinate System

The isomorphism \(\phi: F^n \to V\) which to each \(n\)-tuple \((\lambda_1, \dotsc, \lambda_n) \in F^n\) assigns the vector \(\sum_{k = 1}^{n} \lambda_k \mathbf{b}_k\) is known as the coordinate system with respect to \(B\).