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Linear operator

by , PhD

In linear algebra the term "linear operator" most commonly refers to linear maps (i.e., functions $f:S
ightarrow T$ preserving vector addition and scalar multiplication) that have the added peculiarity of mapping a vector space into itself (i.e., $f:S
ightarrow S$). The term may be used with a different meaning in other branches of mathematics.

Table of Contents

Definition

Before providing a definition of linear operator, we need to remember that a function $f:S
ightarrow T$ that associates one and only one element of a vector space $T$ to each element of another vector space $S$ is said to be a linear map if and only if[eq1]for any two scalars $lpha _{1}$ and $lpha _{2}$ and any two vectors $s_{1},s_{2}in S$.

Linear operators are defined analogously.

Definition Let $S$ be a vector space. A function $f:S
ightarrow S$ is said to be a linear operator if and only if[eq2]for any two scalars $lpha _{1}$ and $lpha _{2}$ and any two vectors $s_{1},s_{2}in S$.

Let us provide a simple example.

Example Consider the space $S$ of all $2	imes 1$ column vectors having real entries. Suppose the function $f:S
ightarrow S$ associates to each vector [eq3]a vector[eq4]Choose any two vectors $s_{1},s_{2}in S$ and any two scalars $lpha _{1}$ and $lpha _{2}$. By repeatedly applying the definitions of vector addition and scalar multiplication, we obtain[eq5]Therefore, $f$ is a linear operator.

Properties inherited from linear maps

Since a linear operator is a special kind of linear map, it inherits all the properties of linear maps. For convenience, we report here the most important of these inherited properties, but if you are already familiar with linear maps, you can safely skip this section.

A linear operator is completely defined by its values on a basis

A linear operator $f:S
ightarrow S$ is completely determined by its values on a basis of $S$.

Proposition Let $S$ be a linear space, [eq6] a basis for $S$, and [eq7] a set of K elements of $S$. Then, there is a unique linear operator $f:S
ightarrow S$ such that[eq8]for $k=1,ldots ,K$.

Proof

See the proof in the lecture on linear maps.

Square matrices define linear operators

Multiplication of vectors by a square matrix defines a linear operator.

Proposition Let $S$ be the linear space of all Kx1 column vectors. Let A be a $K	imes K$ matrix. Let $f:S
ightarrow S$ be defined, for any $sin S$, by [eq9]where $As$ denotes the matrix product between A and $s$. Then $f$ is a linear operator.

Proof

See the proof provided in the lecture on linear maps.

Proposition Let $S$ be the linear space of all $1	imes K$ row vectors. Let A be a $K	imes K$ matrix. Consider the transformation $f:S
ightarrow S$ defined, for any $sin S$, by [eq10]where $sA$ denotes the matrix product between $s$ and A. Then $f$ is a linear operator.

Proof

As before, see the proof in the lecture on linear maps.

Combinations of multiple terms

The "linearity preserving" property extends to linear combinations involving more than two terms.

Proposition Let [eq11] be n scalars and let [eq12] be K elements of a linear space $S$. If $f:S
ightarrow S$ is a linear operator, then[eq13]

Proof

Also in this case, see the proof in the lecture on linear maps.

The matrix of a linear operator is square

Remember that every linear map $f:S
ightarrow T$ between two finite-dimensional vector spaces can be represented by a matrix [eq14], called the matrix of the linear map. The notation [eq15] indicates that the matrix depends on the choice of two bases: a basis $B$ for the space $S$ and a basis $C$ for the space $T$.

The matrix is constructed as follows: [eq16]where the columns are the coordinate vectors of the transformations [eq17] of the vectors belonging to the basis [eq18].

The number of columns of [eq15] is equal to the number of elements in the basis $B$, while the number of rows of [eq20] is equal to the number of elements in the basis $C$.

In the case of a linear operator, the codomain $T$ coincides with the domain $S$, that is, $f:S
ightarrow S$. There are two important consequences of this fact.

First, any two bases $B$ and $C$ of $S$ have the same number of elements (by the dimension theorem). Therefore, the matrix [eq15] of a linear operator is square. Hence, we can apply to linear operators the rich set of theoretical tools that can be applied exclusively to square matrices (e.g., the concepts of inverse, trace, determinant, eigenvalues and eigenvectors).

Second, we can (although we are not obliged to) use a unique basis $B$ for both the domain and codomain. When we choose this kind of simplification, the matrix of the linear map is [eq22]which we can also simply denote by [eq23].

Example Let $S$ be a linear space spanned by the basis [eq24]. Suppose $f:S
ightarrow S$ is a linear operator such that[eq25]Then, the coordinate vectors needed to form the matrix of the linear operator are[eq26]and[eq27]Thus, the matrix of the linear operator with respect to $B$ is the square matrix[eq28]

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let $S$ be a linear space spanned by the basis [eq29]. Suppose $f:S
ightarrow S$ is a linear operator such that[eq30]

Find the matrix [eq23] of the linear operator $f$.

Solution

After applying the linear operator, the coordinate vectors of the elements of the basis become[eq32]and[eq33]and[eq34]Thus, the matrix of the linear operator with respect to $B$ is the square matrix[eq35]

Exercise 2

Use the matrix [eq23] found in the previous exercise to compute how the operator $f$ transforms the coordinates of the vector $sin S $ such that[eq37]

Solution

The transformation can be computed by performing a matrix multiplication:[eq38]

How to cite

Please cite as:

Taboga, Marco (2017). "Linear operator", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/linear-operator.

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