Proofs for linear maps – Serlo

We will give here a proof structure that shows how to prove linearity of a map.

General procedure

Recap: Definition of a linear map

We recall that a linear map (or homomorphism) is a structure-preserving map of a $K$-vector space $V$ into a $K$-vector space $W$. That is, for the map $f\colon V\to W$, the following two conditions must hold:

1. $f$ must be additive, i.e., for $v,w\in V$ we have that: $f(v+w)=f(v)+f(w)$
2. $f$ must be homogeneous, i.e., for $v\in V,\lambda \in K$ we have that: $f(\lambda \cdot v)=\lambda \cdot f(v)$.

So for a linear map it doesn't matter if we first do the addition or scalar multiplication in the vector space $V$ and then map the sum into the vector space $W$, or first map the vectors $v,\,w$ into the vector space $W$ and perform the addition or scalar multiplication there, using the images of the map.

Proving that a map is linear

The proof that a map is linear can be done according to the following structure. First, we assume that a map $f\colon V\to W$ is given between vector spaces. That is, $V$ and $W$ are $K$-vector spaces and $f$ is well-defined. Then for the linearity of $f$ we have to show:

1. additivity: $\forall v,\,w\in V:\quad f(v+w)=f(v)+f(w)$
2. homogeneity: $\forall v\in V\,\forall \lambda \in K:\quad f(\lambda \cdot v)=\lambda \cdot f(v)$

Exercise (Introductory example)

We consider the following map

$f\colon \mathbb {R} ^{2}\to \mathbb {R} ,\quad f{\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}:=2v_{1}+v_{2}$

and show that it is linear.

Proof (Introductory example)

First, $\mathbb {R} ^{2}$ and $\mathbb {R}$ are vector spaces over the field $\mathbb {R}$. Moreover, the map $f$ is well-defined.

Proof step: proving additivity

Let ${\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}},\,{\begin{pmatrix}w_{1}\\w_{2}\end{pmatrix}}\in \mathbb {R} ^{2}$.

${\begin{aligned}f\left({\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}+{\begin{pmatrix}w_{1}\\w_{2}\end{pmatrix}}\right)&=f{\begin{pmatrix}v_{1}+w_{1}\\v_{2}+w_{2}\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&=\ 2(v_{1}+w_{1})+(v_{2}+w_{2})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law}}\right.}\\[0.3em]&=\ 2v_{1}+2w_{1}+v_{2}+w_{2}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutative and associative law }}\right.}\\[0.3em]&=\ (2v_{1}+v_{2})+(2w_{1}+w_{2})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&=\ f{\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}+f{\begin{pmatrix}w_{1}\\w_{2}\end{pmatrix}}\end{aligned}}$

Thus we have proved the additivity of $f$.

Proof step: proving homogeneity

Let ${\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}\in \mathbb {R} ^{2}$ and $\lambda \in \mathbb {R}$. Then, we have

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

Thus we have proved the homogeneity of $f$.

The map to zero

The map to zero is the map which sends every vector to zero. For instance, the map to zero of $\mathbb {R}$ to $\mathbb {R} ^{3}$ looks as follows:

$f\colon \mathbb {R} \to \mathbb {R} ^{3},\quad x\mapsto {\begin{pmatrix}0\\0\\0\end{pmatrix}}$

Exercise (The map to zero linear)

Show that the map $f\colon \mathbb {R} \to \mathbb {R} ^{3},\quad x\mapsto {\begin{pmatrix}0\\0\\0\end{pmatrix}}$ is linear.

Proof (The map to zero linear)

We already know that $\mathbb {R}$ and $\mathbb {R} ^{3}$ are both $\mathbb {R}$-vector spaces and that the map to zero is well-defined.

Proof step: additivity

For all $x,y\in \mathbb {R}$ we have that

${\begin{aligned}f(x+y)&={\begin{pmatrix}0\\0\\0\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additive neutral element}}\right.}\\[0.3em]&={\begin{pmatrix}0\\0\\0\end{pmatrix}}+{\begin{pmatrix}0\\0\\0\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&=f(x)+f(y)\end{aligned}}$

Proof step: homogeneity

For all $x\in \mathbb {R} ,\lambda \in \mathbb {R}$ we have that

${\begin{aligned}f(\lambda \cdot x)&={\begin{pmatrix}0\\0\\0\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{scalar multiplication }}\right.}\\[0.3em]&=\lambda \cdot {\begin{pmatrix}0\\0\\0\end{pmatrix}}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f\right.}\\[0.3em]&=\lambda \cdot f(x)\end{aligned}}$

Thus, the map to zero is linear.

An example in $\mathbb {R} ^{2}$

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.

A linear map in the vector space of sequences

Next, we consider the space of all sequences of real numbers. This space is infinite-dimensional, because there are not finitely many sequences generating this sequence space. But it is a vector space, as we have shown in the chapter about sequence spaces.

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.

Abstract example

In this chapter, we deal with somewhat more abstract vectors. Let $M,\,N$ be arbitrary sets; $K$ a field and $V$ a $K$-vector space. We now consider the set of all maps/ functions of the set $M$ into the vector space $V$ and denote this set with ${\text{Fun}}(M,V)$. Furthermore, we also consider the set of all maps of the set $N$ into the vector space $V$ and denote this set with ${\text{Fun}}(N,V)$. The addition of two maps is defined for $f,g\in {\text{Fun}}(M,V)$ by

$(f+g)(m)=f(m)+g(m)$

Die scalar multiplication is defined for $\lambda \in K$ via

$(\lambda \cdot f)(m)=\lambda \cdot f(m)$

Analogously, we define addition scalar multiplication for ${\text{Fun}}(N,V)$.

Exercise (The set ${\text{Fun}}(M,V)$ is a $K$-vector space)

Show that ${\text{Fun}}(M,V)$ is a $K$-vector space.

How to get to the proof? (The set ${\text{Fun}}(M,V)$ is a $K$-vector space)

Simply check the vector space axioms.

We now show that the precomposition with a mapping $t\in {\text{Fun}}(N,M)$ is a linear map from ${\text{Fun}}(M,V)$ to ${\text{Fun}}(N,V)$.

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.