The Dimension of a Vector Space

Math 303: Section 24

Dr. Janssen

Intro

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

Preview Activity 24.1 Discussion

Upshot: Every basis for a vector space \(V\) has the same number of vectors in it.

Finite Dimensional Vector Spaces

Definition

Definition 1 A finite-dimensional vector space is a vector space that can be spanned by a finite number of vectors. The dimension of a non-trivial finite-dimensional vector space is the number of vectors in a basis for \(V\). The dimension of the trivial space is defined to be 0. We denote the dimension of \(V\) by \(\dim(V)\).

note that not every space is finite dimensional! An example is \(\P\): the collection of all powers of \(t\) form a linearly independent set, which implies that \(\P\) can contain no finite basis. We say \(\P\) is infinite-dimensional.

Activity 24.1

Find the dimensions of each of the following spaces. Justify your answers.

• \(\P_1\)
• \(\P_2\)
• \(\P_n\)

• \(\M_{2\times 3}\)
• \(\M_{3\times 4}\)
• \(\M_{k\times n}\)

upshot: to find the dimension of a space, we need to find a basis.

Activity 24.2

Let \(W = \setof{(a+b) + (a-b) t + (2a+3b)t^2}{a,b\in\R}\).

• Find a finite set of polynomials in \(W\) that span \(W\).
• Determine if the spanning set you found is linearly independent or dependent. Clearly explain your process.
• What is \(\dim(W)\)? Explain.

It’s 2!

Note that the polynomials \(1+t +2t^2\) and \(1-t + 3t^2\) span \(W\). If \(a(1+t +2t^2) + b(1-t+3t^2) = 0\), then the constant terms imply \(a = -b\); but then the linear term resolves to \(2at\), which need not be zero.

Do we want to do another one?

Activity 24.3

Let \(V\) be a finite dimensional vector space of dimension \(n\) and let \(W\) be a subspace of \(V\). Explain why \(W\) cannot have dimension larger than \(\dim(V)\), and if \(W\ne V\), then \(\dim(W) < \dim(V)\).

Conditions for a Basis

Typically, we need to verify two properties to confirm that a given set \(S\) of vectors forms a basis for a vector space \(W\):

• \(W = \span(S)\)
• \(S\) is linearly independent.

But! If we know \(\dim(W)\) ahead of time, things are a bit simpler.

Activity 24.4

For all that follows, let \(W\) be a subspace of a vector space \(V\) with \(\dim(W) = k\).

Suppose first that \(S\) is a subset of \(W\) containing \(k\) vectors, and is linearly independent. Let’s show that \(\span(S) = W\).

• Suppose \(S\) does not span \(W\). Explain why this implies that \(W\) contains a set of \(k+1\) linearly independent vectors.
• Explain why the result of the previous part tells us that \(S\) is a basis for \(W\).

Now suppose that \(S\) is a subset of \(W\) with \(k\) vectors that spans \(W\). Let’s show \(S\) is linearly independent.

• If \(S\) is not linearly independent, explain why we can then find a proper subset of \(S\) that is linearly independent and has the same span as \(S\).
• Explain why the result of the previous part tells us that \(S\) is a basis for \(W\).

Summary: Theorem 24.4

Theorem 1 Let \(W\) be a subspace of dimension \(k\) of a vector space \(V\) and let \(S\) be a subset of \(W\) containing exactly \(k\) vectors.

• If \(S\) is linearly independent, then \(S\) is a basis for \(W\).
• If \(S\) spans \(W\), then \(S\) is a basis for \(W\).

Application: Principal Component Analysis