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Linear Algebra: Companion notes

Abhay Soman

Section 2.1 Definition of a vector space

Let \(V\) be an abelian group. We write the binary operation on \(V\) additively. Thus \((V,+)\) have the following properties:
  1. For \(v,w\in V\text{,}\) \(v+w=w+v\text{.}\)
  2. For \(v,w,z\in V\text{,}\) \(v+(w+z)=(v+w)+z\text{.}\)
  3. There exists a unique element \(0\in V\) such that for any \(v\in V\text{,}\) \(0+v=v+0=v\text{.}\)
  4. For every \(v\in V\) there exists a unique element (which is denoted by \(-v\)) such that \(v+(-v)=(-v)+v=0\text{.}\)
Now we give the definition of a vector space over a field.

Definition 2.1.1. (Vector space over a field).

Let \(F\) be a field and let \(V\) be an abelian group. We call \(V\) a vector space over \(F\) or an \(F\)-vector space if there is a map, called scalar multiplication
\begin{equation*} F\times V\to V \end{equation*}
satisfying the following conditions. First, for a given \(\alpha\in F\) and \(v\in V\text{,}\) let us denote the image of \((\alpha,v)\) under this map simply by \(\alpha\cdot v\text{.}\)
  1. Let \(1\) be the unity in \(F\text{.}\) Then,
    \begin{equation*} 1\cdot v= v \end{equation*}
    for every \(v\in V\text{.}\)
  2. For \(\alpha,\beta\in F\) and \(v\in V\text{,}\) we have
    \begin{equation*} (\alpha\beta)\cdot v=\alpha\cdot(\beta\cdot v). \end{equation*}
  3. For \(\alpha\in F\) and \(v,w\in V\) we have
    \begin{equation*} \alpha\cdot(v+w)=\alpha\cdot v+\alpha\cdot w. \end{equation*}
  4. For \(\alpha,\beta\in F\) and \(v\in V\text{,}\) we have
    \begin{equation*} (\alpha+\beta)\cdot v=\alpha\cdot v+\beta\cdot v. \end{equation*}
The elements of \(F\) are called scalars and elements of \(V\) are called vectors.

Remark 2.1.2.

Often we omit "\(\cdot\)" for scalar multiplication and simply write \(\alpha v\text{.}\)

Remark 2.1.3.

Please try not to confuse the terminology of a vector that you have studied in 12th, viz., a vector is something that has a ’length’, ’base’, and it is denoted by an arrow, etc.. The definition above is what we will use throughout!
Let \(\R\) be the field of real numbers. Consider \(\R^2=\{(x,y):x,y\in\R\}\text{.}\) The componentwise addition turns \(\R^2\) into an abelian group. We define a scalar multiplication as follows:
\begin{equation*} \alpha\cdot(x,y)=(\alpha x,\alpha y), \end{equation*}
where \(\alpha,x,y\in\R\text{.}\) Note that \(\alpha x\) (resp., \(\alpha y\)) is the usual multiplication in \(\R\text{,}\) while \(\alpha\cdot(x,y)\) is a scalar multiplication. Check that \(\R^2\) is a vector space over \(\R\text{.}\)
If we fix \((1,1)\in\R^2\) then, for given \(\alpha\in\R\) we get
\begin{equation*} \alpha\cdot(1,1)=(\alpha,\alpha). \end{equation*}
In this way, we may say that \(\R\) sits inside \(\R^2\text{.}\)
Let \(n\in\N\text{.}\) Consider \(\R^n=\{(x_1,x_2,\ldots,x_n):x_i\in\R\}\text{.}\) Show that \(\R^n\) is a vector space over \(\R\text{.}\)
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