## Vladimir Dobrushkin

http://math.uri.edu/~dobrush/

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the appendix entitled GNU Free Documentation License.

# Bases and Dimension

A **basis** β for a vector space *V* is a linearly independent subset of *V*
that generates or span *V*. If β is a basis for *V*, we also say that elements of β
form a basis for *V*. This means that every vector from *V* is a finite linear combination of elements
from the basis.

Recall that a set of vectors β is said to generate or span a vector space *V* if every element from
*V* can be represented as a linear combination of vectors from β.

*n*-dimensional real space, and it is called the

**standard basis**. Its dimension is

*n*.

*i*-th row and

*j*-th column. Then the set \( {\bf M}_{i,j} \ : \ 1 \le i \le m , \ 1 \le j \le n \) is a basis for the set of all such real matrices. Its dimension is

*mn*.

*n*. It has dimension

*n*+1. ■

Theorem: Let *V* be a vector space and
\( \beta = \left\{ {\bf u}_1 , {\bf u}_2 , \ldots , {\bf u}_n \right\} \) be a subset of
*V*. Then β is a basis for *V* if and only if each vector *v* in *V* can be uniquely
decomposed into a linear combination of vectors in β, that is, can be uniquely expressed in the form

for unique scalars \( \alpha_1 , \alpha_2 , \ldots , \alpha_n . \)

*V*. If \( {\bf v} \in V , \) then

**v**belongs to the span of the basis β because β is a basis for

*V*, so it generates this vector space. Thus,

**v**is a linear combination of the elments of β. Suppose that

**v**. Subtracting the second equation from the first gives

**v**is uniquely expressible as a linear combination of the elements of β.

Now we prove the converse that if every vector from *V*

**u**

_{1}is expressed as a linear combination of other elements:

*V*will have two distinct linear combination representations:

If the vectors \( \left\{ {\bf u}_1 , {\bf u}_2 , \ldots , {\bf u}_n \right\} \)
form a basis for a vector space *V*, then every vector in *V* can be uniquely expressed in the form

**v**determines a unqiue

*n*-tuple of scalars \( \left[ \alpha_1 , \alpha_2 , \ldots , \alpha_n \right] \) and, conversely, each

*n*-tuple of scalars determines a unique vector \( {\bf v} \in V \) by using the entries of the

*n*-tuple as the coefficients of a linear combination of \( {\bf u}_1 , {\bf u}_2 , \ldots , {\bf u}_n . \) This fact suggests that

*V*is like the

*n*-dimensional vector space \( \mathbb{R}^n , \) where

*n*is the number of vectors in the basis for

*V*.

Theorem: Let *S* be a linearly independent subset of a vector space *V*,
and let **v** be an element of *V* that is not in *S*. Then
\( S \cup \{ {\bf v} \} \) is linearly dependent if and only if **v**
belongs to the span of the set *S*.

*S*is linearly independent, one of the

**u**

_{i}'s, say

**u**

_{1}, equals

**v**. Thus, \( a_1 {\bf v} + a_2 {\bf u}_2 + \cdots + a_n {\bf u}_n = {\bf 0} , \) and so

**u**

_{1}, ... ,

**u**

_{n}, which are elements of

*S*, we conclude that

**v**belongs to the span of

*S*.

Conversely, let \( {\bf v} \in \mbox{span}(S) . \) Then there exist vectors
**v**_{1}, ... , **v**_{m} in *S* and scalars b_{1},
b_{2}, ... , b_{m} such that \( {\bf v} = b_1 {\bf v}_1 + b_2 {\bf v}_2 + \cdots +
b_m {\bf v}_m . \) Hence,

**v**in this linear combination is nonzero, and so the set \( \left\{ {\bf v} , {\bf v}_1 , {\bf v}_2 , \ldots , {\bf v}_m \right\} \) is linearly dependent. Therefore, \( S \cup \{ {\bf v} \} \) is linearly dependent.

Theorem: If a vector space *V* is generated by a finite set *S*, then
some subset of *S* is a basis for *V*

*S*that is a basis for

*V*. Otherewise,

*S*contains a nonzero element

**u**

_{1}. Recall that

The next example demonstrates how *Mathematica* can determine the basis or set of linearly independent
vectors from the given set. Note that basis is not unique and even changing the order of vectors, a software can provide
you another set of linearly independent vectors.

```
MatrixRank[m =
{{1, 2, 0, -3, 1, 0},
{1, 2, 2, -3, 1, 2},
{1, 2, 1, -3, 1, 1},
{3, 6, 1, -9, 4, 3}}]
```

Then each of the following scripts determine a subset of linearly independent vectors:

```
m[[ Flatten[ Position[#, Except[0, _?NumericQ], 1, 1]& /@
```

Last @ QRDecomposition @ Transpose @ m ] ]]

or, using subroutine

```
MinimalSublist[x_List] :=
```

Module[{tm, ntm, ytm, mm = x}, {tm = RowReduce[mm] // Transpose,

ntm = MapIndexed[{#1, #2, Total[#1]} &, tm, {1}],

ytm = Cases[ntm, {___, ___, d_ /; d == 1}]};

Cases[ytm, {b_, {a_}, c_} :> mm[[All, a]]] // Transpose]

we apply it to our set of vectors.
```
m1 = {{1, 2, 0, -3, 1, 0}, {1, 2, 1, -3, 1, 2}, {1, 2, 0, -3, 2,
1}, {3, 6, 1, -9, 4, 3}};
```

MinimalSublist[m1]

`{{1, 1, 1, 3}, {0, 1, 0, 1}, {1, 1, 2, 4}}`

One can use also the standard *Mathematica*command:

**IndependenceTest**. ■

A vector space is called **finite-dimensional** if it has a basis consisting of a finite number of
elements. The unique number of elements in each basis for *V* is called the **dimension** of
*V* and is denoted by dim(*V*). A vector space that is not finite-dimensional is called
**infinite-dimensional**.

Over the field of real numbers, the vector space \( \mathbb{C} \) of all complex numbers has dimension 2 because its basis consists of two elements \( \{ 1, {\bf j} \} . \)