Vector space of a linear map – Serlo

Here, we consider the vector space of linear mappings between two given vector spaces.

The vector space of linear maps

Forming the vector space $\operatorname {Hom} _{K}(V,W)\quad$

Let $K$ be a field and $V,W$ two $K$-vector spaces. We want to investigate the set of all $K$-linear mappings from $V$ to $W$. We give this set $\operatorname {Hom} _{K}(V,W)$ (like "homeomorphism", which is another word for "linear map"). So:

$\operatorname {Hom} _{K}(V,W):=\{\,L:V\to W\,|\,L{\text{ is }}K{\text{-linear}}\,\}.$

We will now prove that this set is again a $K$-vector space, so it has some very convenient properties. More precisely, $\operatorname {Hom} _{K}(V,W)$ is a subspace of the set of functions from $V$ to $W$, called $\operatorname {Fun} (V,W)$.

Theorem

Let $K$ be a field and $V$ and $W$ two $K$-vector spaces. Then the set of linear maps $\operatorname {Hom} _{K}(V,W)$ from $V$ to $W$ is again a $K$-vector space.

How to get to the proof?

It is sufficient to show that $\operatorname {Hom} _{K}(V,W)$ is a $K$-subspace of $\operatorname {Fun} (V,W)$. For this we have to show:

1. $\operatorname {Hom} _{K}(V,W)\neq \emptyset$
2. If $L_{1},L_{2}\in \operatorname {Hom} _{K}(V,W)$ then also $L_{1}+L_{2}\in \operatorname {Hom} _{K}(V,W)$.
3. If $L\in \operatorname {Hom} _{K}(V,W)$ and $\lambda \in K$ then also $\lambda \cdot L\in \operatorname {Hom} _{K}(V,W)$.

Proof

We show that $\operatorname {Hom} _{K}(V,W)$ is a subspace of $\operatorname {Fun} (V,W)$.

Proof step: $\operatorname {Hom} _{K}(V,W)\neq \emptyset$

The map $L\colon V\to W,\quad v\mapsto 0_{W}$ is $K$-linear, so $\operatorname {Hom} _{K}(V,W)\neq \emptyset$.

Proof step: If $L_{1},L_{2}\in \operatorname {Hom} _{K}(V,W)$ then also $L_{1}+L_{2}\in \operatorname {Hom} _{K}(V,W)$.

We have to show that the map

$(L_{1}+L_{2})\colon V\to W$

is $K$-linear.

Proof step: Additivity of $(L_{1}+L_{2})$

Let $v_{1},v_{2}\in V$. Then

${\begin{aligned}&(L_{1}+L_{2})(v_{1}+v_{2})=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}L_{1}+L_{2}\right.}\\[0.3em]=\ &L_{1}(v_{1}+v_{2})+L_{2}(v_{1}+v_{2})=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additivity of }}L_{1}{\text{ and }}L_{2}\right.}\\[0.3em]=\ &L_{1}(v_{1})+L_{1}(v_{2})+L_{2}(v_{1})+L_{2}(v_{2})=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutativity of addition in }}W\right.}\\[0.3em]=\ &L_{1}(v_{1})+L_{2}(v_{1})+L_{1}(v_{2})+L_{2}(v_{2})=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}L_{1}+L_{2}\right.}\\[0.3em]=\ &(L_{1}+L_{2})(v_{1})+(L_{1}+L_{2})(v_{2}).\end{aligned}}$

Proof step: Homogeneity of $(L_{1}+L_{2})$

Let $v\in V$ and $\lambda \in K$. Then

${\begin{aligned}&(L_{1}+L_{2})(\lambda \cdot v)=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}L_{1}+L_{2}\right.}\\[0.3em]=\ &L_{1}(\lambda v)+L_{2}(\lambda v)=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{homogeneity of }}L_{1}{\text{ and }}L_{2}\right.}\\[0.3em]=\ &\lambda \cdot L_{1}(v)+\lambda \cdot L_{2}(v)=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributivity of scalar multiplication}}\right.}\\[0.3em]=\ &\lambda \cdot (L_{1}(v)+L_{2}(v))=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}L_{1}+L_{2}\right.}\\[0.3em]=\ &\lambda \cdot (L_{1}+L_{2})(v).\end{aligned}}$

Proof step: If $L\in \operatorname {Hom} _{K}(V,W)$ and $\mu \in K$ then $\mu \cdot L\in \operatorname {Hom} _{K}(V,W)$.

Let $L:V\to W$ be a linear map and $\mu \in K$. We must show that the map

$\mu \cdot L\colon V\to W,\,v\mapsto \mu \cdot L(v)$

is $K$-linear.

Proof step: Additivity of $\mu \cdot L$

Let $v_{1},v_{2}\in V$. Then

${\begin{aligned}&(\mu \cdot L)(v_{1}+v_{2})=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\mu \cdot L\right.}\\[0.3em]=\ &\mu \cdot L(v_{1}+v_{2})=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additivity of }}L\right.}\\[0.3em]=\ &\mu (L(v_{1})+L(v_{2}))=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributivity of scalar multiplication}}\right.}\\[0.3em]=\ &\mu \cdot L(v_{1})+\mu \cdot L(v_{2})=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\mu \cdot L\right.}\\[0.3em]=\ &(\mu \cdot L)(v_{1})+(\mu \cdot L)(v_{2}).\end{aligned}}$

Proof step: Homogeneity of $\mu \cdot L$

Let $v\in V$ and $\lambda \in K$. Then

${\begin{aligned}&(\mu \cdot L)(\lambda \cdot v)=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\mu \cdot L\right.}\\[0.3em]=\ &\mu \cdot L(\lambda v)=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{linearity of }}L\right.}\\[0.3em]=\ &\mu \cdot (\lambda \cdot L(v))=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of scalar multiplication}}\right.}\\[0.3em]=\ &(\mu \cdot \lambda )\cdot L(v)=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutativity of multiplication in }}K\right.}\\[0.3em]=\ &(\lambda \cdot \mu )\cdot L(v)=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\mu \cdot L\right.}\\[0.3em]=\ &\lambda \cdot (\mu \cdot L)(v).\end{aligned}}$

Therefore, $\operatorname {Hom} _{K}(V,W)$ is a subspace of $\operatorname {Abb} (V,W)$.

The dimension of the vector space of linear mappings

In the following, we calculate the dimension of the vector space $\operatorname {Hom} _{K}(V,W)$ of linear maps between two finite-dimensional vector spaces.

Theorem (Dimension of the vector space of linear maps)

Let $K$ be a field and $V,W$ two $K$ vector spaces with $n=\dim _{K}(V)<\infty ,m=\dim _{K}(W)<\infty$.

Then $\dim _{K}(\operatorname {Hom} _{K}(V,W))=n\cdot m$.

How to get to the proof? (Dimension of the vector space of linear maps)

To find the dimension of $\operatorname {Hom} _{K}(V,W)$, we have to construct a basis of this space.

If we want to define a map from $V$ to $W$, we have to define its image in $W$ for $n$ different basis vectors of $V$. This can again be represented as a linear combination of the $m$ different basis vectors of $W$.

Proof (Dimension of the vector space of linear maps)

At first, we chose two bases:

Let ${\mathcal {B}}_{V}=\{\,v_{1},\dots ,v_{n}\,\}\subset V$ be a basis of $V$,

Let ${\mathcal {B}}_{W}=\{\,w_{1},\dots ,w_{m}\,\}\subset W$ be a basis of $W$.

For each $i\in \{\,1,\dots ,n\,\},j\in \{\,1,\dots ,m\,\}$ we want to define a linear map $f_{i,j}\colon V\to W$. By the principle of linear continuation, we can define these uniquely by their values on the basis ${\mathcal {B}}_{V}$:

${\begin{aligned}f_{i,j}\colon V&\to W\\v_{k}&\mapsto {\begin{cases}0&{\text{if }}k\neq i,\\w_{j}&{\text{if }}k=i.\end{cases}}\end{aligned}}$

Let ${\mathcal {B}}:=\{\,f_{i,j}\,|\,i\in \{\,1,\dots ,n\,\},j\in \{\,1,\dots ,m\,\}\,\}\subset \operatorname {Hom} _{K}(V,W)$.

This set has $|{\mathcal {B}}|=m\cdot n$ elements.

To prove the statement of the theorem, we must therefore justify that ${\mathcal {B}}$ is a basis of $\operatorname {Hom} _{K}(V,W)$.

To do sop, we have to show that ${\mathcal {B}}$ is linearly independent and a generating system.

Proof step: ${\mathcal {B}}$ is linearly independent

Let $\lambda _{i,j}\in K$, such that $\sum _{i=1}^{n}\sum _{j=1}^{m}\lambda _{i,j}\cdot f_{i,j}=0_{\operatorname {Hom} _{K}(V,W)}$.

We must now show that $\lambda _{i,j}=0_{K}$ for all $i,j$.

Let $k\in \{\,1,\dots ,n\,\}$.

Then:

${\begin{aligned}0_{W}&=0_{\operatorname {Hom} _{K}(V,W)}(v_{k})\\&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\lambda _{i,j}\right.}\\[0.3em]&=\left(\sum _{i=1}^{n}\sum _{j=1}^{m}\lambda _{i,j}\cdot f_{i,j}\right)(v_{k})\\&{\color {OliveGreen}\left\downarrow \ {\text{definition of the sum and scalar multiplication of linear maps}}\right.}\\[0.3em]&=\sum _{i=1}^{n}\sum _{j=1}^{m}\lambda _{i,j}\cdot f_{i,j}(v_{k})\\&{\color {OliveGreen}\left\downarrow \ {\text{splitting the sum}}\right.}\\[0.3em]&=\sum _{j=1}^{m}\lambda _{k,j}\cdot \underbrace {f_{k,j}(v_{k})} _{=w_{j}}+\sum _{i\neq k}\sum _{j=1}^{m}\lambda _{i,j}\cdot \underbrace {f_{i,j}(v_{k})} _{=0_{W}}\\&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f_{i,j}\right.}\\[0.3em]&=\sum _{j=1}^{m}\lambda _{k,j}\cdot w_{j}.\end{aligned}}$

As the $w_{j}$ form a basis of $W$, they are linearly independent. Therefore, $\lambda _{k,j}=0_{K}$ already follows for all $j$ and our fixed $k$.

Since this $k$ was chosen arbitrarily, $\lambda _{k,j}=0_{K}$ now follows for all $k,j$.

The $f_{i,j}$ are therefore linearly independent.

Proof step: ${\mathcal {B}}$ is a generating system

Let $g\in \operatorname {Hom} _{K}(V,W)$.

For every $i\in \{\,1,\dots ,n\,\}$ we have $f(v_{i})\in W$. Since ${\mathcal {B}}_{W}$ is a basis of $W$, there is a unique decomposition $g(v_{i})=\sum _{j=1}^{m}\alpha _{i,j}\cdot w_{j}$ such that $\alpha _{i,j}\in K$.

We will now show that

$g=\sum _{i=1}^{n}\sum _{j=1}^{m}\alpha _{i,j}\cdot f_{i,j}.$

Due to the linearity, this can be verified on the basis vectors ${\mathcal {B}}_{V}$. Let $k\in \{\,1,\dots ,n\,\}$ be arbitrary. Then,

${\begin{aligned}g(v_{k})&=\\&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\alpha _{k,j}\right.}\\[0.3em]&=\sum _{j=1}^{m}\alpha _{k,j}\cdot w_{j}\\&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f_{k,j}\right.}\\[0.3em]&=\sum _{j=1}^{m}\alpha _{k,j}\cdot f_{k,j}(v_{k})\\&{\color {OliveGreen}\left\downarrow \ f_{i,j}(v_{k})=0_{W}{\text{ für alle }}i\neq k\right.}\\[0.3em]&=\sum _{j=1}^{m}\alpha _{k,j}\cdot f_{k,j}(v_{k})+\underbrace {\sum _{i\neq k}\sum _{j=1}^{m}\alpha _{i,j}\cdot \underbrace {f_{i,j}(v_{k})} _{=0_{W}}} _{=0_{W}}\\&{\color {OliveGreen}\left\downarrow \ {\text{concluding the sums}}\right.}\\[0.3em]&=\sum _{i=1}^{n}\sum _{j=1}^{m}\alpha _{i,j}\cdot f_{i,j}(v_{k})\\&{\color {OliveGreen}\left\downarrow \ {\text{definition of the sum and scalar multiplication of linear mappings}}\right.}\\[0.3em]&=\left(\sum _{i=1}^{n}\sum _{j=1}^{m}\alpha _{i,j}\cdot f_{i,j}\right)(v_{k}).\end{aligned}}$

Therefore, ${\mathcal {B}}$ is a generatin system.

This proves the statement of the theorem.

Hint

A similar statement also applies to infinite-dimensional vector spaces:

If $\dim _{K}(V)=\infty$ or $\dim _{K}(W)=\infty$, and $V\neq 0\neq W$, then we also have $\dim _{K}(\operatorname {Hom} _{K}(V,W))=\infty$.

However, if $V=0$ and $W$ is arbitrary (or vice versa), then we always get $\operatorname {Hom} _{K}(V,W)=0$.