Vector space of a linear map – "Math for Non-Geeks"

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

The vector space of linear maps

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 ${\mathrm{Hom}}_K(V,W)$ (like "homeomorphism", which is another word for "linear map"). So:

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

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

Theorem

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

How to get to the proof?

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

1. ${\mathrm{Hom}}_K(V,W)\ne \mathrm{\varnothing }$
2. If $L_1,L_2\in {\mathrm{Hom}}_K(V,W)$ then also $L_1+L_2\in {\mathrm{Hom}}_K(V,W)$.
3. If $L\in {\mathrm{Hom}}_K(V,W)$ and $\lambda \in K$ then also $\lambda \cdot L\in {\mathrm{Hom}}_K(V,W)$.

Proof

We show that ${\mathrm{Hom}}_K(V,W)$ is a subspace of $\mathrm{Fun}(V,W)$.

Proof step: ${\mathrm{Hom}}_K(V,W)\ne \mathrm{\varnothing }$

The map $L:V\to W,v\mapsto 0_W$ is $K$-linear, so ${\mathrm{Hom}}_K(V,W)\ne \mathrm{\varnothing }$.

Proof step: If $L_1,L_2\in {\mathrm{Hom}}_K(V,W)$ then also $L_1+L_2\in {\mathrm{Hom}}_K(V,W)$.

We have to show that the map

$(L_1+L_2):V\to W$

is $K$-linear.

Proof step: Additivity of $(L_1+L_2)$

Let $v_1,v_2\in V$. Then

$\begin{array}{cc}& (L_1+L_2)(v_1+v_2)=\\ & \downarrow \text{definition of }{L}_1+L_2\\ =& L_1(v_1+v_2)+L_2(v_1+v_2)=\\ & \downarrow \text{additivity of }{L}_1\text{ and }{L}_2\\ =& L_1(v_1)+L_1(v_2)+L_2(v_1)+L_2(v_2)=\\ & \downarrow \text{commutativity of addition in }W\\ =& L_1(v_1)+L_2(v_1)+L_1(v_2)+L_2(v_2)=\\ & \downarrow \text{definition of }{L}_1+L_2\\ =& (L_1+L_2)(v_1)+(L_1+L_2)(v_2).\end{array}$

Proof step: Homogeneity of $(L_1+L_2)$

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

$\begin{array}{cc}& (L_1+L_2)(\lambda \cdot v)=\\ & \downarrow \text{definition of }{L}_1+L_2\\ =& L_1(\lambda v)+L_2(\lambda v)=\\ & \downarrow \text{homogeneity of }{L}_1\text{ and }{L}_2\\ =& \lambda \cdot L_1(v)+\lambda \cdot L_2(v)=\\ & \downarrow \text{distributivity of scalar multiplication}\\ =& \lambda \cdot (L_1(v)+L_2(v))=\\ & \downarrow \text{definition of }{L}_1+L_2\\ =& \lambda \cdot (L_1+L_2)(v).\end{array}$

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

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

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

is $K$-linear.

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

Let $v_1,v_2\in V$. Then

$\begin{array}{cc}& (\mu \cdot L)(v_1+v_2)=\\ & \downarrow \text{definition of }\mu \cdot L\\ =& \mu \cdot L(v_1+v_2)=\\ & \downarrow \text{additivity of }L\\ =& \mu (L(v_1)+L(v_2))=\\ & \downarrow \text{distributivity of scalar multiplication}\\ =& \mu \cdot L(v_1)+\mu \cdot L(v_2)=\\ & \downarrow \text{definition of }\mu \cdot L\\ =& (\mu \cdot L)(v_1)+(\mu \cdot L)(v_2).\end{array}$

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

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

$\begin{array}{cc}& (\mu \cdot L)(\lambda \cdot v)=\\ & \downarrow \text{definition of }\mu \cdot L\\ =& \mu \cdot L(\lambda v)=\\ & \downarrow \text{linearity of }L\\ =& \mu \cdot (\lambda \cdot L(v))=\\ & \downarrow \text{associativity of scalar multiplication}\\ =& (\mu \cdot \lambda )\cdot L(v)=\\ & \downarrow \text{commutativity of multiplication in }K\\ =& (\lambda \cdot \mu )\cdot L(v)=\\ & \downarrow \text{definition of }\mu \cdot L\\ =& \lambda \cdot (\mu \cdot L)(v).\end{array}$

Therefore, ${\mathrm{Hom}}_K(V,W)$ is a subspace of $\mathrm{Abb}(V,W)$.

The dimension of the vector space of linear mappings

In the following, we calculate the dimension of the vector space ${\mathrm{Hom}}_K(V,W)$ 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 $n={\dim }_K(V)<\mathrm{\infty },m={\dim }_K(W)<\mathrm{\infty }$.

Then ${\dim }_K({\mathrm{Hom}}_K(V,W))=n\cdot m$.

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

To find the dimension of ${\mathrm{Hom}}_K(V,W)$, 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 ${ℬ}_V=\{v_1,\dots ,v_n\}\subset V$ be a basis of $V$,

Let ${ℬ}_W=\{w_1,\dots ,w_m\}\subset W$ be a basis of $W$.

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

$\begin{array}{cc}{f}_{i,j}:V& \to W\\ v_k& \mapsto \{\begin{array}{cc}0& \text{if }k\ne i,\\ w_j& \text{if }k=i.\end{array}\end{array}$

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

This set has $|ℬ|=m\cdot n$ elements.

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

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

Proof step: $ℬ$ is linearly independent

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

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

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

Then:

$\begin{array}{cc}{0}_W& =0_{{\mathrm{Hom}}_K(V,W)}(v_k)\\ & \downarrow \text{definition of }{\lambda }_{i,j}\\ & =(\sum _{i=1}^n\sum _{j=1}^m{\lambda }_{i,j}\cdot f_{i,j})(v_k)\\ & \downarrow \text{definition of the sum and scalar multiplication of linear maps}\\ & =\sum _{i=1}^n\sum _{j=1}^m{\lambda }_{i,j}\cdot f_{i,j}(v_k)\\ & \downarrow \text{splitting the sum}\\ & =\sum _{j=1}^m{\lambda }_{k,j}\cdot \underset{=w_j}{\underbrace{f_{k,j}(v_k)}}+\sum _{i\ne k}\sum _{j=1}^m{\lambda }_{i,j}\cdot \underset{=0_W}{\underbrace{f_{i,j}(v_k)}}\\ & \downarrow \text{definition of }{f}_{i,j}\\ & =\sum _{j=1}^m{\lambda }_{k,j}\cdot w_j.\end{array}$

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: $ℬ$ is a generating system

Let $g\in {\mathrm{Hom}}_K(V,W)$.

For every $i\in \{1,\dots ,n\}$ we have $f(v_i)\in W$. Since ${ℬ}_W$ 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 ${ℬ}_V$. Let $k\in \{1,\dots ,n\}$ be arbitrary. Then,

$\begin{array}{cc}g(v_k)& =\\ & \downarrow \text{definition of }{\alpha }_{k,j}\\ & =\sum _{j=1}^m{\alpha }_{k,j}\cdot w_j\\ & \downarrow \text{definition of }{f}_{k,j}\\ & =\sum _{j=1}^m{\alpha }_{k,j}\cdot f_{k,j}(v_k)\\ & \downarrow f_{i,j}(v_k)=0_W\text{ for all }i\ne k\\ & =\sum _{j=1}^m{\alpha }_{k,j}\cdot f_{k,j}(v_k)+\underset{=0_W}{\underbrace{\sum _{i\ne k}\sum _{j=1}^m{\alpha }_{i,j}\cdot \underset{=0_W}{\underbrace{f_{i,j}(v_k)}}}}\\ & \downarrow \text{concluding the sums}\\ & =\sum _{i=1}^n\sum _{j=1}^m{\alpha }_{i,j}\cdot f_{i,j}(v_k)\\ & \downarrow \text{definition of the sum and scalar multiplication of linear mappings}\\ & =(\sum _{i=1}^n\sum _{j=1}^m{\alpha }_{i,j}\cdot f_{i,j})(v_k).\end{array}$

Therefore, $ℬ$ is a generating system.

This proves the statement of the theorem.

Hint

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

If ${\dim }_K(V)=\mathrm{\infty }$ or ${\dim }_K(W)=\mathrm{\infty }$, and $V\ne 0\ne W$, then we also have ${\dim }_K({\mathrm{Hom}}_K(V,W))=\mathrm{\infty }$.

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