Serlo: EN: Image of a linear map

{{#invoke:Mathe für Nicht-Freaks/Seite|oben}} The image of a linear map $f:V\to W$ is the set of all vectors in $W$ that are "hit by $f$". This set of vectors forms a subspace of $W$ and can be used to make the linear map $f$ surjective.

Derivation

Image of the linear map $f:{ℝ}^2\to {ℝ}^3;$$(x,y{)}^T\mapsto (x,y,-0,5x{)}^T$

Visualization of the linear map $f:{ℝ}^2\to {ℝ}^2;$$(x,y{)}^T\mapsto (x+y,0{)}^T$

We consider a linear map $f:V\to W$ between two $K$-vector spaces $V$ and $W$. A vector $v\in V$ is transformed by $f$ into a vector $f(v)\in W$. The mapping $f$ does not necessarily hit all elements from $W$, because $f$ is not necessarily surjective. The mapped vectors $f(v)$ form a subset $\{f(v)|v\in V\}\subseteq W$. This set is called image of $f$.

Since $f$ is linear, $f$ preserves the structure of the vector spaces $V$ and $W$. Therefore, we conjecture that $f$ maps the vector space $V$ into a vector space. Consequently, the image of $f$, i.e., the set $\{f(v)|v\in V\}$ should be a subspace of $W$. We will indeed prove this in a theorem below.

Definition

Mathe für Nicht-Freaks: Vorlage:Definition Mathe für Nicht-Freaks: Vorlage:Hinweis In the derivation we already considered that $\mathrm{im}(f)$ should be a subspace of $W$. We now prove this as a theorem.

Image and surjectivity

We already know that a mapping $f:V\to W$ is surjective if and only if the mapping "hits" all elements of $W$. Formally, this means that $f:V\to W$ is surjective if and only if $\mathrm{im}(f)=W$. Now if $f$ is a linear map, then $\mathrm{im}(f)$ is a subspace of $W$. In particular, if $W$ is finite-dimensional, then $f$ is surjective exactly if $\dim W=\dim (\mathrm{im}(f))$.

Mathe für Nicht-Freaks: Vorlage:Beispiel Sometimes it is useful to show the surjectivity of $f$ by proving $\dim (\mathrm{im}(f))=\dim W$.

The relationship between image and generating system

We have seen in the article on epimorphisms, that a linear map $f:V\to W$ preserves generators of $V$ if and only if it is surjective. In this case, the image of each generator of $V$ generates the entire vector space $W$. In particular, the image of each generator of $V$ generates the image $\mathrm{im}(f)$ of $f$. The last statement holds also for non-surjective linear maps:

Image and linear system Vorlage:Anker

Let $A$ be an $(n\times m)$ matrix and $b\in K^n$. The associated system of linear equations is $Ax=b$. We can also interpret the matrix $A$ as a linear map $f_A:K^m\to K^n,x\mapsto Ax$. In particular, the image $\mathrm{im}(f_A)$ of $f_A$ is a subset of $K^n$.

If $b\in \mathrm{im}(f_A)$, there is some $x_0\in K^m$ such that $f_A(x_0)=b$. By definition of $f_A$ we have $A{x}_0=b$. Thus, the linear system of equations $Ax=b$ is solvable. Conversely, if $Ax=b$ is solvable, then there exists an $x_0\in K^m$ with $A{x}_0=b$. For this $x_0$, we now have $f_A(x_0)=b$. Thus $b\in \mathrm{im}(f_A)$.

So the image gives us a criterion for the solvability of systems of linear equations: A linear system of equations $Ax=b$ is solvable if and only if $b$ lies in the image of $f_A$. However, the criterion makes no statement about the uniqueness of solutions. For this, one can use the kernel.

Examples

We will now look at how to determine the image of a linear map.

After considering two examples in finite-dimensional vector spaces, we can venture to an example with an infinite-dimensional vector space. We consider the same function in the examples for determining the kernel of a linear map.

When solving systems of linear equations, we will see many more examples. We will also learn a methodical way of solving for the determination of images.

Making linear maps "epic"

We now want to construct a surjective linear map from a given linear map $f:V\to W$. If we consider $f$ to be a mapping of sets, we already know how to accomplish this: We restrict the target set of $f$ to $\mathrm{im}(f)$ and get some restricted mapping $f^{\prime }:V\to \mathrm{im}(f);v\mapsto f(v)$. Now, we just need to check that $f^{\prime }$ is linear. But this is clear because $\mathrm{im}(f)\subseteq W$ is a subspace of $W$. So all we need to do to make $f$ surjective (i.e., an epi-morphism) is to restrict the objective of $f$ to $\mathrm{im}(f)$.

This method also gives us an approach for making functions between other structures surjective: We need to check that the restriction on the image preserves the structure. For example, for a group homomorphism $\phi :G\to H$ we can show that $\mathrm{im}(\phi )$ is again a group and ${\phi }^{\prime }:G\to \mathrm{im}(\phi );g\mapsto \phi (g)$ is again a group homomorphism.

Outlook: How surjective is a linear map? - The cokernel

In the article about the kernel we see that the kernel "stores" exactly that information which a linear map $f:V\to W$ "eliminates". Further, $f$ is injective if and only if $\ker (f)=0$ and the kernel intuitively represents a "measure of the non-injectivity" of $f$.

We now want to construct a similar measure of the surjectivity of $f$. The image of $f$ is not sufficient for this purpose: For example, the images of $g:{ℝ}^2\to {ℝ}^2;(x,y{)}^T\mapsto (x,y{)}^T$ and $h:{ℝ}^2\to {ℝ}^3;(x,y)\mapsto (x,y,0)$ are isomorphic, but $g$ is surjective and $h$ is not. From the image alone, no conclusions can be drawn as of whether $f$ is surjective, because surjectivity also depends on the target space $W$. To measure "non-surjectivity," on the other hand, we need a vector space that measures, which part of $W$ is not hit by $f$.

The space $\mathrm{im}(f)$ contains the information, which vectors are hit by $f$. The goal is to "remove this information" from $W$. We have already realized this "removal of information" in the article on the factor space by taking the quotient space $W/\mathrm{im}(f)$. We call this space $W/\mathrm{im}(f)$ the cokernel of $f$. It is indeed suitable for characterizing the non-surjectivity of $f$, because $W/\mathrm{im}(f)$ is equal to the null space $\{0\}$ if and only if $f$ is surjective: A vector in $W$ that is not hit by $f$ yields a nontrivial element in $W/\mathrm{im}(f)$ and, conversely, a nontrivial element in $W/\mathrm{im}(f)$ yields an element in $W$ that is not hit by $f$.

The kokernel even measures how non-surjective $f$ is exactly: if $W/\mathrm{im}(f)$ is larger, more vectors are not hit by $W$. If $W$ is finite dimensional, we can measure the size of $W/\mathrm{im}(f)$ using the dimension. Thus, $\dim (W/\mathrm{im}(f))=\dim (W)-\dim (\mathrm{im}(f))$ is a number we can use to quantify how non-surjective $f$ is. However, unlike $W/\mathrm{im}(f)$, this number does not allow us to reconstruct the exact vectors that are not hit by $f$.

Exercises

<section begin=zuordnung_abbildung_bild /> Mathe für Nicht-Freaks: Vorlage:Aufgabe<section end=zuordnung_abbildung_bild /> <section begin=surjektivität_dimension/> Mathe für Nicht-Freaks: Vorlage:Aufgabe<section end=surjektivität_dimension /> <section begin=bild_einer_matrix /> Mathe für Nicht-Freaks: Vorlage:Gruppenaufgabe<section end=bild_einer_matrix />

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