Bases for Vector Spaces

Math 303: Section 23

Dr. Janssen

Intro

\[ \def\R{{\mathbb R}} \def\P{{\mathbb P}} \def\b{{\mathbf{b}}} \def\x{{\mathbf{x}}} \def\y{{\mathbf{y}}} \def\u{{\mathbf{u}}} \def\v{{\mathbf{v}}} \def\w{{\mathbf{w}}} \def\z{{\mathbf{z}}} \DeclareMathOperator{\null}{Nul} \DeclareMathOperator{\span}{Span} \newcommand{\set}[1]{\left\{ {#1} \right\}} \newcommand{\setof}[2]{{\left\{#1\,\colon\,#2\right\}}} \]

Linear (In)dependence

Definition 1 A set of vectors \(\set{\v_1, \v_2, \ldots, \v_k}\) in a vector space \(V\) is linearly independent if the vector equation

\[ x_1 \v_1 + x_2 \v_2 + \cdots + x_k \v_k = \mathbf{0} \]

for scalars \(x_1, x_2, \ldots, x_k\) has only the trivial solution

\[ x_1 = x_2 = x_3 = \cdots = x_k = 0. \]

If a set of vectors is not linearly independent, then the set is linearly dependent.

Preview Activity 23.1

(in)dependence of sets of polynomials in \(\P_1\), or the space of \(2\times 2\) matrices.

Consequences

Theorem 1 A set \(\set{\v_1, \ldots, \v_k}\) of vectors in a vector space \(V\) is linearly dependent if and only if at least one of the vectors in the set can be written as a linear combination of the remaining vectors in the set.

Theorem 2 A set \(\set{\v_1, \ldots, \v_k}\) of vectors in a vector space \(V\) is linearly independent if and only if none of the vectors in the set can be written as a linear combination of the remaining vectors in the set.

Upshot is that if a set is linearly independent, we can throw a vector out and the span won’t change.

Spans Preserved

Theorem 3 Let \(\set{\v_1, \v_2, \ldots, \v_k}\) be a set of vectors in a vector space \(V\). If for some \(i\) between \(1\) and \(k\), \(\v_i\in \span\set{\v_1, \v_2, \ldots, \v_k}\), then

\[ \span\set{\v_1, \v_2, \ldots, \v_k} = \span\set{\v_1, \v_2, \ldots, \v_{i-1}, \v_{i+1}, \ldots, \v_k}. \]

Bases

Basis = minimal spanning set.

If we have a spanning set \(S\) for a vsp \(V\), then if \(S\) is linearly dependent, we can throw a vector out. This means \(S\) can’t be minimal. But if \(S\) is linearly dependent, we can’t throw a vector out–all vectors in the set are needed to form the span. So we get the following formal definition.

Definition

Definition 2 A basis for a vector space \(V\) is a subset \(S\) of \(V\) if

• \(\span S = V\)
• \(S\) is linearly independent

Key implications

• A basis for a vector space \(V\) is a minimal spanning set for \(V\)
• A basis for \(V\) is a subset \(S\) of \(V\) for which \(\span S = V\) and \(S\) is linearly independent
• No vector in a basis can be written as a linear combination of the other vectors in the basis.
• If a subset \(S\) of a vector space \(V\) has the property that one of the vectors in \(S\) is a linear combination of the other vectors in \(S\), then \(S\) is not a basis for \(V\).

Activity 23.1

• Is \(S = \set{1+t, t, 1-t}\) a basis for \(\P_1\)? Explain.
• Explain why the set \(S = \set{1, t, t^2, \ldots, t^n}\) is a basis for \(\P_n\). This basis is called the standard basis for \(\P_n\).
• Show that the set

\[ \set{ \left[\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}\right], \left[\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}\right], \left[\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}\right], \left[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right]} \]

is a basis for \(\mathcal{M}_{2\times 2}\).

Note that not every vector space has a finite basis! And being able to ‘use’ this ‘fact’ requires a somewhat controversial set theory axiom. But it’s only somewhat controversial, and not with me, so we will use it.

Application: Wavelets and Image Compression

Intro