Exercises Linear Maps – Serlo

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

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

Showing linearity of a mapping

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

Exercise (Linear map into a field)

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

Proof step: Additivity

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

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

Thus $f$ is additive.

Proof step: Homogeneity

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

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

Thus $f$ is homogeneous and $f$ is linear.

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

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

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

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

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

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

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

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

Aktuelles Ziel: Additivity

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

Aktuelles Ziel: Scaling

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

Exercise (Linearity of the embedding)

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

Solution (Linearity of the embedding)

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

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

So $f$ is linear.

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

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

Exercise (Linearity of $f$)

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

Proof (Linearity of $f$)

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

Proof step: additivity

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

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

Proof step: homogeneity

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

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

Thus the map is linear.

Important special cases

Exercise (The identity is a linear map)

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

Proof (The identity is a linear map)

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

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

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

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

Exercise (The map to zero is a linear map)

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

Proof (The map to zero is a linear map)

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

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

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

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

Thus, the map to zero is linear

Linear maps between function spaces

Exercise (Mapping on a function space)

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

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

Show that $\phi$ is linear.

Solution (Mapping on a function space)

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

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

and

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

Thus, we have established linearity.

Exercise (The precomposition with a map is linear.)

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

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

Show that $\Theta$ is linear.

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

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

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

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

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

Proof (The precomposition with a map is linear.)

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

Proof step: additivity

For all $n\in N$ we have that

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

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

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

Proof step: homogeneity

For all $n\in N$ we have that

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

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

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

Exercise (Sequence space)

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

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

is linear.

How to get to the proof? (Sequence space)

To show linearity, two properties need to be checked:

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

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

Proof (Sequence space)

Proof step: additivity

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

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

It follows that $f$ is additive.

Proof step: homogeneity

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

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

So $f$ is homogeneous.

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

Construction of a linear map from given values

Exercise (Construction of a linear map)

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

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

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

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

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

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

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

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

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

Exercise (Linear maps under some conditions)

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

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

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

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

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

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

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

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

Solution (Linear maps under some conditions)

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

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

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

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

So there is no such $f$.

Linear independence of two preimages

Exercise

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

How to get to the proof?

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

$w=\rho \cdot w$

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

Solution

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

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

Thus

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

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

Exercises: Isomorphisms

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

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

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

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

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

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

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

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

Now only one step is missing:

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

Let $v\in V$ be arbitrary.

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

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

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

Exercise (Isomorphic coordinate spaces)

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

Solution (Isomorphic coordinate spaces)

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

Exercise (Isomorphism criteria for endomorphisms)

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

(i) $f$ is an isomorphism.

(ii) $f$ is injective.

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

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

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

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

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

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

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

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

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

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

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

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

If we now apply $f$ to this equation, we get:

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

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

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

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

Exercise (Function spaces)

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

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 $\operatorname {dim} (\operatorname {fun} (X,K))=n=\operatorname {dim} (K^{n})$ holds.

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

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

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

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

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

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

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

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

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

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

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

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

Exercises: Images

Exercise (Associating image spaces to figures)

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

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

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 (Image of a matrix)

1. 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 $\operatorname {im} (f)$?
2. 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

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

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.

Exercises: Kernel

Exercise

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

Solution

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

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

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

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

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

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

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

The kernel of f

We have already seen that

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

Now we determine the image of $f$ by applying $f$ to the canonical basis.

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

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

• 
  The image of f
• 
  Image and kernel of f together

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

Exercise

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

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

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

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

Solution

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

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

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

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

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

Exercise (Injectivity and dimension of $V$ and $W$)

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

How to get to the proof? (Injectivity and dimension of $V$ and $W$)

To prove equivalence, we need to show two implications. For the execution, we use that every monomorphism $f\colon V\to W$ preserves linear independence: If $\{b_{1},\ldots ,b_{n}\}\subseteq V$ is a basis of $V$, then the $n$ vectors $f(b_{1}),\ldots ,f(b_{n})\in W$ are linearly independent. For the converse direction, we need to construct a monomorphism from $V$ to $W$ using the assumption $\dim V\leq \dim W$. To do this, we choose bases in $V$ and $W$ and then use the principle of linear continuation to define a monomorphism by the images of the basis vectors.

Solution (Injectivity and dimension of $V$ and $W$)

Proof step: There is a monomorphism $\implies \dim(V)\leq \dim(W)$

Let $f:V\to W$ be a monomorphism and $\{v_{1},...,v_{n}\}$ a basis of $V$. Then $\{v_{1},...,v_{n}\}$ is in particular linearly independent and therefore $\{f(v_{1}),...,f(v_{n})\}$ is linearly independent. Thus, it follows that $\dim(W)\geq n=\dim(V)$. So $\dim(W)\geq \dim(V)$ is a necessary criterion for the existence of a monomorphism from $V$ to $W$.

Proof step: $\dim(V)\leq \dim(W)\implies$ there is a monomorphism

Conversely, in the case $\dim(V)\leq \dim(W)$ we can construct a monomorphism: Let $\{v_{1},\dots ,v_{n}\}$ be a basis of $V$ and $\{w_{1},\dots ,w_{m}\}$ be a basis of $W$. Then $n=\dim(V)\leq \dim(W)=m$. We define a linear map $f\colon V\to W$ by setting

$f(v_{i})=w_{i}$

for all $i=1,\ldots ,n$. According to the principle of linear continuation, such a linear map exists and is uniquely determined. We now show that $f$ is injective by proving that $\ker(f)=\{0_{V}\}$ holds. Let $x\in \ker(f)$. Because $\{v_{1},\dots ,v_{n}\}$ is a basis of $V$, there exist some $\lambda _{1},\ldots ,\lambda _{n}\in K$ with

$x=\sum _{i=1}^{n}\lambda _{i}v_{i}.$

Thus, we get

${\begin{aligned}0_{V}=f(x)&=f\left(\sum _{i=1}^{n}\lambda _{i}v_{i}\right)\\[0.3em]&\ {\color {OliveGreen}\left\downarrow \ f{\text{ is linear}}\right.}\\[0.3em]&=\sum _{i=1}^{n}\lambda _{i}f(v_{i})\\[0.3em]&\ {\color {OliveGreen}\left\downarrow \ f(v_{i})=w_{i}\right.}\\[0.3em]&=\sum _{i=1}^{n}\lambda _{i}w_{i}\\[0.3em]&\ {\color {OliveGreen}\left\downarrow \ \lambda _{i}=0{\text{ for }}i>n\right.}\\[0.3em]&=\sum _{i=1}^{m}\lambda _{i}w_{i}\end{aligned}}$

Since $\{w_{1},\dots ,w_{m}\}$ are linearly independent, $\lambda _{i}=0_{K}$ must hold for all $i=1,\ldots ,n$. So it follows for $x$ that

$x=\sum _{i=1}^{n}\lambda _{i}v_{i}=\sum _{i=1}^{n}0_{K}\cdot v_{i}=0_{V}.$

We have shown that $\ker(f)=\{0_{V}\}$ holds and thus $f$ is a monomorphism.