A linear map (or function, or transformation) transforms elements of a linear space (the domain) into elements of another linear space (the codomain).

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

A formal definition of range follows.

Definition Let and be two vector spaces. Let be a linear map. The setis called the range (or image) of .

Some examples follow.

Example Let and respectively be the spaces of all and column vectors having real entries. Let be the linear map defined by the matrix productwhere Denote by the two columns of and by the two entries of an arbitrarily chosen . The product can be written as a linear combination of the columns of with coefficients taken from the vector :The coefficients and are scalars that can be chosen at will from the set of real numbers by appropriately choosing the corresponding vectorIn other words, as varies over the domain , we build all the possible linear combinations of the two columns and . But the set of all linear combinations of two vectors is their linear span. To conclude, we have that the range of isNote that and are linearly independent because they are not multiples of each other. Therefore, the dimension of is equal to 2, less than the dimension of the space of all 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 is a proper subset of .

Example In the lecture on linear maps we have explained that a linear transformation is completely specified by the values taken by in correspondence of a basis of . Let and be bases for and , respectively. Let us define byAny vector can be represented in terms of the basis aswhere are scalars. By the linearity of , we have thatAs varies over the domain , the coefficients take on any possible value in the set of real numbers . To be even more precise, the triplet takes on any possible value in (otherwise would not be the space spanned by the basis ). As a consequence, the two coefficientstake on any possible value in (we can, for example, set and let and vary at will). Hence, as varies over , its transformationsare all the possible linear combinations of and . But is a basis for . Therefore, the said linear combinations span all of . To summarize,

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 and be two vector spaces. Let be a linear map. Then, the range is a subspace of .

Proof

We need to check that any linear combination of elements of still belongs to (i.e., that the definition of subspace holds). Choose any two vectors and any two scalars and . Then, there exists two vectors such thatMoreover, by the linearity of the map , we have thatTherefore, an arbitrary linear combination is associated by to the vectorwhich belongs to . Thus, any linear combination of elements of is the transformation of some element of through the function , that is, it still belongs to . 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 and note that, by the linearity of , we have that

Below you can find some exercises with explained solutions.

Let be the space of all column vectors having real entries. Let be the linear map defined by where

Find the image of .

Solution

For any , denote by and the two entries of , so thatAs varies over , the scalar can take on any real value. Thus, the image of is the span of the vectorIn other words,

Let be the space of all column vectors having real entries. Let be the linear map defined by the matrix productwhere

Is the range of a proper subspace of ?

Solution

The product can be written aswhere are the three entries of and are the three columns of . The latter are linearly independent: by looking at , we can see that none of its columns can be written as a linear combination of the others. Thus, the columns span a space of dimension 3, which coincides with the space of all column vectors. Thus,As a consequence, is not a proper subspace of , as it coincides with .

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