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Range of a linear map

by , PhD

A linear map (or function, or transformation) $f:S ightarrow T$ transforms elements of a linear space $S$ (the domain) into elements of another linear space $T$ (the codomain).

The range (or image) of a linear transformation is the subset of the codomain $T$ formed by all the values taken by the map $fleft( s ight) $ as its argument $s$ varies over the domain $S$.

Table of Contents

Definition of range

A formal definition of range follows.

Definition Let $S$ and $T$ be two vector spaces. Let $f:S ightarrow T$ be a linear map. The set[eq1]is called the range (or image) of $f$.

Some examples follow.

Example Let $S$ and $T$ respectively be the spaces of all $2 imes 1$ and $3 imes 1$ column vectors having real entries. Let $f:S ightarrow T$ be the linear map defined by the matrix product[eq2]where [eq3]Denote by $A_{1},A_{2}$ the two columns of A and by $s_{1},s_{2}$ the two entries of an arbitrarily chosen $sin S$. The product $As$ can be written as a linear combination of the columns of A with coefficients taken from the vector $s$:[eq4]The coefficients $s_{1}$ and $s_{2}$ are scalars that can be chosen at will from the set of real numbers by appropriately choosing the corresponding vector[eq5]In other words, as $s$ varies over the domain $S$, we build all the possible linear combinations of the two columns $A_{1}$ and $A_{2}$. But the set of all linear combinations of two vectors is their linear span. To conclude, we have that the range of $f$ is[eq6]Note that $A_{1}$ and $A_{2}$ are linearly independent because they are not multiples of each other. Therefore, the dimension of $QTR{rm}{range}f$ is equal to 2, less than the dimension of the space $T$ of all $3 imes 1$ column vectors, which is equal to 3 (these facts are explained in the lecture on the dimension of a linear space). Hence, in this case, the image of the function $f$ is a proper subset of $T$.

Example In the lecture on linear maps we have explained that a linear transformation $f:S ightarrow T$ is completely specified by the values taken by $f$ in correspondence of a basis of $S$. Let [eq7] and [eq8] be bases for $S$ and $T$, respectively. Let us define $f$ by[eq9]Any vector $sin S$ can be represented in terms of the basis $B$ as[eq10]where [eq11] are scalars. By the linearity of $f$, we have that[eq12]As $s$ varies over the domain $S$, the coefficients [eq13] take on any possible value in the set of real numbers R. To be even more precise, the triplet [eq14] takes on any possible value in $U{211d} ^{3}$ (otherwise $S$ would not be the space spanned by the basis $B$). As a consequence, the two coefficients[eq15]take on any possible value in $U{211d} ^{2}$ (we can, for example, set $lpha _{2}=0$ and let $lpha _{1}$ and $lpha _{3}$ vary at will). Hence, as $s$ varies over $S$, its transformations[eq16]are all the possible linear combinations of $c_{1}$ and $c_{2}$. But [eq17] is a basis for $T$. Therefore, the said linear combinations span all of $T$. To summarize,[eq18]

The range is a linear subspace of the codomain

As you might have guessed from the previous examples, the range is always a subspace of the codomain (i.e., it is a subset of the codomain which is closed with respect to linear combinations).

Proposition Let $S$ and $T$ be two vector spaces. Let $f:S ightarrow T$ be a linear map. Then, the range $QTR{rm}{range}f$ is a subspace of $T$.

Proof

We need to check that any linear combination of elements of $QTR{rm}{range}f$ still belongs to $QTR{rm}{range}f$ (i.e., that the definition of subspace holds). Choose any two vectors [eq19] and any two scalars $lpha _{1}$ and $lpha _{2}$. Then, there exists two vectors $s_{1},s_{2}in S$ such that[eq20]Moreover, by the linearity of the map $f$, we have that[eq21]Therefore, an arbitrary linear combination [eq22] is associated by $f$ to the vector[eq23]which belongs to $S$. Thus, any linear combination of elements of $QTR{rm}{range}f$ is the transformation of some element of $S$ through the function $f $ , that is, it still belongs to $QTR{rm}{range}f$. This is what we needed to prove.

Interestingly, the zero vector always belongs to the range. To see why this is the case, choose any $sin S$ and note that, by the linearity of $f$, we have that[eq24]

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let $S=T$ be the space of all $2 imes 1$ column vectors having real entries. Let $f:S ightarrow T$ be the linear map defined by [eq25]where [eq26]

Find the image of $f$.

Solution

For any $sin S$, denote by $s_{1}$ and $s_{2}$ the two entries of $s$, so that[eq27]As $s$ varies over $S$, the scalar $s_{1}$ can take on any real value. Thus, the image of $f$ is the span of the vector[eq28]In other words,[eq29]

Exercise 2

Let $S=T$ be the space of all $3 imes 1$ column vectors having real entries. Let $f:S ightarrow T$ be the linear map defined by the matrix product[eq30]where [eq31]

Is the range of $f$ a proper subspace of $T$?

Solution

The product $As$ can be written as[eq32]where $s_{1},s_{2},s_{3}$ are the three entries of $s$ and $A_{1},A_{2},A_{3}$ are the three columns of A. The latter are linearly independent: by looking at A, we can see that none of its columns can be written as a linear combination of the others. Thus, the columns [eq33] span a space of dimension 3, which coincides with the space of all $3 imes 1$ column vectors. Thus,[eq34]As a consequence, $QTR{rm}{range}f$ is not a proper subspace of $T$, as it coincides with $T$.

How to cite

Please cite as:

Taboga, Marco (2021). "Range of a linear map", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/range-of-a-linear-map.

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