# Serlo: EN: Image of a linear map

* ↳ Project "Serlo"
* ↳ Linear algebra Contents "Linear algebra" Vector spaces  Linear combinations, generators and bases  Linear maps  Linear map  Properties of linear maps  Linear continuation  Proofs for linear maps  Monomorphisms  Epimorphisms  Isomorphisms  Endomorphism and Automorphism  Image of a linear map  Kernel of a linear map  Vector space of a linear map  Dual space  Exercises  Matrices
The image of a linear map $f\colon V\to W$ 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

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

Definition (Image of a linear map)
Let $V$ and $W$ be two $K$-vector spaces and $f:V\to W$ a linear map. Then we call $\operatorname {im} (f):=\lbrace f(v)|v\in V\rbrace$ the image of $f$.
Hint
In the literature, the notation $f(V)$ is also often used instead of $\operatorname {im} (f)$ for the image of $f$.
In the derivation we already considered that $\operatorname {im} (f)$ should be a subspace of $W$. We now prove this as a theorem.
Theorem (The image is a subspace)
Let $f:V\rightarrow W$ a linear map between the $K$-vector spaces $V$ and $W$. Then $\operatorname {im} (f)$ is a subspace of $W$.
Proof (The image is a subspace)
To show that $\operatorname {im} (f)$ is a subspace, we need to check the subspace criteria:
* $\operatorname {im} (f)\subseteq W$
* $0_{W}\in \operatorname {im} (f)$
* For all $w_{1},w_{2}\in \operatorname {im} (f)$ we have $w_{1}+w_{2}\in \operatorname {im} (f)$.
* For all $w\in \operatorname {im} (f)$ and for all $\rho \in K$ we have $\rho \cdot w\in \operatorname {im} (f)$.
Proof step: $\operatorname {im} (f)\subseteq W$
For every $v\in V$ we have $f(v)\in W$. So $\operatorname {im} (f)=\{f(v)|v\in V\}\subseteq W$.
Proof step: $0_{W}\in \operatorname {im} (f)$
Since $f$ is a linear map, it holds that $f(0_{V})=0_{W}$ . Thus $0_{W}\in \operatorname {im} (f)$.
Proof step: For all $w_{1},w_{2}\in \operatorname {im} (f)$ we have $w_{1}+w_{2}\in \operatorname {im} (f)$.
Consider $w_{1},w_{2}\in \operatorname {im} (f)$ as given. That means, we can choose vectors $v_{1}$ and $v_{2}$ from $V$ with $f(v_{1})=w_{1}$ and $f(v_{2})=w_{2}$. We now show that $w_{1}+w_{2}\in \operatorname {im} (f)$ . To do this, we need to find a vector in $V$ that is mapped by $f$ to $w_{1}+w_{2}$. Now
${\begin{aligned}&w_{1}+w_{2}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f(v_{1})=w_{1}{\text{ and }}f(v_{2})=w_{2}\right.}\\[0.3em]=\ &f(v_{1})+f(v_{2})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f{\text{ is linear}}\right.}\\[0.3em]=\ &f(v_{1}+v_{2})\end{aligned}}$
As $f(v_{1}+v_{2})=w_{1}+w_{2}$ and $v_{1}+v_{2}\in V$ we have that $w_{1}+w_{2}$ is inside the image of $f$.
Proof step: For all $w\in \operatorname {im} (f)$ and for all $\rho \in K$ we have $\rho \cdot w\in \operatorname {im} (f)$.
Let $w\in \operatorname {im} (f)$ and $\rho \in K$. Then there is a vector $v\in V$ with $f(v)=w$. We need to show that there is a vector in $V$ that is mapped to $\rho \cdot w$. It holds:
${\begin{aligned}&\rho \cdot w\\[0.3em]&{\color {OliveGreen}\left\downarrow \ w=f(v)\right.}\\[0.3em]=\ &\rho \cdot f(v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f{\text{ is linear}}\right.}\\[0.3em]=\ &f(\rho \cdot v)\end{aligned}}$
Now, since $\rho \cdot v\in V$ we have that $\rho \cdot w\in \operatorname {im} (f)$.

## 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 $\operatorname {im} (f)=W$. Now if $f$ is a linear map, then $\operatorname {im} (f)$ is a subspace of $W$. In particular, if $W$ is finite-dimensional, then $f$ is surjective exactly if $\dim {W}=\dim({\operatorname {im} (f)})$.
Example
The identity $\operatorname {id} \colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},(x,y)\mapsto (x,y)$ is a linear map. It is surjective, because every element $(x,y)^{T}\in \mathbb {R} ^{2}$ has the preimage $(x,y)^{T}\in \mathbb {R} ^{2}$. Hence, we have $\operatorname {im} (\operatorname {id} )=\mathbb {R} ^{2}$ and in particular $\dim({\operatorname {im} (\operatorname {id} )})=2=\dim(\mathbb {R} ^{2})$.
The map $f\colon \mathbb {R} ^{3}\to \mathbb {R} ^{2},(x,y,z)\mapsto (x,y)$ is also linear. Further, each element $(x,y)^{T}\in \mathbb {R} ^{2}$ has a preimage, for example $(x,y,0)^{T}\in \mathbb {R} ^{3}$. Thus we have shown $\operatorname {im} (f)=\mathbb {R} ^{2}$ and thus, $f$ is surjective. In particular $\dim({\operatorname {im} (f)})=2=\dim(\mathbb {R} ^{2})$.
The embedding $e\colon \mathbb {R} ^{2}\to \mathbb {R} ^{3},(x,y)\to (x,y,0)$ is also linear, but not surjective. The vector $(0,0,1)^{T}$ is not contained in $\operatorname {im} (f)=\{(x,y,0)\mid x,y\in \mathbb {R} \}$. Thus $\dim({\operatorname {im} (e)})<\dim(\mathbb {R} ^{3})$ must hold. And indeed $\dim({\operatorname {im} (e)})=2<3=\dim(\mathbb {R} ^{3})$.
Sometimes it is useful to show the surjectivity of $f$ by proving $\dim(\operatorname {im} (f))=\dim W$.
Example
We consider the linear map $f\colon \mathbb {R} ^{3}\to \mathbb {R} ^{2};(x,y,z)\mapsto (2x-7y+3z,2y-3z)$ and ask if $f$ is surjective. We want to answer the question by determining the dimension of $\operatorname {im} (f)$ and comparing it with $\dim(\mathbb {R} ^{2})=2$. To do this, we first look for linearly independent vectors in the image of $f$. The vectors $f((1,0,0)^{T})=(2,0)^{T}$ and $f((0,1,0^{T}))=(-7,2)$ are linearly independent. Therefore, $\dim(\operatorname {im} (f))\geq 2=\dim(\mathbb {R} ^{2})$. Now $\operatorname {im} (f)\subseteq \mathbb {R} ^{2}$ from which we get $\dim(\operatorname {im} (f))\leq \dim(\mathbb {R} ^{2})$. Thus, we obtain $\dim(\operatorname {im} (f))=\dim(\mathbb {R} ^{2})$ and $f$ is surjective.

## The relationship between image and generating system

We have seen in the article on epimorphisms, that a linear map $f\colon V\to W$ 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 $\operatorname {im} (f)$ of $f$. The last statement holds also for non-surjective linear maps:
Theorem (The image is the span of the images of a generating system)
Let $f\colon V\to W$ be a linear map between two $K$-vector spaces $V$ and $W$. Let $E\subseteq V$ be a generator of $V$. Then:
$\operatorname {span} (f(E))=\operatorname {im} (f).$
Proof (The image is the span of the images of a generating system)
We show the two inclusions.
Proof step: $\subseteq$
Let $w\in \operatorname {span} (f(E))$. Then there are $n\in \mathbb {N}$, $b_{1},\dots ,b_{n}\in f(E)$ and coefficients $\lambda _{1},\dots ,\lambda _{n}\in K$, such that
${\begin{aligned}w=\sum _{i=1}^{n}\lambda _{i}b_{i}.\end{aligned}}$
Since the $b_{i}$ are in $f(E)$, there exist some $e_{i}\in E$ with $f(e_{i})=b_{i}$ for $1\leq i\leq n$. Then, because of the linearity of $f$, we have
${\begin{aligned}w=\sum _{i=1}^{n}\lambda _{i}b_{i}=\sum _{i=1}^{n}\lambda _{i}f(e_{i})=f\left(\sum _{i=1}^{n}\lambda _{i}e_{i}\right)\in \operatorname {im} (f).\end{aligned}}$
Proof step: $\supseteq$
Let $w\in \operatorname {im} (f)$. Then there is a $v\in V$ with $f(v)=w$. Since $E$ is a generator of $V$, there are an $n\in \mathbb {N}$, $e_{1},\dots ,e_{n}\in E$ and coefficients $\lambda _{1},\dots ,\lambda _{n}\in K$, such that
${\begin{aligned}v=\sum _{i=1}^{n}\lambda _{i}e_{i}.\end{aligned}}$
Now linearity of $f$ finally implies:
${\begin{aligned}w=f(v)=\sum _{i=1}^{n}\lambda _{i}\underbrace {f(e_{i})} _{\in f(E)}\in \operatorname {span} (f(E)).\end{aligned}}$

## Image and linear system

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 $\operatorname {im} (f_{A})$ of $f_{A}$ is a subset of $K^{n}$.
If $b\in \operatorname {im} (f_{A})$, there is some $x_{0}\in K^{m}$ such that $f_{A}(x_{0})=b$. By definition of $f_{A}$ we have $Ax_{0}=b$. 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 $Ax_{0}=b$. For this $x_{0}$, we now have $f_{A}(x_{0})=b$. Thus $b\in \operatorname {im} (f_{A})$.
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.
Example
Let us consider the linear map
$f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\quad {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}\mapsto {\begin{pmatrix}x_{1}\\0\end{pmatrix}}.$
This is a projection to the $x$ axis. Intuitively, then, the image of $f$ should be the $x$-axis, i.e.
$\operatorname {im} (f)=\left\{{\begin{pmatrix}x\\0\end{pmatrix}};x\in \mathbb {R} \right\}.$
We now want to prove this:
If $(x_{1},x_{2})^{T}\in \operatorname {im} (f)$, then there exists some $(z_{1},z_{2})^{T}\in \mathbb {R} ^{2}$ with $(x_{1},x_{2})^{T}=f((z_{1},z_{2})^{T})=(z_{1},0)^{T}$. So $x_{2}=0$.
Conversely, because $f((x_{1},0)^{T})=(x_{1},0)^{T}$ every vector of the form $(x_{1},0)^{T}$ has a preimage under $f$. So every such vector lies in $\operatorname {im} (f)$.
This proves the desired statement.
Example
Let $K$ be a field. We consider the linear map
$f\colon K^{2}\to K^{3},\quad {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x+y\\-x\\y\end{pmatrix}}.$
We want to determine the image of $f$. To do this, we exploit the fact that $\{e_{1}=(1,0)^{T},e_{2}=(0,1)^{T}\}$ is a basis of $K^{2}$ , so in particular it is a generator. We have seen in the last section that then $\operatorname {im} (f)=\operatorname {span} (f(e_{1}),f(e_{2}))$.
We can specify this space explicitly by calculating the span:
${\begin{aligned}\operatorname {im} (f)&=\operatorname {span} (f(e_{1}),f(e_{2}))\\[1em]&{\color {OliveGreen}\left\downarrow \ {\text{Insert definition of the span}}\right.}\\[1em]&=\left\{\lambda \cdot {\begin{pmatrix}1\\-1\\0\end{pmatrix}}+\mu \cdot {\begin{pmatrix}1\\0\\1\end{pmatrix}};\lambda ,\mu \in K\right\}\\[1em]&{\color {OliveGreen}\left\downarrow \ {\text{Add up}}\right.}\\[1em]&=\left\{{\begin{pmatrix}\lambda +\mu \\-\lambda \\\mu \end{pmatrix}};\lambda ,\mu \in K\right\}\\[1em]&{\color {OliveGreen}\left\downarrow \ {\text{ replace }}\lambda {\text{ by }}-x{\text{ and }}\mu {\text{ by }}y\right.}\\[1em]&=\left\{{\begin{pmatrix}y-x\\x\\y\end{pmatrix}};x,y\in K\right\}\end{aligned}}$
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.
Example
Our goal is to determine the image of the linear map of the derivative $d$ of polynomials over $\mathbb {R}$. The set $\lbrace 1,X,X^{2},X^{3},\dots \rbrace$ is a basis of $\mathbb {R} \lbrack X\rbrack$. The derivative function $d:\mathbb {R} \lbrack X\rbrack \rightarrow \mathbb {R} \lbrack X\rbrack$ is defined by $d(X^{i}):=i\cdot X^{i-1}$ for all $i\in \mathbb {N}$.
We now want to know whether $d$ is surjective. To do this, we note that $d({\tfrac {1}{i+1}}\cdot X^{i+1})=X^{i}$ holds for every $i\geq 0$. Thus every basis element of $\mathbb {R} [X]$ is hit. So $\operatorname {im} (f)\supseteq \operatorname {span} (1,X,X^{2},\dots )=\mathbb {R} [X]$, and $d$ is indeed surjective.
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.
link as soon as it is written.

## Making linear maps "epic"

We now want to construct a surjective linear map from a given linear map $f\colon V\to W$. 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 $\operatorname {im} (f)$ and get some restricted mapping $f'\colon V\to \operatorname {im} (f);v\mapsto f(v)$. Now, we just need to check that $f'$ is linear. But this is clear because $\operatorname {im} (f)\subseteq W$ 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 $\operatorname {im} (f)$.
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 $\varphi \colon G\to H$ we can show that $\operatorname {im} (\varphi )$ is again a group and $\varphi '\colon G\to \operatorname {im} (\varphi );g\mapsto \varphi (g)$ 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 $f\colon V\to W$ "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 $g\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2};(x,y)^{T}\mapsto (x,y)^{T}$ and $h\colon \mathbb {R} ^{2}\to \mathbb {R} ^{3};(x,y)\mapsto (x,y,0)$ 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 $\operatorname {im} (f)$ 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 $W/\operatorname {im} (f)$. We call this space $W/\operatorname {im} (f)$ the cokernel of $f$. It is indeed suitable for characterizing the non-surjectivity of $f$, because $W/\operatorname {im} (f)$ 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 $W/\operatorname {im} (f)$ and, conversely, a nontrivial element in $W/\operatorname {im} (f)$ yields an element in $W$ that is not hit by $f$.
The kokernel even measures how non-surjective $f$ is exactly: if $W/\operatorname {im} (f)$ is larger, more vectors are not hit by $W$. If $W$ is finite dimensional, we can measure the size of $W/\operatorname {im} (f)$ using the dimension. Thus, $\dim(W/\operatorname {im} (f))=\dim(W)-\dim(\operatorname {im} (f))$ is a number we can use to quantify how non-surjective $f$ is. However, unlike $W/\operatorname {im} (f)$, this number does not allow us to reconstruct the exact vectors that are not hit by $f$.

## Exercises

Exercise (Associating image spaces to figures)
We consider the following four subspaces from the vector space $\mathbb {R} ^{2}$, given as images of the linear maps
* $f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)^{T}\mapsto (2(x+y),x-3y)^{T}$
* $g\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)^{T}\mapsto (x,2x)^{T}$
* $h\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)^{T}\mapsto (-3(x-y),(x-y))^{T}$
* $k\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)^{T}\mapsto (x,0)^{T}$
Match these four subspaces to the subspaces $U_{1},U_{2},U_{3},U_{4}$ shown in the figures below.
* $U_{1}$: The subspace spanned by $(1,2)^{T}$ in $\mathbb {R} ^{2}$
* $U_{2}$: The line spanned by $(3,-1)^{T}$
* $U_{3}$: A plane covering all of the two-dimensional space
* $U_{4}$: A line spanned by $(1,0)^{T}$
Solution (Associating image spaces to figures)
First we look for the image of $f$: To find $\operatorname {im} (f)$, we can apply a theorem from above: If $E$ is a generator of $\mathbb {R} ^{2}$, then $\operatorname {im} (f)=\operatorname {span} (f(E))$ holds. We take the standard basis $\{(1,0)^{T},(0,1)^{T}\}$ as the generator of $\mathbb {R} ^{2}$. Then
$\operatorname {im} (f)=\operatorname {span} \left(f{\begin{pmatrix}1\\0\end{pmatrix}},f{\begin{pmatrix}0\\1\end{pmatrix}}\right).$
Now we apply $f$ to the standard basis
${\begin{aligned}f{\begin{pmatrix}1\\0\end{pmatrix}}&={\begin{pmatrix}2\\1\end{pmatrix}}\\f{\begin{pmatrix}0\\1\end{pmatrix}}&={\begin{pmatrix}2\\-3\end{pmatrix}}\end{aligned}}$
The vectors $(2,1)^{T},(2,-3)^{T}$ generate the image of $f$. Moreover, they are linearly independent and thus a basis of $\mathbb {R} ^{2}$. Therefore $\operatorname {im} (f)=\mathbb {R} ^{2}$. So $\operatorname {im} (f)=U_{3}$.
Next, we want to find the image of $g$. However, it is also possible to compute the image $\operatorname {im} (g)$ directly by definition, which we will demonstrate here.
${\begin{aligned}\operatorname {im} (g)&=\left\{g{\begin{pmatrix}x\\y\end{pmatrix}}\mid {\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}\right\}\\&=\left\{{\begin{pmatrix}x\\2x\end{pmatrix}}\mid {\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}\right\}\\&=\left\{x{\begin{pmatrix}1\\2\end{pmatrix}}\mid {\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}\right\}\\&{\color {OliveGreen}\left\downarrow \ {\text{The left side does not depend on }}y\right.}\\[0.3em]&=\left\{x{\begin{pmatrix}1\\2\end{pmatrix}}\mid x\in \mathbb {R} \right\}\\&=\operatorname {span} \left({\begin{pmatrix}1\\2\end{pmatrix}}\right)\end{aligned}}$
So the image of $g$ is spanned by the vector $(1,2)^{T}$. Thus $\operatorname {im} (g)=U_{1}$.
Now we determine the image of $h$ using, for example, the same method as for $f$. That means we apply $h$ to the standard basis:
${\begin{aligned}h{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}-3\\1\end{pmatrix}}\\h{\begin{pmatrix}0\\1\end{pmatrix}}={\begin{pmatrix}3\\-1\end{pmatrix}}\end{aligned}}$
Both vectors are linearly dependent. So it follows that $\operatorname {im} (h)=\operatorname {span} ((-3,1)^{T})$ and thus $\operatorname {im} (h)=U_{2}$.
Finally, we determine the image of $k$. For this we proceed for example as with $g$.
${\begin{aligned}\operatorname {im} (k)&=\left\{k{\begin{pmatrix}x\\y\end{pmatrix}}\mid {\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}\right\}\\&=\left\{{\begin{pmatrix}x\\0\end{pmatrix}}\mid {\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}\right\}\\&=\left\{x{\begin{pmatrix}1\\0\end{pmatrix}}\mid {\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}\right\}\\&{\color {OliveGreen}\left\downarrow \ {\text{The left side does not depend on }}y\right.}\\[0.3em]&=\left\{x{\begin{pmatrix}1\\0\end{pmatrix}}\mid x\in \mathbb {R} \right\}\\&=\operatorname {span} \left({\begin{pmatrix}1\\0\end{pmatrix}}\right)\end{aligned}}$
So the image of $k$ is spanned by the vector $(1,0)^{T}$. Thus $\operatorname {im} (k)$ is the $x$-axis, so $\operatorname {im} (k)=U_{4}$.
Exercise (Surjectivity and dimension of $V$ and $W$)
Let $V$ and $W$ be two finite-dimensional vector spaces. Show that there exists a surjective linear map $f\colon V\to W$ if and only if $\dim(V)\geq \dim(W)$.
How to get to the proof? (Surjectivity and dimension of $V$ and $W$)
We want to estimate the dimensions of $V$ and $W$ against each other. The dimension is defined as the cardinality of a basis. That is, if $b_{1},\dots ,b_{n}$ is a basis of $V$ and $c_{1},\dots ,c_{m}$ is a basis of $W$, we must show that $n\geq m$ holds if and only if there exists a surjective linear map. "if and only if" means that we need to establish two directions ($\Rightarrow ,\Leftarrow$).
Given a surjective linear map $f\colon V\to W$, we must show that the dimension of $V$ is at least $m$. Now bases are maximal linearly independent subsets. That is, to estimate the dimension from below, we need to construct a linearly independent subset with $m$ elements. In the figure, we have already a linearly independent subset with $m$ elements, which is the basis $c_{1},\dots ,c_{m}$. Because $f$ is surjective, we can lift these to vectors ${\hat {c}}_{1},\dots ,{\hat {c}}_{m}\in V$ with $f({\hat {c}}_{i})=c_{i}$. Now we need to verify that ${\hat {c}}_{1},\dots ,{\hat {c}}_{m}$ are linearly independent in $V$. We see this, by converting a linear combination $\lambda _{1}{\hat {c}}_{1}+\dots \lambda _{m}{\hat {c}}_{m}=0$ via $f$ into a linear combination $0=f(\lambda _{1}{\hat {c}}_{1}+\dots \lambda _{m}{\hat {c}}_{m})=\lambda _{1}c_{1}+\dots \lambda _{m}c_{m}$ and exploiting the linear independence of $c_{1},\dots ,c_{m}$.
Conversely, if $n\geq m$ holds, we must construct a surjective linear map $f\colon V\to W$. Following the principle of linear continuation, we can construct the linear map $f$ by specifying how $f$ acts on a basis of $V$. For this we need elements of $W$ on which we can send $b_{1},\dots ,b_{n}$. We have already chosen a basis of $W$ above. Therefore, it is convenient to define $f$ as follows:
$f(b_{i})={\begin{cases}c_{i}&i\leq m\\0&i>m\end{cases}}$
Then the image of $f$ is spanned by the vectors $f(b_{1})=c_{1},\dots ,f(b_{m})=c_{m},f(b_{m+1})=0,\dots ,f(b_{m})=0$. However, these vectors also span all of $W$ and thus $f$ is surjective.
Solution (Surjectivity and dimension of $V$ and $W$)
Proof step: "$\Rightarrow$"
Suppose there is a suitable surjective mapping $f$. We show that the dimension of $\operatorname {im} (f)=f(V)$ cannot be larger than the dimension of $V$ (this is true for any linear map). Because of the surjectivity of $f$, it follows that $\dim(V)\geq \dim(\operatorname {im} (f))=\dim(W)$.
So let $w_{1},\ldots ,w_{n}\in \operatorname {im} (f)$ be linearly independent. There exists $v_{1},\ldots ,v_{n}\in V$ with $f(v_{i})=w_{i}$ for $i\in \{1,\ldots ,n\}$. We show that $v_{1},\ldots ,v_{n}$ are also linearly independent: Let $\lambda _{1},\ldots ,\lambda _{n}\in K$ with $\sum _{i=1}^{n}\lambda _{i}v_{i}=0$. Then we also have that
$0=f(\sum _{i=1}^{n}\lambda _{i}v_{i})=\sum _{i=1}^{n}\lambda _{i}f(v_{i})=\sum _{i=1}^{n}\lambda _{i}w_{i},$
By linear independence of $w_{1},\ldots ,w_{n}$, it follows that $\lambda _{1}=\ldots =\lambda _{n}=0$. So $v_{1},\ldots ,v_{n}$ are also linearly independent. Overall, we have shown that
$w_{1},\ldots ,w_{n}\in \operatorname {im} (f){\text{ linearly independent }}\implies v_{1},\ldots ,v_{n}{\text{ linearly independent for any choice of preimages }}v_{i}\in f^{-1}(w_{i}).$
In particular, it holds that a basis of $V$ (a maximal linearly independent subset of $V$) must contain at least as many elements as a basis of $\operatorname {im} (f)$, that is, $\dim(V)\geq \dim(\operatorname {im} (f))$.
Proof step: "$\Leftarrow$"
Assume that $\dim(V)\geq \dim(W)$. We use that a linear map is already uniquely determined by the images of the basis vectors. Let $\{v_{1},\ldots ,v_{m}\}$ be a basis of $V$ and $\{w_{1},\ldots ,w_{n}\}$ be a basis of $W$. Define the surjective linear map $f\colon V\to W$ by
$f(v_{i})={\begin{cases}w_{i}&{\text{ if }}i\leq n\\0&{\text{ else.}}\end{cases}}$
This works, since by assumption, $m\geq n$ holds. The mapping constructed in this way is surjective, since by construction, $\{w_{1},\ldots ,w_{n}\}\subseteq \operatorname {im} (f)$. As the image of $f$ is a subspace of $W$, the subspace generated by these vectors, i.e., $W$, also lies in the image of $f$. Accordingly, $W\subseteq \operatorname {im} (f)\subseteq W$ holds and $f$ is surjective.
Exercise (Image of a matrix)
* Consider the matrix $(1,2)\in \mathbb {R} ^{1\times 2}$ and the mapping $f\colon \mathbb {R} ^{2}\to \mathbb {R} ,x\mapsto (1,2)x$ induced by it. What is the image $\operatorname {im} (f)$?
* Now let $A=(a_{1},\ldots ,a_{m})\in K^{n\times m}$ be any matrix over a field $K$, where $a_{1},\ldots ,a_{m}\in K^{n}$ denote the columns of $A$. Consider the mapping $f_{A}\colon K^{m}\to K^{n},x\mapsto Ax$ induced by $A$. Show that $\operatorname {im} (f_{A})=\operatorname {span} \{a_{1},\ldots ,a_{m}\}$ holds. So the image of a matrix is the span of its columns.
Solution (Image of a matrix)
Solution sub-exercise 1:
We know that the image $\operatorname {im} (f)$ of the linear map $f$ is a subspace of $\mathbb {R}$. Since the $\mathbb {R}$-vector space $\mathbb {R}$ has dimension $1$, a subspace can only have dimension $0$ or $1$. In the first case the subspace is the null vector space, in the second case it is already all of $\mathbb {R}$. So $\mathbb {R}$ has only the two subspaces $\{0\}$ and $\mathbb {R}$. Since $(1,2)(1,0)^{T}=1\neq 0$ holds, we have that $\operatorname {im} (f)\neq \{0\}$. Thus, $\operatorname {im} (f)=\mathbb {R}$.
Solution sub-exercise 2:
Proof step: "$\subseteq$"
Let $y\in \operatorname {im} (f_{A})$. Then, there is some $x=(x_{1},\ldots ,x_{m})^{T}\in K^{m}$ with $Ax=y$. We can write $x$ as $x=\sum _{i=1}^{m}x_{i}e_{i}$. Plugging this into the equation $Ax=y$, we get.
${\begin{aligned}y&=Ax\\&{\color {OliveGreen}\left\downarrow \ x=\sum _{i=1}^{m}x_{i}e_{i}\right.}\\[0.3em]&=A\left(\sum _{i=1}^{m}x_{i}e_{i}\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{application of }}A{\text{ is linear}}\right.}\\[0.3em]&=\sum _{i=1}^{m}x_{i}Ae_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ Ae_{i}=a_{i}{\text{, the }}i{\text{-th column of }}A\right.}\\[0.3em]&=\sum _{i=1}^{m}x_{i}a_{i}.\end{aligned}}$
Since $\sum _{i=1}^{m}x_{i}a_{i}\in \operatorname {span} \{a_{1},\ldots ,a_{n}\}$, we obtain $y\in \operatorname {span} \{a_{1},\ldots ,a_{n}\}$.
Proof step: "$\supseteq$"
Let $y=\sum _{i=1}^{m}y_{i}\cdot a_{i}\in \operatorname {span} (f_{A})$ with $y_{i}\in K$ for $i=1,\ldots ,m$. We want to find $x\in K^{m}$ with $Ax=y$. So let us define $x:=\sum _{i=1}^{m}y_{i}e_{i}$. The same calculation as in the first step of the proof then shows
$Ax=A\left(\sum _{i=1}^{m}y_{i}e_{i}\right)=\sum _{i=1}^{m}y_{i}Ae_{i}=\sum _{i=1}^{m}y_{i}a_{i}=y.$
Kernel of a linear map →
Den Bereich zur Analysis 1 gibt es jetzt auch als Buch! Bestelle dir dein Exemplar oder lade dir das Buch gleich kostenlos als PDF herunter:
Feedback? Do you want to join?
If you have questions concerning the content, or didn't understand something, the feel free to contact us! We would love to answer your questions! Also we are thankful for critics and/or comments! If you share our vision to explain university math in an comprehensible way, then contact us under:
E-Mail: en@serlo.org
This article is licensed under the free license CC-BY-SA 3.0. With that you can use it, modify it or share it freely, as long as you name „Serlo“ as source and put you changes under the same CC-BY-SA 3.0 oder an compatible license. On the page „Kopier uns!“ we explain you what you have to pay attention to, when using our texts, picture or videos.