## 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.

# Matrix Spaces

There are two kinds of matrix spaces: one is defined by a set of matrices of special type (considered previously in section), and another is associated with every matrix. This section is devoted to the latter case.
Let **A** be an *m*-by-*n* matrix that maps \( \mathbb{R}^n \) into
\( \mathbb{R}^m . \) So for every (colomn) vector
\( {\bf x} = \left[ x_1 , x_2 , \ldots , x_n \right]^{\mathrm{T}} \)
corresponds a vector \( {\bf A}\,{\bf x} \in \mathbb{R}^m . \)

Let *V* and *U* be vector spaces and
\( T:\,U \to V \) be linear transformation. Then the set of all vectors
of the form \( {\bf y} = {\bf A}\,{\bf x} \in V \)
is called the **range** (or image) of *T*.

**Theorem:** Let *V* and *U* be vector spaces and
\( T:\,U \to V \) be linear transformation. Then the range of
*T* is a subspace of *V*.

**0**

_{U}and

**0**

_{V}to denote the zero vectors of

*U*and

*V*, respectively. We also denote by

*R(T)*the range of linear transformation

*T*.

Because \( T\left( **0**_{U} \right) = **0**_{V}, \)
we have that \( **0**_{V} \in R(T) . \) Now let us choose two vectors
\( {\bf x}, \, {\bf y} \in R(T) \) from the range and a scalar *k*.
Then there exist two vectors **u** and **v** such that
\( T({\bf u}) = {\bf x} \) and \( T({\bf v}) = {\bf y} . \)
Hence, \( T({\bf u} + {\bf v}) = T({\bf u}) + T({\bf v}) = {\bf x} + {\bf y} \) and
\( T(k{\bf u}) = k\, T({\bf u}) = k\, {\bf x} . \) Thus,
\( T({\bf u} + {\bf v}) = T({\bf u}) + T({\bf v}) = {\bf x} + {\bf y} \in R(T) \) and \( T(k\,{\bf u}) = k \, T({\bf x}) = k\,{\bf x} \in R(T) , \) so *R(T)* is a subspace of *V*.

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

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*.

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

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**.

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**. ■