# Exercises Linear Maps – Serlo

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We have compiled some tasks on linear maps here. The proof structures can help you to solve other similar tasks. As a reminder, here is the definition of a linear map:

Definition (Linear map)

Let ${\displaystyle f\colon {V}\to {W}}$ be a mapping between the two vector spaces ${\displaystyle {V}}$ and ${\displaystyle {W}}$. We call ${\displaystyle f}$ a linear map from ${\displaystyle {V}}$ to ${\displaystyle {W}}$ if the following two properties are satisfied:

1. Additivity: For all ${\displaystyle v_{1},v_{2}\in V}$ we have that
${\displaystyle f\left(v_{1}+v_{2}\right)=f(v_{1})+f(v_{2})}$
2. Homogeneity: For all ${\displaystyle v\in V}$ and ${\displaystyle \lambda \in K}$ we have that
${\displaystyle f(\lambda \cdot v)=\lambda \cdot f(v)}$

## Showing linearity of a mapping

### Linear maps from ${\displaystyle \mathbb {R} ^{n}}$ to ${\displaystyle \mathbb {R} ^{m}}$

Exercise (Linear map into a field)

Let ${\displaystyle f\colon \mathbb {R} ^{3}\to \mathbb {R} }$ be defined by ${\displaystyle f((x,y,z)^{T}):=5x+y-3z}$. Show that the map ${\displaystyle f}$ is linear.

How to get to the proof? (Linear map into a field)

First you have to show the additivity and then homogeneity of the map.

Solution (Linear map into a field)

For this step, let ${\displaystyle v={\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\in \mathbb {R} ^{3}}$ and ${\displaystyle w={\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}\in \mathbb {R} ^{3}}$.

{\displaystyle {\begin{aligned}f(v+w)&=\\[0.3em]&=\ f\left({\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}+{\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}\right)=f{\begin{pmatrix}v_{1}+w_{1}\\v_{2}+w_{2}\\v_{3}+w_{3}\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&=\ 5(v_{1}+w_{1})+(v_{2}+w_{2})-3(v_{3}+w_{3})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributivity in }}\mathbb {R} \right.}\\[0.3em]&=\ 5v_{1}+5w_{1}+v_{2}+w_{2}-3v_{3}-3w_{3}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutativity and associativity in }}\mathbb {R} \right.}\\[0.3em]&=\ (5v_{1}+v_{2}-3v_{3})+(5w_{1}+w_{2}-3w_{3})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&=\ f{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}+f{\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}=f(v)+f(w)\end{aligned}}}

Thus ${\displaystyle f}$ is additive.

Proof step: Homogeneity

Let ${\displaystyle v={\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\in \mathbb {R} ^{3}}$ and ${\displaystyle \lambda \in \mathbb {R} }$.

{\displaystyle {\begin{aligned}f(\lambda \cdot v)&=f\left(\lambda \cdot {\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\right)=f{\begin{pmatrix}\lambda v_{1}\\\lambda v_{2}\\\lambda v_{3}\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&=\ 5(\lambda v_{1})+(\lambda v_{2})-3(\lambda v_{3})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of the multiplication in }}\mathbb {R} \right.}\\[0.3em]&=\ \lambda \cdot (5v_{1})+\lambda \cdot (v_{2})-\lambda \cdot (3v_{3})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributivity in }}\mathbb {R} \right.}\\[0.3em]&=\ \lambda \cdot (5v_{1}+v_{2}-3v_{3})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&=\ \lambda \cdot f{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}=\lambda \cdot f(v)\end{aligned}}}

Thus ${\displaystyle f}$ is homogeneous and ${\displaystyle f}$ is linear.

Exercise (Linear map from ${\displaystyle \mathbb {R} ^{3}}$ to ${\displaystyle \mathbb {R} ^{2}}$)

Show that the map ${\displaystyle L\colon \mathbb {R} ^{3}\to \mathbb {R} ^{2}}$ with ${\displaystyle L{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}:={\begin{pmatrix}v_{2}-v_{3}\\3v_{1}+5v_{3}\end{pmatrix}}}$ is linear.

How to get to the proof? (Linear map from ${\displaystyle \mathbb {R} ^{3}}$ to ${\displaystyle \mathbb {R} ^{2}}$)

You have to show that for ${\displaystyle v=(v_{1},v_{2},v_{3})^{T}}$ and ${\displaystyle w=(w_{1},w_{2},w_{3})^{T}}$ it holds true that

${\displaystyle L\left({\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}+{\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}\right)=L{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}+L{\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}}$

And you have to show that for ${\displaystyle \rho \in \mathbb {R} }$, it holds true that

${\displaystyle L\left(\rho \cdot {\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\right)=\rho \cdot L{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}}$

Solution (Linear map from ${\displaystyle \mathbb {R} ^{3}}$ to ${\displaystyle \mathbb {R} ^{2}}$)

{\displaystyle {\begin{aligned}L\left({\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}+{\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}\right)&=L{\begin{pmatrix}v_{1}+w_{1}\\v_{2}+w_{2}\\v_{3}+w_{3}\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}L\right.}\\[0.3em]&=\ {\begin{pmatrix}(v_{2}+w_{2})-(v_{3}+w_{3})\\3(v_{1}+w_{1})+5(v_{3}+w_{3})\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutative, associative, distributive in }}\mathbb {R} \right.}\\[0.3em]&=\ {\begin{pmatrix}(v_{2}-v_{3})+(w_{2}-w_{3})\\(3v_{1}-5v_{3})+(3w_{1}-5w_{3})\end{pmatrix}}\\[0.3em]&=\ {\begin{pmatrix}(v_{2}-v_{3})\\(3v_{1}-5v_{3})\end{pmatrix}}+{\begin{pmatrix}(w_{2}-w_{3})\\(3w_{1}-5w_{3})\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}L\right.}\\[0.3em]&=\ L{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}+L{\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}\end{aligned}}}

Aktuelles Ziel: Scaling

{\displaystyle {\begin{aligned}L\left(\rho \cdot {\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\right)&=L{\begin{pmatrix}\rho v_{1}\\\rho v_{2}\\\rho v_{3}\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}L\right.}\\[0.3em]&=\ {\begin{pmatrix}\rho v_{2}-\rho v_{3}\\3(\rho v_{1})+5(\rho v_{3})\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutative, associative, distributive in }}\mathbb {R} \right.}\\[0.3em]&=\ {\begin{pmatrix}\rho (v_{2}-v_{3})\\(\rho (3v_{1}+5v_{3})\end{pmatrix}}\\[0.3em]&=\ \rho \cdot {\begin{pmatrix}(v_{2}-v_{3})\\(3v_{1}+5v_{3})\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}L\right.}\\[0.3em]&=\ \rho \cdot L{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\end{aligned}}}

Exercise (Linearity of the embedding)

Show that for ${\displaystyle m\geq n}$, the map ${\displaystyle f\colon \mathbb {R} ^{n}\to \mathbb {R} ^{m}:\quad (x_{1},x_{2},\ldots ,x_{n})^{T}\mapsto (x_{1},x_{2},\ldots ,x_{n},\underbrace {0,\ldots ,0} _{(m-n){\text{ times}}})^{T}}$ is linear.

Solution (Linearity of the embedding)

Let ${\displaystyle v=(v_{1},\ldots ,v_{n})^{T}\in \mathbb {R} ^{n}}$ and ${\displaystyle w=(w_{1},\ldots ,w_{n})^{T}\in \mathbb {R} ^{n}}$, as well as ${\displaystyle \lambda ,\mu \in \mathbb {R} }$. By definition of the map ${\displaystyle f}$, we have that

${\displaystyle f(\lambda v+\mu w)=f\left({\begin{pmatrix}\lambda v_{1}+\mu w_{1}\\\vdots \\\lambda v_{n}+\mu w_{n}\end{pmatrix}}\right)={\begin{pmatrix}\lambda v_{1}+\mu w_{1}\\\vdots \\\lambda v_{n}+\mu w_{n}\\0\\\vdots \\0\end{pmatrix}}=\lambda \cdot {\begin{pmatrix}v_{1}\\\vdots \\v_{n}\\0\\\vdots \\0\end{pmatrix}}+\mu \cdot {\begin{pmatrix}w_{1}\\\vdots \\w_{n}\\0\\\vdots \\0\end{pmatrix}}=\lambda f(v)+\mu f(w).}$

So ${\displaystyle f}$ is linear.

We consider an example for a linear map of ${\displaystyle \mathbb {R} ^{2}}$ to ${\displaystyle \mathbb {R} ^{2}}$:

${\displaystyle f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2}}$ with ${\displaystyle f{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}\\x_{1}-5x_{2}\end{pmatrix}}}$

Exercise (Linearity of ${\displaystyle f}$)

Show that the map ${\displaystyle f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}\mapsto {\begin{pmatrix}x_{1}+x_{2}\\x_{1}-5x_{2}\end{pmatrix}}}$ is linear.

Proof (Linearity of ${\displaystyle f}$)

${\displaystyle \mathbb {R} ^{2}}$ is an ${\displaystyle \mathbb {R} }$-vector space. In addition, the map is well-defined.

Let ${\displaystyle (x_{1},x_{2})^{T}}$ and ${\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\displaystyle \mathbb {R} ^{2}}$. Then, we have:

{\displaystyle {\begin{aligned}f\left({\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}+{\begin{pmatrix}y_{1}\\y_{2}\end{pmatrix}}\right)&=f{\begin{pmatrix}x_{1}+y_{1}\\x_{2}+y_{2}\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&={\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\(x_{1}+y_{1})-5\cdot (x_{2}+y_{2})\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law}}\right.}\\[0.3em]&={\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{separate vectors}}\right.}\\[0.3em]&={\begin{pmatrix}x_{1}+x_{2}\\x_{1}-5x_{2}\end{pmatrix}}+{\begin{pmatrix}y_{1}+y_{2}\\y_{1}-5y_{2}\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&=f{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}+f{\begin{pmatrix}y_{1}\\y_{2}\end{pmatrix}}\end{aligned}}}

Proof step: homogeneity

Let ${\displaystyle \lambda \in \mathbb {R} }$ and ${\displaystyle (x_{1},x_{2})^{T}\in \mathbb {R} ^{2}}$. Then:

{\displaystyle {\begin{aligned}f\left(\lambda \cdot {\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}\right)&=f{\begin{pmatrix}\lambda \cdot x_{1}\\\lambda \cdot x_{2}\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&={\begin{pmatrix}\lambda x_{1}+\lambda x_{2}\\\lambda x_{1}-5\lambda x_{2}\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law}}\right.}\\[0.3em]&={\begin{pmatrix}\lambda (x_{1}+x_{2})\\\lambda (x_{1}-5x_{2})\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{scalar multiplication}}\right.}\\[0.3em]&=\lambda \cdot {\begin{pmatrix}(x_{1}+x_{2})\\(x_{1}-5x_{2})\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&=\lambda \cdot f{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}\\[0.3em]\end{aligned}}}

Thus the map is linear.

### Important special cases

Exercise (The identity is a linear map)

Let ${\displaystyle V}$ be a ${\displaystyle K}$-vector space. Prove that the identity ${\displaystyle \operatorname {id} :V\to V}$ with ${\displaystyle \operatorname {id} (v)=v}$ is a linear map.

Proof (The identity is a linear map)

The identity is additive: Let ${\displaystyle v,w\in V}$, then.

${\displaystyle \operatorname {id} (v+w)=v+w=\operatorname {id} (v)+\operatorname {id} (w)}$

The identity is homogeneous: Let ${\displaystyle \lambda \in K}$ and ${\displaystyle v\in V}$, then

${\displaystyle \operatorname {id} (\lambda \cdot v)=\lambda \cdot v=\lambda \cdot \operatorname {id} (v)}$

Exercise (The map to zero is a linear map)

Let ${\displaystyle V,W}$ be two ${\displaystyle K}$-vector spaces. Show that the map to zero ${\displaystyle f:V\to W}$, which maps all vectors ${\displaystyle v\in V}$ to the zero vector ${\displaystyle 0_{{}_{W}}}$, is linear.

Proof (The map to zero is a linear map)

${\displaystyle f}$ is additive: let ${\displaystyle v_{1},v_{2}}$ be vectors in ${\displaystyle V}$. Then

${\displaystyle f(v_{1}+v_{2})=0_{{}_{W}}=0_{{}_{W}}+0_{{}_{W}}=f(v_{1})+f(v_{2})}$

${\displaystyle f}$ is homogeneous: Let ${\displaystyle v\in V}$ and let ${\displaystyle \lambda \in K}$. Then

${\displaystyle f(\lambda \cdot v)=0_{{}_{W}}=\lambda \cdot 0_{{}_{W}}=\lambda \cdot f(v)}$

Thus, the map to zero is linear

### Linear maps between function spaces

Exercise (Mapping on a function space)

Consider the function space ${\displaystyle \operatorname {Fun} ([0,1],\mathbb {R} )}$ of all functions from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} }$, as well as the map

{\displaystyle {\begin{aligned}\phi \colon \operatorname {Fun} ([0,1],\mathbb {R} )&\to \mathbb {R} ,\\f&\mapsto f(0).\end{aligned}}}

Show that ${\displaystyle \phi }$ is linear.

Solution (Mapping on a function space)

The operations on the function space are defined element-wise in each case. That means: for ${\displaystyle f,g\in \operatorname {Fun} ([0,1],\mathbb {R} )}$, ${\displaystyle \lambda \in \mathbb {R} }$ and ${\displaystyle x\in [0,1]}$ we have that ${\displaystyle (f+g)(x)=f(x)+g(x)}$ and ${\displaystyle (\lambda f)(x)=\lambda f(x)}$. In particular, this is true for ${\displaystyle x=0}$, which implies

{\displaystyle {\begin{aligned}\phi (f+g)=(f+g)(0)=f(0)+g(0)=\phi (f)+\phi (g)\end{aligned}}}

and

{\displaystyle {\begin{aligned}\phi (\lambda f)=(\lambda f)(0)=\lambda f(0)=\lambda \phi (f)\end{aligned}}}

Thus, we have established linearity.

Exercise (The precomposition with a map is linear.)

Let ${\displaystyle V}$ be a vector space, let ${\displaystyle M,N}$ be sets, and let ${\displaystyle {\text{Fun}}(M,V)}$ or ${\displaystyle {\text{Fun}}(N,V)}$ be the vector space of functions from ${\displaystyle M}$ or ${\displaystyle N}$ to ${\displaystyle V}$. Let ${\displaystyle t\in {\text{Fun}}(N,M)}$ be arbitrary but fixed. We consider the mapping

{\displaystyle {\begin{aligned}\Theta :{\text{Fun}}(M,V)&\to {\text{Fun}}(N,V)\\g&\mapsto g\circ t\end{aligned}}}

Show that ${\displaystyle \Theta }$ is linear.

It is important that you exactly follow the definitions. Note that ${\displaystyle \Theta }$ is a map that assigns to every map of ${\displaystyle M}$ to ${\displaystyle V}$ a map of ${\displaystyle N}$ to ${\displaystyle V}$. These maps, which are elements of ${\displaystyle {\text{Fun}}(M,V)}$ and ${\displaystyle {\text{Fun}}(N,V)}$ respectively, need not themselves be linear, since there is no vector space structure on the sets ${\displaystyle M}$ and ${\displaystyle N}$.

Summary of proof (The precomposition with a map is linear.)

In order to prove the linearity of ${\displaystyle \Theta }$, we need to check the two properties again:

1. ${\displaystyle \Theta }$ is additive: ${\displaystyle \Theta (g+h)=\Theta (g)+\Theta (h)}$ for all ${\displaystyle g,h\in {\text{Fun}}(M,V)}$
2. ${\displaystyle \Theta }$ is homogeneous: ${\displaystyle \Theta (\lambda \cdot g)=\lambda \cdot \Theta (g)}$ for all ${\displaystyle g\in {\text{Fun}}(M,V)}$ and ${\displaystyle \lambda \in K}$

So at both points an equivalence of maps ${\displaystyle N\to V}$ is to be shown. For this we evaluate the maps at every m element ${\displaystyle y\in N}$.

Proof (The precomposition with a map is linear.)

Let ${\displaystyle g,h\in {\text{Fun}}(M,V)}$.

For all ${\displaystyle n\in N}$ we have that

{\displaystyle {\begin{aligned}\Theta (g+h)(n)&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\Theta \right.}\\[0.3em]&=\ ((g+h)\circ t)(n)\\[0.3em]&=\ (g+h)(t(n))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{vector addition on Fun}}(M,V)\right.}\\[0.3em]&=\ g(t(n))+h(t(n))\\[0.3em]&=\ (g\circ t)(n)+(h\circ t)(n)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\Theta \right.}\\[0.3em]&=\ \Theta (g)(n)+\Theta (h)(n)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{vector addition on Fun}}(N,V)\right.}\\[0.3em]&=\ (\Theta (g)+\Theta (h))(n)\end{aligned}}}

Thus we have shown ${\displaystyle \Theta (g+h)=\Theta (g)+\Theta (h)}$, i.e., ${\displaystyle \Theta }$ is additive.

Let ${\displaystyle g\in {\text{Fun}}(M,V)}$ and ${\displaystyle \lambda \in K}$.

Proof step: homogeneity

For all ${\displaystyle n\in N}$ we have that

{\displaystyle {\begin{aligned}\Theta (\lambda \cdot g)(n)&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\Theta \right.}\\[0.3em]&=\ ((\lambda \cdot g)\circ t)(n)\\[0.3em]&=\ (\lambda \cdot g)(t(n))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{scalar multiplication on Fun}}(M,V)\right.}\\[0.3em]&=\ \lambda \cdot g(t(n))\\[0.3em]&=\ \lambda \cdot (g\circ t)(n)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\Theta \right.}\\[0.3em]&=\ \lambda \cdot \Theta (g)(n)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{scalar multiplication on Fun}}(N,V)\right.}\\[0.3em]&=\ (\lambda \cdot \Theta (g))(n)\end{aligned}}}

Thus we have shown ${\displaystyle \Theta (\lambda \cdot g)=\lambda \cdot \Theta (g)}$, i.e., ${\displaystyle \Theta }$ is homogeneous.

Now, additivity and homogeneity of ${\displaystyle \Theta }$ implies that ${\displaystyle \Theta }$ is a linear map.

Exercise (Sequence space)

Let ${\displaystyle V}$ be the ${\displaystyle \mathbb {R} }$-vector space of all real-valued sequences. Show that the map

{\displaystyle {\begin{aligned}f:V&\to V\\(a_{0},a_{1},a_{2},\ldots )&\mapsto (a_{1},a_{2},a_{3},\ldots )\end{aligned}}}

is linear.

How to get to the proof? (Sequence space)

To show linearity, two properties need to be checked:

1. ${\displaystyle f}$ is additive: ${\displaystyle f(v+w)=f(v)+f(w)}$ for all ${\displaystyle v,w\in V}$
2. ${\displaystyle f}$ is homogeneous: ${\displaystyle f(\lambda \cdot v)=\lambda \cdot f(v)}$ for all ${\displaystyle v\in V}$ and ${\displaystyle \lambda \in \mathbb {R} }$

The vectors ${\displaystyle v}$ and ${\displaystyle w}$ are sequences of real numbers, i.e. they are of the form ${\displaystyle v=(a_{0},a_{1},a_{2},\ldots )}$ and ${\displaystyle w=(b_{0},b_{1},b_{2},\ldots )}$ with ${\displaystyle a_{k},b_{k}\in \mathbb {R} }$ for all ${\displaystyle k\in \mathbb {N} _{0}}$.

Proof (Sequence space)

Let ${\displaystyle v=(a_{0},a_{1},a_{2},\ldots )\in V}$ and ${\displaystyle w=(b_{0},b_{1},b_{2},\ldots )\in V}$. Then, we have

{\displaystyle {\begin{aligned}f(v+w)&=f((a_{0},a_{1},a_{2},\ldots )+(b_{0},b_{1},b_{2},\ldots ))\\[0.3em]&=\ f(a_{0}+b_{0},a_{1}+b_{1},a_{2}+b_{2},\ldots )\\[0.3em]&=\ (a_{1}+b_{1},a_{2}+b_{2},a_{3}+b_{3},\ldots )\\[0.3em]&=\ (a_{1},a_{2},a_{3},\ldots )+(b_{1},b_{2},b_{3},\ldots )\\[0.3em]&=\ f(a_{0},a_{1},a_{2},\ldots )+f(b_{0},b_{1},b_{2},\ldots )\\[0.3em]&=\ f(v)+f(w)\end{aligned}}}

It follows that ${\displaystyle f}$ is additive.

Proof step: homogeneity

Let ${\displaystyle v=(a_{0},a_{1},a_{2},\ldots )\in V}$ and ${\displaystyle \lambda \in \mathbb {R} }$. Then, we have

{\displaystyle {\begin{aligned}f(\lambda \cdot v)&=f(\lambda \cdot (a_{0},a_{1},a_{2},\ldots ))\\[0.3em]&=\ f(\lambda a_{0},\lambda a_{1},\lambda a_{2},\ldots )\\[0.3em]&=\ (\lambda a_{1},\lambda a_{2},\lambda a_{3},\ldots )\\[0.3em]&=\ \lambda \cdot (a_{1},a_{2},a_{3},\ldots )\\[0.3em]&=\ \lambda \cdot f(a_{0},a_{1},a_{2},\ldots )\\[0.3em]&=\ \lambda \cdot f(v)\end{aligned}}}

So ${\displaystyle f}$ is homogeneous.

Thus it was proved that ${\displaystyle f}$ is a ${\displaystyle \mathbb {R} }$-linear map.

## Construction of a linear map from given values

Exercise (Construction of a linear map)

Let ${\displaystyle a_{1}=(1,0,0)^{T},a_{2}=(0,1,0)^{T},a_{3}=(1,1,0)^{T},a_{4}=(0,0,1)^{T}\in \mathbb {R} ^{3}}$.

Further, consider ${\displaystyle b_{1}=(1,5)^{T},b_{2}=(5,3)^{T},b_{3}=(6,8)^{T},b_{4}=(0,1)^{T}\in \mathbb {R} ^{2}}$.

Find a linear map ${\displaystyle f\colon \mathbb {R} ^{3}\to \mathbb {R} ^{2}}$ with ${\displaystyle f(a_{i})=b_{i}}$ for all ${\displaystyle i\in \{1,2,3,4\}}$.

How to get to the proof? (Construction of a linear map)

Hint: Use the principle of linear continuation.

Solution (Construction of a linear map)

We see that ${\displaystyle (a_{1},a_{2},a_{4})}$ is a basis of ${\displaystyle \mathbb {R} ^{3}}$, namely the standard basis.

According to the theorem of linear continuation, we can construct a linear map

${\displaystyle f\colon \mathbb {R} ^{3}\to \mathbb {R} ^{2}}$ defined by ${\displaystyle f(a_{1}):=b_{1},f(a_{2}):=b_{2},f(a_{4}):=b_{4}}$

Now we only have to check if ${\displaystyle f(a_{3})=b_{3}}$ is satisfied. It is true that ${\displaystyle a_{3}=a_{1}+a_{2}}$, so

${\displaystyle f(a_{3})=f(a_{1}+a_{2})=f(a_{1})+f(a_{2})=b_{1}+b_{2}=(1,5)^{T}+(5,3)^{T}=(6,8)^{T}=b_{3}}$

Thus the condition ${\displaystyle f(a_{i})=b_{i}}$ is satisfied for each ${\displaystyle i\in \{1,2,3,4\}}$. The mapping ${\displaystyle f}$ is linear by definition, so we are done.

Exercise (Linear maps under some conditions)

Let ${\displaystyle u=(1,0,-1)^{T},\,v=(0,1,2)^{T}}$ and ${\displaystyle w=(1,2,3)^{T}}$. Is there an ${\displaystyle \mathbb {R} }$-linear map ${\displaystyle f:\mathbb {R} ^{3}\mathbb {R} ^{2}}$ that satisfies ${\displaystyle f(u)=(0,1)^{T},\,f(v)=(1,-1)^{T},\,f(w)=(2,1)^{T}}$?

How to get to the proof? (Linear maps under some conditions)

First you should check if the vectors ${\displaystyle u,v,w}$ are linearly independent. If this is the case, ${\displaystyle \{u,v,w\}}$ is a basis of ${\displaystyle \mathbb {R} ^{3}}$ because of ${\displaystyle \operatorname {dim} (\mathbb {R} ^{3})=3}$. Using the principle of linear continuation, the existence of such a linear map would follow ${\displaystyle f}$. Let thus ${\displaystyle \lambda _{1},\lambda _{2},\lambda _{3}\in \mathbb {R} }$:

${\displaystyle \lambda _{1}u+\lambda _{2}v+\lambda _{3}w={\begin{pmatrix}\lambda _{1}+\lambda _{3}\\\lambda _{2}+2\lambda _{3}\\-\lambda _{1}+2\lambda _{2}+3\lambda _{3}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}.}$

But then also ${\displaystyle \lambda _{1}=-\lambda _{3},\,\lambda _{2}=-2\lambda _{3}}$ and so ${\displaystyle 2\lambda _{1}=\lambda _{2}}$ must be fulfilled. However, this equation has not only the "trivial" solution ${\displaystyle \lambda _{1}=\lambda _{2}=\lambda _{3}=0}$. In fact, the upper equation is satisfied for ${\displaystyle \lambda _{1}=1,\,\lambda _{2}=2,\,\lambda _{3}=-1}$. Thus, one obtains

{\displaystyle {\begin{aligned}u+2v=w.\end{aligned}}}

For such a map ${\displaystyle f}$, the relation ${\displaystyle f(u)+2f(v)=f(w)}$ would then have to hold, which is a contradiction to

{\displaystyle {\begin{aligned}f(u)+2f(v)=(2,-1)^{T},\quad f(w)=(2,1)^{T}\end{aligned}}}

Solution (Linear maps under some conditions)

Let us first assume that such a linear map ${\displaystyle f}$ would exist. By the following calculation

{\displaystyle {\begin{aligned}u+2v={\begin{pmatrix}1\\0\\-1\end{pmatrix}}+{\begin{pmatrix}0\\2\\4\end{pmatrix}}={\begin{pmatrix}1\\2\\3\end{pmatrix}}=w\end{aligned}}}

we see that ${\displaystyle f(u)+2f(v)=f(w)}$ should hold. But this is a contradiction to the other conditions, because those would imply

{\displaystyle {\begin{aligned}f(u)+2f(v)=(0,1)^{T}+2(1,-1)^{T}=(2,-1)^{T}\neq (2,1)^{T}=f(w)\end{aligned}}}

So there is no such ${\displaystyle f}$.

## Linear independence of two preimages

Exercise

Let ${\displaystyle L\colon V\to W}$ be a linear map and let ${\displaystyle v_{1}}$ and ${\displaystyle v_{2}}$ be two distinct vectors from ${\displaystyle V}$, both mapped to a vector ${\displaystyle w\in W}$ with ${\displaystyle w\neq 0_{W}}$. Prove that ${\displaystyle v_{1}}$ and ${\displaystyle v_{2}}$ are linearly independent.

How to get to the proof?

We show that the two vectors cannot be linearly dependent. So assume that ${\displaystyle v_{1},v_{2}}$ were linearly dependent. Then there would be a ${\displaystyle \rho \in K}$ such that ${\displaystyle v_{1}=\rho \cdot v_{2}}$. We now map these two dependent vectors into the vector space ${\displaystyle W}$ using the linear map ${\displaystyle L}$. This yields

${\displaystyle w=\rho \cdot w}$

Since by premise, ${\displaystyle w\neq 0}$, this is a contradiction and our assumption of linear dependence must be false.

Solution

Assume that ${\displaystyle v_{1}}$ and ${\displaystyle v_{2}}$ were linearly dependent. Then there would be a ${\displaystyle \rho \in K}$ with ${\displaystyle v_{1}=\rho \cdot v_{2}}$ and ${\displaystyle \rho \neq 1}$. Since the map ${\displaystyle L}$ is linear, it follows that

${\displaystyle w=L(v_{2})=L(v_{1})=L(\rho \cdot v_{2})=\rho \cdot L(v_{2})=\rho \cdot w}$

Thus

${\displaystyle w=\rho \cdot w\Rightarrow w-\rho \cdot w=(1_{K}-\rho )\cdot w=0_{W}\Rightarrow 1_{K}-\rho =0_{K}\,\lor \,w=0_{W}}$

Since by assumption ${\displaystyle w\neq 0_{W}}$, we must have ${\displaystyle 1_{K}-\rho =0_{K}}$. But this contradicts our assumption ${\displaystyle \rho \neq 1_{K}}$. Thus we get a contradiction to our assumption of linear dependence. So the vectors ${\displaystyle v_{1},v_{2}\in V}$ are linearly independent.

## Exercises: Isomorphisms

Exercise (complex ${\displaystyle \mathbb {R} }$-vector spaces)

Let ${\displaystyle V}$ be a finite-dimensional ${\displaystyle \mathbb {C} }$-vector space. Show that ${\displaystyle V\cong \mathbb {R} ^{2\operatorname {dim} _{\mathbb {C} }(V)}}$ (interpreted as ${\displaystyle \mathbb {R} }$-vector spaces).

Solution (complex ${\displaystyle \mathbb {R} }$-vector spaces)

Set ${\displaystyle n:=\operatorname {dim} _{\mathbb {C} }(V)}$. We choose a ${\displaystyle \mathbb {C} }$ basis ${\displaystyle {\mathcal {B}}=\{b_{1},\dots ,b_{n}\}}$ of ${\displaystyle V}$. Define ${\displaystyle c_{j}:=i\cdot b_{j}}$ for all ${\displaystyle 1\leq j\leq n}$.

We have to show that ${\displaystyle \{b_{1},\dots ,b_{n},c_{1},\dots ,c_{n}\}}$ is an ${\displaystyle \mathbb {R} }$-basis of ${\displaystyle V}$. Then, ${\displaystyle \operatorname {dim} _{\mathbb {R} }(V)=2n=\operatorname {dim} _{\mathbb {R} }(\mathbb {R} ^{2n})}$. According to a theorem above, we have ${\displaystyle V\cong \mathbb {R} ^{2n}}$ as ${\displaystyle \mathbb {R} }$-vector spaces.

We now show ${\displaystyle \mathbb {R} }$-linear independence.

Proof step: ${\displaystyle \{b_{1},\dots ,b_{n},c_{1},\dots ,c_{n}\}}$ is ${\displaystyle \mathbb {R} }$-linearly independent

Let ${\displaystyle \beta _{1},\dots ,\beta _{n},\gamma _{1},\dots ,\gamma _{n}\in \mathbb {R} }$ and assume that ${\displaystyle \sum _{j=1}^{n}\beta _{j}\cdot b_{j}+\sum _{j=1}^{n}\gamma _{j}\cdot c_{j}=0}$. We substitute the definition for ${\displaystyle c_{j}}$, conclude the sums and obtain ${\displaystyle \sum _{j=1}^{n}(\beta _{j}+i\cdot \gamma _{j})\cdot b_{j}=0}$. By ${\displaystyle \mathbb {C} }$-linear independence of ${\displaystyle b_{j}}$ we obtain ${\displaystyle \beta _{j}+i\cdot \gamma _{j}=0}$ for all ${\displaystyle j\in \{1,\dots ,n\}}$. Thus, ${\displaystyle \beta _{j}=\gamma _{j}=0}$ for all ${\displaystyle j\in \{1,\dots ,n\}}$. This establishes the ${\displaystyle \mathbb {R} }$-linear independence.

Now only one step is missing:

Proof step: ${\displaystyle \{b_{1},\dots ,b_{n},c_{1},\dots ,c_{n}\}}$ is a generator with respect to ${\displaystyle \mathbb {R} }$

Let ${\displaystyle v\in V}$ be arbitrary.

Since ${\displaystyle {\mathcal {B}}}$ is a ${\displaystyle \mathbb {C} }$-basis of ${\displaystyle V}$ , we can find some ${\displaystyle \lambda _{1},\dots ,\lambda _{n}\in \mathbb {C} }$, such that ${\displaystyle v=\sum _{j=1}^{n}\lambda _{j}\cdot b_{j}}$ . We write ${\displaystyle \lambda _{j}=\beta _{j}+\gamma _{j}i}$ with ${\displaystyle \beta _{j},\gamma _{j}\in \mathbb {R} }$ for all ${\displaystyle j}$. Then we obtain

{\displaystyle {\begin{aligned}v&=\sum _{j=1}^{n}\lambda _{j}\cdot b_{j}\\&=\sum _{j=1}^{n}(\beta _{j}+\gamma _{j}i)\cdot b_{j}\\&=\sum _{j=1}^{n}(\beta _{j}\cdot b_{j}+\gamma _{j}\cdot (i\cdot b_{j}))\\&=\sum _{j=1}^{n}(\beta _{j}\cdot b_{j}+\gamma _{j}\cdot c_{j})\\&=\sum _{j=1}^{n}\beta _{j}\cdot b_{j}+\sum _{j=1}^{n}\gamma _{j}\cdot c_{j}.\end{aligned}}}

So ${\displaystyle v}$ is inside the ${\displaystyle \mathbb {R} }$-span of ${\displaystyle \{b_{1},\dots ,b_{n},c_{1},\dots ,c_{n}\}}$. This establishes the assertion.

Exercise (Isomorphic coordinate spaces)

Let ${\displaystyle K}$ be a field and consider ${\displaystyle n,m\in \mathbb {N} _{0}}$. Prove that ${\displaystyle K^{n}\cong K^{m}}$ holds if and only if ${\displaystyle m=n}$.

Solution (Isomorphic coordinate spaces)

We know that ${\displaystyle \operatorname {dim} (K^{k})=k}$ for all ${\displaystyle k\in \mathbb {N} _{0}}$. We use the theorem above, which states that finite-dimensional vector spaces are isomorphic exactly if their dimensions coincide. So ${\displaystyle K^{n}\cong K^{m}}$ holds if and only if ${\displaystyle n=\operatorname {dim} (K^{n})=\operatorname {dim} (K^{m})=m}$ .

Exercise (Isomorphism criteria for endomorphisms)

Let ${\displaystyle K}$ be a field, ${\displaystyle V}$ a finite-dimensional ${\displaystyle K}$-vector space and ${\displaystyle f:V\to V}$ a ${\displaystyle K}$-linear map. Prove that the following three statements are equivalent:

(i) ${\displaystyle f}$ is an isomorphism.

(ii) ${\displaystyle f}$ is injective.

(iii) ${\displaystyle f}$ is surjective.

(Note: For this task, it may be helpful to know the terms kernel and image of a linear map. Using the dimension theorem, this exercise becomes much easier. However, we give a solution here, which works without the dimension theorem).

Solution (Isomorphism criteria for endomorphisms)

(i)${\displaystyle \implies }$(ii) and (iii): According to the definition of an isomorphism, ${\displaystyle f}$ is bijective, i.e. injective and surjective. Therefore (ii) and (iii) hold.

(ii)${\displaystyle \implies }$(i): Let ${\displaystyle f}$ be an injective mapping. We need to show that ${\displaystyle f}$ is also surjective. The image ${\displaystyle \mathrm {im} (f):=\{f(v)~|~v\in V\}}$ of ${\displaystyle f}$ is a subspace of ${\displaystyle V}$. This can be verified by calculation. We now define a mapping ${\displaystyle f'}$ that does the same thing as ${\displaystyle f}$, except that it will be surjective by definition. This mapping is defined as follows:

{\displaystyle {\begin{aligned}f':V&\to \mathrm {im} (f)\\v&\mapsto f(v)\end{aligned}}}

The surjectivity comes from the fact that every element ${\displaystyle w\in \mathrm {im} (f)}$ can be written as ${\displaystyle w=f(v')}$, for a suitable ${\displaystyle v'\in V}$. Moreover, the mapping ${\displaystyle f'}$ is injective and linear. This is because ${\displaystyle f}$ already has these two properties. So ${\displaystyle V}$ and ${\displaystyle \mathrm {im} (f)}$ are isomorphic. Therefore, ${\displaystyle V}$ and ${\displaystyle \mathrm {im} (f)}$ have the same finite dimension. Since ${\displaystyle \mathrm {im} (f)}$ is a subspace of ${\displaystyle V}$, ${\displaystyle \mathrm {im} (f)=V}$ holds. This can be seen by choosing a basis in ${\displaystyle \mathrm {im} (f)}$, for instance the basis given by the vectors ${\displaystyle v_{1},\dots ,v_{n}\in \mathrm {im} (f)}$. These ${\displaystyle v_{1},\dots ,v_{n}}$ are also linearly independent in ${\displaystyle V}$, since ${\displaystyle \mathrm {im} (f)\subseteq V}$. And since ${\displaystyle V}$ and ${\displaystyle \mathrm {im} (f)}$ have the same dimension, the ${\displaystyle v_{1},\dots ,v_{n}}$ are also a basis in ${\displaystyle V}$. So the two vector spaces ${\displaystyle V}$ and ${\displaystyle \mathrm {im} (f)}$ must now be the same, because all elements from them are ${\displaystyle K}$-linear combinations formed with the ${\displaystyle v_{1},\dots ,v_{n}}$. Thus we have shown that ${\displaystyle f}$ is surjective.

(iii)${\displaystyle \implies }$(i): Now suppose ${\displaystyle f}$ is surjective. We need to show that ${\displaystyle f}$ is also injective. Let ${\displaystyle \mathrm {ker} (f):=\{v\in V~|~f(v)=0\}}$ be the kernel of the mapping ${\displaystyle f}$. You may convince yourself by calculation, that this kernel is a subspace of ${\displaystyle V}$. Let ${\displaystyle v_{1},\dots ,v_{k}}$ be a basis of ${\displaystyle \mathrm {ker} (f)}$. We can complete this (small) basis to a (large) basis of ${\displaystyle V}$, by including the additional vectors ${\displaystyle v_{k+1},\dots ,v_{n}}$. We will now show that ${\displaystyle f(v_{k+1}),\dots ,f(v_{n})}$ are linearly independent. So let coefficients ${\displaystyle \lambda _{k+1},\dots ,\lambda _{n}\in K}$ be given such that

{\displaystyle {\begin{aligned}\lambda _{k+1}f(v_{k+1})+\dots +\lambda _{n}f(v_{n})=0.\end{aligned}}}

By linearity of ${\displaystyle f}$ we conclude: ${\displaystyle f(\lambda _{k+1}v_{k+1}+\dots \lambda _{n}v_{n})=0}$. This means that the linear combination

{\displaystyle {\begin{aligned}\lambda _{k+1}v_{k+1}+\dots +\lambda _{n}v_{n}\end{aligned}}}

is in the kernel of ${\displaystyle f}$. But we already know a basis of ${\displaystyle \mathrm {ker} (f)}$. Therefore there are coefficients ${\displaystyle \lambda _{1},\dots ,\lambda _{k}\in K}$, such that

{\displaystyle {\begin{aligned}\lambda _{k+1}v_{k+1}+\dots +\lambda _{n}v_{n}=\lambda _{1}v_{1}+\dots +\lambda _{k}v_{k}.\end{aligned}}}

Because of the linear independence of ${\displaystyle v_{1},\dots ,v_{n}}$ it now follows that ${\displaystyle \lambda _{1},\dots ,\lambda _{n}=0}$. Therefore, the ${\displaystyle f(v_{k+1}),\dots ,f(v_{n})}$ are linearly independent. Next, we will show that these vectors also form a basis of ${\displaystyle V}$. To do this, we show that each vector in ${\displaystyle V}$ can be written as a linear combination of the ${\displaystyle f(v_{k+1}),\dots ,f(v_{n})}$. Let ${\displaystyle w\in V}$. Because of the surjectivity of ${\displaystyle f}$, there is a ${\displaystyle v\in V}$, with ${\displaystyle w=f(v)}$. Since the ${\displaystyle v_{1},\dots ,v_{n}}$ form a basis of ${\displaystyle V}$, there are coefficients ${\displaystyle \lambda _{1},\dots ,\lambda _{n}\in K}$ such that

{\displaystyle {\begin{aligned}v=\lambda _{1}v_{1}+\dots +\lambda _{n}v_{n}\end{aligned}}}

If we now apply ${\displaystyle f}$ to this equation, we get:

{\displaystyle {\begin{aligned}w=f(v)=\lambda _{1}\underbrace {f(v_{1})} _{=0}+\dots +\lambda _{k}\underbrace {f(v_{k})} _{=0}+\lambda _{k+1}f(v_{k+1})+\dots +\lambda _{n}f(v_{n}).\end{aligned}}}

Here we used the linearity of ${\displaystyle f}$. Since the first ${\displaystyle k}$ elements of our basis are in the kernel, their images are ${\displaystyle 0}$. So we get the desired representation of ${\displaystyle w}$:

{\displaystyle {\begin{aligned}w=f(v)=\lambda _{k+1}f(v_{k+1})+\dots +\lambda _{n}f(v_{n}).\end{aligned}}}

Thus we have shown that ${\displaystyle f(v_{k+1}),\dots ,f(v_{n})}$ forms a linearly independent generator of ${\displaystyle V}$. So these vectors form a basis of ${\displaystyle V}$. Now if ${\displaystyle k}$ were not ${\displaystyle 0}$, two finite bases in ${\displaystyle V}$ would not contain equally many elements. This cannot be the case. Therefore, ${\displaystyle k=0}$, so ${\displaystyle \mathrm {ker} (f)}$ is the trivial vector space and ${\displaystyle f}$ is indeed injective.

Exercise (Function spaces)

Let ${\displaystyle X}$ be a finite set with ${\displaystyle n\in \mathbb {N} }$ elements and let ${\displaystyle K}$ be a field. We have seen that the set of functions from ${\displaystyle X}$ to ${\displaystyle K}$ forms a ${\displaystyle K}$-vector space, denoted by ${\displaystyle \operatorname {Fun} (X,K)}$. Show that ${\displaystyle \operatorname {Fun} (X,K)\cong K^{n}}$.

Solution (Function spaces)

We already know according to a theorem above that two finite dimensional vector spaces are isomorphic exactly if they have the same dimension. So we just need to show that ${\displaystyle \operatorname {dim} (\operatorname {fun} (X,K))=n=\operatorname {dim} (K^{n})}$ holds.

To show this, we first need a basis of ${\displaystyle \operatorname {Fun} (X,K)}$. For this, let ${\displaystyle x_{1},\dots ,x_{n}}$ be the elements of the set ${\displaystyle X}$. We define ${\displaystyle f_{1},\dots ,f_{n}\in \operatorname {Fun} (X,K)}$ by

${\displaystyle f_{j}(x_{i}):=\delta _{i,j}={\begin{cases}1,{\text{ for }}i=j\\0,{\text{ for }}i\neq j.\end{cases}}}$

We now show that the functions ${\displaystyle f_{1},\dots ,f_{n}}$ indeed form a basis of ${\displaystyle \operatorname {Fun} (X,K)}$.

Proof step: ${\displaystyle f_{1},\dots ,f_{n}}$ are linearly independent

Let ${\displaystyle \alpha _{1},\dots ,\alpha _{n}\in K}$ with ${\displaystyle \sum _{k=1}^{n}\alpha _{k}\cdot f_{k}=0}$ being the zero function. If we apply this function to any ${\displaystyle x_{j}}$ with ${\displaystyle j\in \{1,\dots ,n\}}$, then we obtain: ${\displaystyle \sum _{k=1}^{n}\alpha _{k}\cdot f_{k}(x_{j})=0}$. By definition of ${\displaystyle f_{1},\dots ,f_{n}}$ it follows that

${\displaystyle 0=\sum _{k=1}^{n}\alpha _{k}f_{k}(x_{j})=\alpha _{j}\cdot f_{j}(x_{j})=\alpha _{j}\cdot 1=\alpha _{j}}$.

Since ${\displaystyle j}$ was arbitrary and ${\displaystyle \sum _{k=1}^{n}\alpha _{k}\cdot f_{k}(x_{j})=0}$ must hold for all ${\displaystyle x_{j}\in X}$, it follows that ${\displaystyle \alpha _{1}=\dots =\alpha _{n}=0}$. So we have shown that ${\displaystyle f_{1},\dots ,f_{n}}$ are linearly independent.

Proof step: ${\displaystyle f_{1},\dots ,f_{n}}$ generate ${\displaystyle \operatorname {Fun} (X,K)}$

Let ${\displaystyle g\in \operatorname {Fun} (X,K)}$ be arbitrary. We now want to write ${\displaystyle g}$ as a linear combination of ${\displaystyle f_{1},\dots ,f_{n}}$. For this we show ${\displaystyle g=\sum _{j=1}^{n}g(x_{j})\cdot f_{j}}$, i.e., ${\displaystyle g}$ is a linear combination of ${\displaystyle f_{1},\dots ,f_{n}}$ with coefficients ${\displaystyle g(x_{1}),\dots ,g(x_{n})\in K}$. We now verify that ${\displaystyle g(x_{i})=\sum _{j=1}^{n}g(x_{j})\cdot f_{j}(x_{i})}$ for all ${\displaystyle i\in \{1,\dots ,n\}}$. Let ${\displaystyle i\in \{1,\dots ,n\}}$ be arbitrary. By definition of ${\displaystyle f_{1},\dots ,f_{n}}$ we obtain:

${\displaystyle \sum _{j=1}^{n}g(x_{j})\cdot f_{j}(x_{i})=g(x_{i})\cdot f_{i}(x_{i})=g(x_{i})\cdot 1=g(x_{i})}$.

Since equality holds for all ${\displaystyle i}$, the functions agree at every point and are therefore identical. So we have shown that ${\displaystyle f_{1},\dots ,f_{n}}$ generate ${\displaystyle \operatorname {Fun} (X,K)}$.

Thus we have proved that ${\displaystyle f_{1},\dots ,f_{n}}$ is a basis of ${\displaystyle \operatorname {Fun} (X,K)}$. Since we have ${\displaystyle n}$ basis elements of ${\displaystyle \operatorname {Fun} (X,K)}$, it follows that ${\displaystyle \operatorname {dim} (\operatorname {Fun} (X,K))=n=\operatorname {dim} (K^{n})}$.

## Exercises: Images

Exercise (Associating image spaces to figures)

We consider the following four subspaces from the vector space ${\displaystyle \mathbb {R} ^{2}}$, given as images of the linear maps

1. ${\displaystyle f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)^{T}\mapsto (2(x+y),x-3y)^{T}}$
2. ${\displaystyle g\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)^{T}\mapsto (x,2x)^{T}}$
3. ${\displaystyle h\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)^{T}\mapsto (-3(x-y),(x-y))^{T}}$
4. ${\displaystyle k\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)^{T}\mapsto (x,0)^{T}}$

Match these four subspaces to the subspaces ${\displaystyle U_{1},U_{2},U_{3},U_{4}}$ shown in the figures below.

Solution (Associating image spaces to figures)

First we look for the image of ${\displaystyle f}$: To find ${\displaystyle \operatorname {im} (f)}$, we can apply a theorem from above: If ${\displaystyle E}$ is a generator of ${\displaystyle \mathbb {R} ^{2}}$, then ${\displaystyle \operatorname {im} (f)=\operatorname {span} (f(E))}$ holds. We take the standard basis ${\displaystyle \{(1,0)^{T},(0,1)^{T}\}}$ as the generator of ${\displaystyle \mathbb {R} ^{2}}$. Then

${\displaystyle \operatorname {im} (f)=\operatorname {span} \left(f{\begin{pmatrix}1\\0\end{pmatrix}},f{\begin{pmatrix}0\\1\end{pmatrix}}\right).}$

Now we apply ${\displaystyle f}$ to the standard basis

{\displaystyle {\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 ${\displaystyle (2,1)^{T},(2,-3)^{T}}$ generate the image of ${\displaystyle f}$. Moreover, they are linearly independent and thus a basis of ${\displaystyle \mathbb {R} ^{2}}$. Therefore ${\displaystyle \operatorname {im} (f)=\mathbb {R} ^{2}}$. So ${\displaystyle \operatorname {im} (f)=U_{3}}$.

Next, we want to find the image of ${\displaystyle g}$. However, it is also possible to compute the image ${\displaystyle \operatorname {im} (g)}$ directly by definition, which we will demonstrate here.

{\displaystyle {\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 ${\displaystyle g}$ is spanned by the vector ${\displaystyle (1,2)^{T}}$. Thus ${\displaystyle \operatorname {im} (g)=U_{1}}$.

Now we determine the image of ${\displaystyle h}$ using, for example, the same method as for ${\displaystyle f}$. That means we apply ${\displaystyle h}$ to the standard basis:

{\displaystyle {\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 ${\displaystyle \operatorname {im} (h)=\operatorname {span} ((-3,1)^{T})}$ and thus ${\displaystyle \operatorname {im} (h)=U_{2}}$.

Finally, we determine the image of ${\displaystyle k}$. For this we proceed for example as with ${\displaystyle g}$.

{\displaystyle {\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 ${\displaystyle k}$ is spanned by the vector ${\displaystyle (1,0)^{T}}$. Thus ${\displaystyle \operatorname {im} (k)}$ is the ${\displaystyle x}$-axis, so ${\displaystyle \operatorname {im} (k)=U_{4}}$.

Exercise (Image of a matrix)

1. Consider the matrix ${\displaystyle (1,2)\in \mathbb {R} ^{1\times 2}}$ and the mapping ${\displaystyle f\colon \mathbb {R} ^{2}\to \mathbb {R} ,x\mapsto (1,2)x}$ induced by it. What is the image ${\displaystyle \operatorname {im} (f)}$?
2. Now let ${\displaystyle A=(a_{1},\ldots ,a_{m})\in K^{n\times m}}$ be any matrix over a field ${\displaystyle K}$, where ${\displaystyle a_{1},\ldots ,a_{m}\in K^{n}}$ denote the columns of ${\displaystyle A}$. Consider the mapping ${\displaystyle f_{A}\colon K^{m}\to K^{n},x\mapsto Ax}$ induced by ${\displaystyle A}$. Show that ${\displaystyle \operatorname {im} (f_{A})=\operatorname {span} \{a_{1},\ldots ,a_{m}\}}$ 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 ${\displaystyle \operatorname {im} (f)}$ of the linear map ${\displaystyle f}$ is a subspace of ${\displaystyle \mathbb {R} }$. Since the ${\displaystyle \mathbb {R} }$-vector space ${\displaystyle \mathbb {R} }$ has dimension ${\displaystyle 1}$, a subspace can only have dimension ${\displaystyle 0}$ or ${\displaystyle 1}$. In the first case the subspace is the null vector space, in the second case it is already all of ${\displaystyle \mathbb {R} }$. So ${\displaystyle \mathbb {R} }$ has only the two subspaces ${\displaystyle \{0\}}$ and ${\displaystyle \mathbb {R} }$. Since ${\displaystyle (1,2)(1,0)^{T}=1\neq 0}$ holds, we have that ${\displaystyle \operatorname {im} (f)\neq \{0\}}$. Thus, ${\displaystyle \operatorname {im} (f)=\mathbb {R} }$.

Solution sub-exercise 2:

Proof step: "${\displaystyle \subseteq }$"

Let ${\displaystyle y\in \operatorname {im} (f_{A})}$. Then, there is some ${\displaystyle x=(x_{1},\ldots ,x_{m})^{T}\in K^{m}}$ with ${\displaystyle Ax=y}$. We can write ${\displaystyle x}$ as ${\displaystyle x=\sum _{i=1}^{m}x_{i}e_{i}}$. Plugging this into the equation ${\displaystyle Ax=y}$, we get.

{\displaystyle {\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 ${\displaystyle \sum _{i=1}^{m}x_{i}a_{i}\in \operatorname {span} \{a_{1},\ldots ,a_{n}\}}$, we obtain ${\displaystyle y\in \operatorname {span} \{a_{1},\ldots ,a_{n}\}}$.

Proof step: "${\displaystyle \supseteq }$"

Let ${\displaystyle y=\sum _{i=1}^{m}y_{i}\cdot a_{i}\in \operatorname {span} (f_{A})}$ with ${\displaystyle y_{i}\in K}$ for ${\displaystyle i=1,\ldots ,m}$. We want to find ${\displaystyle x\in K^{m}}$ with ${\displaystyle Ax=y}$. So let us define ${\displaystyle x:=\sum _{i=1}^{m}y_{i}e_{i}}$. The same calculation as in the first step of the proof then shows

${\displaystyle Ax=A\left(\sum _{i=1}^{m}y_{i}e_{i}\right)=\sum _{i=1}^{m}y_{i}Ae_{i}=\sum _{i=1}^{m}y_{i}a_{i}=y.}$

Exercise (Surjectivity and dimension of ${\displaystyle V}$ and ${\displaystyle W}$)

Let ${\displaystyle V}$ and ${\displaystyle W}$ be two finite-dimensional vector spaces. Show that there exists a surjective linear map ${\displaystyle f\colon V\to W}$ if and only if ${\displaystyle \dim(V)\geq \dim(W)}$.

How to get to the proof? (Surjectivity and dimension of ${\displaystyle V}$ and ${\displaystyle W}$)

We want to estimate the dimensions of ${\displaystyle V}$ and ${\displaystyle W}$ against each other. The dimension is defined as the cardinality of a basis. That is, if ${\displaystyle b_{1},\dots ,b_{n}}$ is a basis of ${\displaystyle V}$ and ${\displaystyle c_{1},\dots ,c_{m}}$ is a basis of ${\displaystyle W}$, we must show that ${\displaystyle n\geq m}$ holds if and only if there exists a surjective linear map. "if and only if" means that we need to establish two directions (${\displaystyle \Rightarrow ,\Leftarrow }$).

Given a surjective linear map ${\displaystyle f\colon V\to W}$, we must show that the dimension of ${\displaystyle V}$ is at least ${\displaystyle m}$. Now bases are maximal linearly independent subsets. That is, to estimate the dimension from below, we need to construct a linearly independent subset with ${\displaystyle m}$ elements. In the figure, we have already a linearly independent subset with ${\displaystyle m}$ elements, which is the basis ${\displaystyle c_{1},\dots ,c_{m}}$. Because ${\displaystyle f}$ is surjective, we can lift these to vectors ${\displaystyle {\hat {c}}_{1},\dots ,{\hat {c}}_{m}\in V}$ with ${\displaystyle f({\hat {c}}_{i})=c_{i}}$. Now we need to verify that ${\displaystyle {\hat {c}}_{1},\dots ,{\hat {c}}_{m}}$ are linearly independent in ${\displaystyle V}$. We see this, by converting a linear combination ${\displaystyle \lambda _{1}{\hat {c}}_{1}+\dots \lambda _{m}{\hat {c}}_{m}=0}$ via ${\displaystyle f}$ into a linear combination ${\displaystyle 0=f(\lambda _{1}{\hat {c}}_{1}+\dots \lambda _{m}{\hat {c}}_{m})=\lambda _{1}c_{1}+\dots \lambda _{m}c_{m}}$ and exploiting the linear independence of ${\displaystyle c_{1},\dots ,c_{m}}$.

Conversely, if ${\displaystyle n\geq m}$ holds, we must construct a surjective linear map ${\displaystyle f\colon V\to W}$. Following the principle of linear continuation, we can construct the linear map ${\displaystyle f}$ by specifying how ${\displaystyle f}$ acts on a basis of ${\displaystyle V}$. For this we need elements of ${\displaystyle W}$ on which we can send ${\displaystyle b_{1},\dots ,b_{n}}$. We have already chosen a basis of ${\displaystyle W}$ above. Therefore, it is convenient to define ${\displaystyle f}$ as follows:

${\displaystyle f(b_{i})={\begin{cases}c_{i}&i\leq m\\0&i>m\end{cases}}}$

Then the image of ${\displaystyle f}$ is spanned by the vectors ${\displaystyle f(b_{1})=c_{1},\dots ,f(b_{m})=c_{m},f(b_{m+1})=0,\dots ,f(b_{m})=0}$. However, these vectors also span all of ${\displaystyle W}$ and thus ${\displaystyle f}$ is surjective.

Solution (Surjectivity and dimension of ${\displaystyle V}$ and ${\displaystyle W}$)

Proof step: "${\displaystyle \Rightarrow }$"

Suppose there is a suitable surjective mapping ${\displaystyle f}$. We show that the dimension of ${\displaystyle \operatorname {im} (f)=f(V)}$ cannot be larger than the dimension of ${\displaystyle V}$ (this is true for any linear map). Because of the surjectivity of ${\displaystyle f}$, it follows that ${\displaystyle \dim(V)\geq \dim(\operatorname {im} (f))=\dim(W)}$.

So let ${\displaystyle w_{1},\ldots ,w_{n}\in \operatorname {im} (f)}$ be linearly independent. There exists ${\displaystyle v_{1},\ldots ,v_{n}\in V}$ with ${\displaystyle f(v_{i})=w_{i}}$ for ${\displaystyle i\in \{1,\ldots ,n\}}$. We show that ${\displaystyle v_{1},\ldots ,v_{n}}$ are also linearly independent: Let ${\displaystyle \lambda _{1},\ldots ,\lambda _{n}\in K}$ with ${\displaystyle \sum _{i=1}^{n}\lambda _{i}v_{i}=0}$. Then we also have that

${\displaystyle 0=f(\sum _{i=1}^{n}\lambda _{i}v_{i})=\sum _{i=1}^{n}\lambda _{i}f(v_{i})=\sum _{i=1}^{n}\lambda _{i}w_{i},}$

By linear independence of ${\displaystyle w_{1},\ldots ,w_{n}}$, it follows that ${\displaystyle \lambda _{1}=\ldots =\lambda _{n}=0}$. So ${\displaystyle v_{1},\ldots ,v_{n}}$ are also linearly independent. Overall, we have shown that

${\displaystyle w_{1},\ldots ,w_{n}\in \operatorname {im} (f){\text{ linearly independent }}\implies v_{1},\ldots ,v_{n}{\text{ linearly independent for any choice of preimages }}v_{i}\in f^{-1}(w_{i}).}$

In particular, it holds that a basis of ${\displaystyle V}$ (a maximal linearly independent subset of ${\displaystyle V}$) must contain at least as many elements as a basis of ${\displaystyle \operatorname {im} (f)}$, that is, ${\displaystyle \dim(V)\geq \dim(\operatorname {im} (f))}$.

Proof step: "${\displaystyle \Leftarrow }$"

Assume that ${\displaystyle \dim(V)\geq \dim(W)}$. We use that a linear map is already uniquely determined by the images of the basis vectors. Let ${\displaystyle \{v_{1},\ldots ,v_{m}\}}$ be a basis of ${\displaystyle V}$ and ${\displaystyle \{w_{1},\ldots ,w_{n}\}}$ be a basis of ${\displaystyle W}$. Define the surjective linear map ${\displaystyle f\colon V\to W}$ by

${\displaystyle f(v_{i})={\begin{cases}w_{i}&{\text{ if }}i\leq n\\0&{\text{ else.}}\end{cases}}}$

This works, since by assumption, ${\displaystyle m\geq n}$ holds. The mapping constructed in this way is surjective, since by construction, ${\displaystyle \{w_{1},\ldots ,w_{n}\}\subseteq \operatorname {im} (f)}$. As the image of ${\displaystyle f}$ is a subspace of ${\displaystyle W}$, the subspace generated by these vectors, i.e., ${\displaystyle W}$, also lies in the image of ${\displaystyle f}$. Accordingly, ${\displaystyle W\subseteq \operatorname {im} (f)\subseteq W}$ holds and ${\displaystyle f}$ is surjective.

## Exercises: Kernel

Exercise

We consider the linear map ${\displaystyle f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)^{T}\mapsto (-3(x-y),x-y)^{T}}$. Determine the kernel of ${\displaystyle f}$.

Solution

We are looking for vectors ${\displaystyle (x,y)^{T}\in \mathbb {R} ^{2}}$ such that ${\displaystyle f\left({\begin{pmatrix}x\\y\end{pmatrix}}\right)={\begin{pmatrix}0\\0\end{pmatrix}}}$. Let ${\displaystyle (x,y)^{T}}$ be any vector in ${\displaystyle \mathbb {R} ^{2}}$ for which ${\displaystyle f\left({\begin{pmatrix}x\\y\end{pmatrix}}\right)={\begin{pmatrix}0\\0\end{pmatrix}}}$ is true. We now examine what properties this vector must have. It holds that

${\displaystyle {\begin{pmatrix}0\\0\end{pmatrix}}=f{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}-3(x-y)\\x-y\end{pmatrix}}}$

So ${\displaystyle -3(x-y)=0}$ and ${\displaystyle x-y=0}$. From this we conclude ${\displaystyle x=y}$. So any vector ${\displaystyle (x,y)^{T}}$ in the kernel of ${\displaystyle f}$ satisfies the condition ${\displaystyle x=y}$. Now take a vector ${\displaystyle (x,x)^{T}}$ with ${\displaystyle x\in \mathbb {R} }$. Then

${\displaystyle f{\begin{pmatrix}x\\x\end{pmatrix}}={\begin{pmatrix}-3(x-x)\\x-x\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}}$

We see that ${\displaystyle (x,x)^{T}\in \ker(f)}$. In total

${\displaystyle \ker(f)=\left\{{\begin{pmatrix}x\\x\end{pmatrix}}|x\in \mathbb {R} \right\}}$

Check your understanding: Can you visualize ${\displaystyle \ker(f)}$ in the plane? What does the image of ${\displaystyle f}$ look like? How do the kernel and the image relate to each other?

${\displaystyle \ker(f)=\left\{{\begin{pmatrix}x\\x\end{pmatrix}}\mid x\in \mathbb {R} \right\}=\operatorname {span} \left({\begin{pmatrix}1\\1\end{pmatrix}}\right)}$

Now we determine the image of ${\displaystyle f}$ by applying ${\displaystyle f}$ to the canonical basis.

{\displaystyle {\begin{aligned}f{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}-3\\1\end{pmatrix}}\\f{\begin{pmatrix}0\\1\end{pmatrix}}={\begin{pmatrix}3\\-1\end{pmatrix}}\end{aligned}}}

So ${\displaystyle \operatorname {im} (f)=\operatorname {span} (f((1,0)^{T}),f((0,1)^{T}))}$ holds. We see that the two vectors are linearly dependent. That is, we can generate the image with only one vector: ${\displaystyle \operatorname {im} (f)=\operatorname {span} ((-3,1)^{T})}$.

In our example, the image and the kernel of the linear map ${\displaystyle f}$ are straight lines through the origin. The two straight lines intersect only at the zero and together span the whole ${\displaystyle \mathbb {R} ^{2}}$.

Exercise

Let ${\displaystyle V}$ be a vector space, ${\displaystyle V\neq \{0\}}$, and ${\displaystyle f\colon V\to V}$ be a nilpotent linear map, i.e., there is some ${\displaystyle n\in \mathbb {N} }$ such that

${\displaystyle f^{n}=\underbrace {f\circ \cdots \circ f} _{n{\text{ times}}}=0}$

is the zero mapping. Show that ${\displaystyle \ker(f)\neq \{0\}}$ holds.

Does the converse also hold, that is, is any linear map ${\displaystyle f\colon V\to V}$ with ${\displaystyle \ker(f)\neq \{0\}}$ nilpotent?

Solution

Proof step: ${\displaystyle f}$ nilpotent ${\displaystyle \implies \ker(f)\neq \{0\}}$

We prove the statement by contraposition. That is we show: If ${\displaystyle \ker(f)=\{0\}}$, then ${\displaystyle f}$ is not nilpotent.

Let ${\displaystyle \ker(f)=\{0\}}$. Then ${\displaystyle f}$ is injective, and as a concatenation of injective functions, ${\displaystyle f\circ f}$ is also injective. By induction it follows that for all ${\displaystyle n\in \mathbb {N} }$ the function ${\displaystyle f^{n}=\underbrace {f\circ \cdots \circ f} _{n{\text{ times}}}}$ is injective. But then also ${\displaystyle \ker(f^{n})=\{0\}}$ for all ${\displaystyle n\in \mathbb {N} }$. Since the kernel of the zero mapping would be all of ${\displaystyle V\neq \{0\}}$, the map ${\displaystyle f^{n}}$ could not be the zero mapping for any ${\displaystyle n\in \mathbb {N} }$. Consequently, ${\displaystyle f}$ is not nilpotent.

Proof step: The converse implication

The converse implication does not hold. There are mappings that are neither injective nor nilpotent. For example we can define

${\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2},\quad {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x\\0\end{pmatrix}}}$

This mapping is not injective, because ${\displaystyle (0,1)^{T}\in \ker(f)}$. But it is also not nilpotent, because we have ${\displaystyle f^{n}((1,0)^{T})=(1,0)\neq 0}$ for all ${\displaystyle n\in \mathbb {N} }$.

Exercise (Injectivity and dimension of ${\displaystyle V}$ and ${\displaystyle W}$)

Let ${\displaystyle V}$ and ${\displaystyle W}$ be two finite-dimensional vector spaces. Show that there exists an injective linear map ${\displaystyle f\colon V\to W}$ if and only if ${\displaystyle \dim(V)\leq \dim(W)}$.

How to get to the proof? (Injectivity and dimension of ${\displaystyle V}$ and ${\displaystyle W}$)

To prove equivalence, we need to show two implications. For the execution, we use that every monomorphism ${\displaystyle f\colon V\to W}$ preserves linear independence: If ${\displaystyle \{b_{1},\ldots ,b_{n}\}\subseteq V}$ is a basis of ${\displaystyle V}$, then the ${\displaystyle n}$ vectors ${\displaystyle f(b_{1}),\ldots ,f(b_{n})\in W}$ are linearly independent. For the converse direction, we need to construct a monomorphism from ${\displaystyle V}$ to