MA20216: Algebra 2A

1.3 Bases

Definitions. Let $\lst {v}1n$ be a list of vectors in a vector space $V$.
- (1) The span of $\lst {v}1n$ is
  $\seteqnumber{0}{1.}{0}$
  \begin{equation*} \Span {\lst {v}1n}:=\set {\lc \lambda {v}1n\st \lambda _i\in \F , 1\leq i\leq n}\leq V. \end{equation*}
- (2) $\lst {v}1n$ span $V$ (or are a spanning list for $V$) if $\Span {\lst {v}1n}=V$.
- (3) $\lst {v}1n$ are linearly independent if, whenever $\lc \lambda {v}1n=0$, then each $\lambda _i=0$, $1\leq i\leq n$, and linearly dependent otherwise.
- (4) $\lst {v}1n$ is a basis for $V$ if they are linearly independent and span $V$.

Definition. A vector space is finite-dimensional if it admits a finite list of vectors as basis and infinite-dimensional otherwise.

If $V$ is finite-dimensional, the dimension of $V$, $\dim V$, is the number of vectors in a (any) basis of $V$.

Terminology. Let $\lst {v}1n$ be a list of vectors.
- (1) A vector of the form $\lc \lambda {v}1n$ is called a linear combination of the $v_i$.
- (2) An equation of the form $\lc \lambda {v}1n=0$ is called a linear relation on the $v_i$.

Recall:

Proposition 1.1 (Algebra 1B, Proposition 1.3.4). $\lst {v}1n$ is a basis for $V$ if and only if any $v\in V$ can be written in the form
$\seteqnumber{0}{1.}{0}$
\begin{equation} \label {eq:1} v=\lc \lambda {v}1n \end{equation}

for unique $\lst \lambda 1n\in \F $. In this case, $\vec \lambda 1n$ is called the coordinate vector of $v$ with respect to $\lst {v}1n$.

1.3.1 Standard bases

In general, finite-dimensional vector spaces have many bases and there is no good reason to prefer any particular one. However, some lucky vector spaces come equipped with a natural basis.

Proposition 1.2. For $\cI $ a set and $i\in \cI $, define $e_i\in \F ^{\cI }$ by
$\seteqnumber{0}{1.}{1}$
\begin{equation*} e_i(j)= \begin{cases} 1&\text {if $i=j$}\\0&\text {if $i\neq j$}, \end {cases} \end{equation*}

for all $j\in \cI $.

If $\cI $ is finite then $(e_i)_{i\in \cI }$ is a basis, called the standard basis, of $\F ^{\cI }$.

In particular, $\dim \F ^{\cI }=\abs {\cI }$.

Proof. For $f\in \F ^{\cI }$, we observe that
$\seteqnumber{0}{1.}{1}$
\begin{equation*} f=\sum _{i\in \cI }f(i)e_i. \end{equation*}

Indeed, for $j\in \cI $,
$\seteqnumber{0}{1.}{1}$
\begin{equation*} \bigl (\sum _{i\in \cI }f(i)e_i\bigr )(j)= \sum _{i\in \cI }f(i)e_i(j)=\sum _{i\neq j}f(i)0+f(j)1=f(j). \end{equation*}

In particular, $(e_i)_{i\in \cI }$ span.

For linear independence, suppose that $\sum _{i\in \cI }\lambda _ie_i=0$ and evaluate both sides at $j\in \cI $ to get
$\seteqnumber{0}{1.}{1}$
\begin{equation*} \lambda _j=0. \end{equation*}

□

Examples.
- • Identify $\F ^n$ with $\F ^{\set {\rng 1n}}$ and then $e_i=(0,\dots ,1,\dots ,0)$ with a single $1$ in the $i$-th place.
- • Similarly, the vector space of column vectors has a standard basis with $\be _{i}$, the column vector with a single $1$ in the $i$-th row:
  $\seteqnumber{0}{1.}{1}$
  \begin{equation*} \be _{i}= \begin{pmatrix} 0\\\vdots \\1\\\vdots \\0 \end {pmatrix}. \end{equation*}
- • Finally, identifying $M_{m\times n}(\F )$ with $\F ^{\set {\rng 1m}\times \set {\rng 1n}}$ yields the standard basis $(e_{(i,j)})_{i,j}$ of $M_{m\times n}(\F )$ where $e_{(i,j)}$ differs from the zero matrix by a single $1$ in the $i$-th row and $j$-th column.

1.3.2 Useful facts

A very useful fact about bases that we shall use many times was proved in Algebra 1B:

Proposition 1.3 (Algebra 1B, Corollary 1.4.7). Any linearly independent list of vectors in a finite-dimensional vector space can be extended to a basis.

Here is another helpful result :

Lemma 1.4 (Algebra 1B, Corollary 1.4.6). Let $V$ be a finite-dimensional vector space and $U\leq V$. Then
$\seteqnumber{0}{1.}{1}$
\begin{equation*} \dim U\leq \dim V \end{equation*}

with equality if and only if $U=V$.