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Linear Algebra

Definition of a Vector Space

For this lesson, we are going to use the membership symbol from set theory:

A vector space is a non-empty set of vectors \(V\) that all have the same dimension and satisfy the following:

  1. Closed under addition: If \(\overrightarrow{v}, \overrightarrow{w} \in V\) then \(\overrightarrow{v} + \overrightarrow{w} \in V.\)
  2. Closed under multiplication with scalars: If \(\overrightarrow{v} \in V\) and \(r\) is a scalar, then \(r\overrightarrow{v} \in V.\)

In general, a vector space has many properties that must be satisfied. These include things like associativity of addition and commutativity of scalars. All the other necessary properties automatically follow as long as we are dealing with real-valued scalars and vectors.

Properties

  1. Vector spaces are closed under subtraction. For example, if \(\overrightarrow{v}, \overrightarrow{w} \in V\) then by closure with respect to multiplication by scalars, \(-1\overrightarrow{w} \in V.\) By closure of addition \(\overrightarrow{v} + (-1)\overrightarrow{w} = \overrightarrow{v} - \overrightarrow{w} \in V.\)
  2. The zero vector \(\overrightarrow{0}\) of dimension \(n\) is in every vector space of \(n\)-dimensional vectors. Since \(V\) is non-empty, there is some element \(\overrightarrow{v} \in V.\) By closure under subtraction, \(\overrightarrow{v} - \overrightarrow{v} = \overrightarrow{0} \in V.\)

Example

Let \(V\) be the set of all \(3\)-dimensional vectors in which the second coordinate is \(0.\) Then \(V\) is a vector space since it is closed under addition and scalar multiplication.

To check, let \(\overrightarrow{v}\) and \(\overrightarrow{w}\) be elements of \(V.\) Then \(\overrightarrow{v} = <v_1, 0, v_2>\) and \(\overrightarrow{w} = <w_1, 0, w_2>\) for some numbers \(v_1, v_2, w_1, w_2.\) Their sum is \[\overrightarrow{v} + \overrightarrow{w} = <v_1,0,v_2>+<w_1,0,w_2> = <v_1+w_1,0,v_2+w_2> \in V\]

Multiplying \(\overrightarrow{v}\) by a scalar \(r\) gives \[r\overrightarrow{v} = r<v_1,0,v_2> = <rv_1,0,rv_2> \in V\]