Finite Dimensional Vector Spaces

Linear Combination

A linear combination of vectors in a vector space \(\setv\) is a vector of the form

\[ \begin{aligned} a_{1}v_{1} + a_{2}v_{2} + \cdots + a_{n}v_{n}\end{aligned} \]

where \(v_{j} \in \setv\) and \(a_{j} \in \field\).

Span

The span of a set of vectors \(v_{1}, \ldots v_{2} \in \setv\) is a set of all vectors which are a linear combination of these vectors

\[ \begin{aligned} span(v_{1}, \ldots, v_{n}) = \{ a_{1}v_{1} + \ldots + a_{n}v_{n} | a_{1}, \ldots a_{n} \in \field \}\end{aligned} \]

The span of an empty list is defined to be the singleton set \(\{ 0 \}\).

The span of a list of vectors is the smalles subspace of \(\setv\) containing all the vectors in the list.

If \(span(v_{1}, \ldots, v_{n}) = \setv\), we say that \(v_{1}, \ldots, v_{n}\) spans \(\setv\).

Finite Dimensional Space

A vector space \(\setv\) is called finite dimensional if some list of vectors in \(\setv\) spans the space. Any space which is not finite dimensional will be called as infinite dimensional space.

Linearly Independent

• Let \(v_{1}, \ldots v_{n}\) be a set of vectors. This set of vectors is said to be linearly independent if \(a_{1}v_{1} + \ldots + a_{n}v_{n} = 0\) for \(a_{1}, \ldots a_{n} \in \field\) only if \(a_{1} = \ldots = a_{n} = 0\). Empty set is also declared to be linearly independent.

• If a set of vectors is removed from a linearly independent list, the remaining vectors in the list are also independent. A list of vectors that is not linearly independent is known as linearly dependent. Any superset of a list of linearly dependent vectors will also be linearly dependent.

• The length of a list of linearly independent vectors \(<\) the length of the list of vectors that span the vector space.

• In a list of linearly dependent vectors, it is possible to express at least one vector as a linear combination of the other vectors in that list. The converse also holds true which can be used to esatablish if a list of vectors is linearly independent or dependent.

• If we express a vector as linear combination of a list of vectors that are linearly independent, the coefficients used in the process are unique. Assume that two representations of the same vector exist

\[ \begin{aligned} v &= \alpha_{1}v_{1} + \cdots + \alpha_{m}v_{m}\newline v &= \gamma_{1}v_{1} + \cdots + \gamma_{m}v_{m}\newline \Rightarrow 0 &= (\alpha_{1} - \gamma_{1})v_{1} + \cdots + (\alpha_{m} - \gamma_{m})v_{m}\newline \Rightarrow (\alpha_{1} - \gamma_{1}) &= \cdots = (\alpha_{m} - \gamma_{m}) = 0 \end{aligned} \]

since the only possible way to get the \(0\) vector is to have all the coefficients \(0\).

Basis

A basis of \(\setv\) is a list of vectors that is linearly independent and spans \(\setv\). For \(\field^{n}\), the list of vectors of length \(n\) such that the \(j^{th}\) element in the list contains all \(0\)s except the element at position \(j\) which is \(1\) is called the standard basis.

For a linear map that is applied on functions, the standard basis is usually the set of polynomials \(1, x, x^{2}, \ldots, x_{n}\) unless stated otherwise.

For a list of vectors \(v_{1}, \ldots, v_{n}\) to be a basis of a vector space \(\setv\), any element \(v \in \setv\) should be expressible as linear combination of the list of vectors \(v = a_{1}v_{1} + \ldots + a_{n}v_{n}\) where \(a_{1}, \ldots a_{n} \in \field\) and are unique. That is, there is no other list of coefficients that will yield vector \(v\) using a linear combination of the vectors in the list.

This follows from the fact that for a linearly independent set of vectors, \(0\) is only expressible by choosing all \(a_{j} = 0\) in the previous equation. If we assume there are two combinations to get \(v\), subtracting them should yield \(0\) which means the list of coefficients is indeed unique.

Every spanning list of a vector space \(\setv\) can be reduced to a basis by removing a set of vectors from that list that may be linearly dependent on the other set of vectors. For any two sets of bases of a vector space, the number of vectors in the basis will always be same.

Dimension

The dimension of a finite dimensional vector space is the length of the basis of the space. It is denoted by dim \(\setv\). For instance, dim \(F^{n} = n\) since the basis has a size \(n\).

It follows that the dimension of a subspace of a vector space \(\setv\) is \(\leq\) the dimension of the vector space \(\setv\). We know that any line passing through origin is a subspace of \(\real^{2}\) and the dim of the line is 1 since all points on the line are just scalar multiples of a vector.

Dimensions of sum of subspaces can be written out as

\[ \begin{aligned} dim(U_{1} + U_{2}) = dim(U_{1}) + dim(U_{2}) - dim(U_{1} \cap U_{2})\end{aligned} \]

analogous to how the size of union of two sets is determined. The same analogy as with sets can be extend to 3 or higher subspaces as well.