Vector space structure on matrices – Serlo

Derivation

Let $m,n\in \mathbb {N}$ and let $V$ be an $n$-dimensional and $W$ an $m$-dimensional $K$-vector space. We have already seen that, after choosing ordered bases, we can represent linear maps from $V$ to $W$ as matrices. So let $B$ be an ordered basis of $V$ and $C$ be an ordered basis of $W$.

The space $\operatorname {Hom} _{K}(V,W)$ of linear maps from $V$ to $W$ is also a $K$-vector space. The representing matrix of a linear map $f\in \operatorname {Hom} _{K}(V,W)$ with respect to the bases $B$ and $C$ is an $(m\times n)$-matrix $M_{C}^{B}(f)\in K^{m\times n}$. We will try now transfer the vector space structure of $\operatorname {Hom} _{K}(V,W)$ to the space $K^{m\times n}$ of $(m\times n)$-matrices over $K$.

So we ask the question: Can we find addition and scalar multiplication on $K^{m\times n}$, such that $M_{C}^{B}(f+g)=M_{C}^{B}(f)+M_{C}^{B}(g)$ and $M_{C}^{B}(\lambda f)=\lambda M_{C}^{B}(f)$ for all linear maps $f,g\colon V\to W$ and all $\lambda \in K$?

On $K^{m\times n}$, is there perhaps even a vector space structure, such that for all finite dimensional vector spaces $V$ and $W$ and all ordered bases $B$ of $V$ and $C$ of $W$, the mapping $\operatorname {Hom} _{K}(V,W)\to K^{m\times n};f\mapsto M_{C}^{B}(f)$ is linear?

It is best to think about these questions yourself. There is an exercise for matrix addition and one for scalar multiplication that can help you with this.

A first step is to answer this question is the following theorem:

Theorem (Bijective maps induce vector space structures)

Let $V$ be a vector space with addition $+$ and scalar multiplication $\cdot$ and $W$ be a set. Let $f:V\to W$ be a bijective mapping. Then there exists exactly one vector space structure $(\oplus ,\odot )$, on $W$, such that $f$ is linear.

Proof (Bijective maps induce vector space structures)

Proof step: Existence

For $w,w'\in W$ and $\lambda \in K$ we define $w\oplus w':=f(f^{-1}(w)+f^{-1}(w'))$, $\lambda \odot w:=f(\lambda \cdot f^{-1}(w))$.

$W$ is closed under these operations, since $f$ always returns us to $W$. That $W$ forms a vector space with these operations follows directly from the vector space structure of $V$. One can view $f$ simply as a renaming of the elements of $V$.

For example, commutativity of the addition on $W$ follows from commutativity of the addition on $V$ as follows: $w\oplus w'=f(f^{-1}(w)+f^{-1}(w'))=f(f^{-1}(w')+f^{-1}(w))=w'\oplus w$.

Associativity of the addition on $W$ also follows from associativity of the addition on $V$:

${\begin{aligned}&w_{1}\oplus (w_{2}\oplus w_{3})\\&\ {\color {OliveGreen}\downarrow {\text{definition of addition on }}{W}}\\&=f(f^{-1}(w_{1})+f^{-1}(w_{2}\oplus w_{3}))\\&\ {\color {OliveGreen}\downarrow {\text{definition of addition on }}{W}}\\&=f(f^{-1}(w_{1})+f^{-1}(f(f^{-1}(w_{2})+f^{-1}(w_{3}))))\\&\ {\color {OliveGreen}\downarrow {f^{-1}\circ f}{\text{ is the identity}}}\\&=f(f^{-1}(w_{1})+(f^{-1}(w_{2})+f^{-1}(w_{3})))\\&\ {\color {OliveGreen}\downarrow {\text{associativity of the addition on }}{V}}\\&=f((f^{-1}(w_{1})+f^{-1}(w_{2}))+f^{-1}(w_{3}))\\&\ {\color {OliveGreen}\downarrow {f^{-1}\circ f}{\text{ is the identity}}}\\&=f(f^{-1}(f((f^{-1}(w_{1})+f^{-1}(w_{2}))+f^{-1}(w_{3}))))\\&\ {\color {OliveGreen}\downarrow {\text{definition of addition on }}{W}}\\&=f(f^{-1}(w_{1}\oplus w_{2})+f^{-1}(w_{3}))\\&\ {\color {OliveGreen}\downarrow {\text{definition of addition on }}{W}}\\&=(w_{1}\oplus w_{2})\oplus w_{3}.\end{aligned}}$

The establishment of the other vector space axioms work analogously. Thus we have found a vector space structure on $W$. Let us now show that $f$ is linear with respect to $(\oplus ,\odot )$. Since $f$ is bijective, it suffices to show that the inverse map with respect to $f$ is linear (see isomorphism ). We have $f^{-1}(w\oplus w')=f^{-1}(f(f^{-1}(w)+f^{-1}(w')))=f^{-1}(w)+f^{-1}(w')$ and $f^{-1}(\lambda \odot w)=f^{1}(f(\lambda \cdot f^{-1}(w')))=\lambda \cdot f^{-1}(w')$. Thus $f^{-1}:(W,\oplus ,\odot )\to (V,+,\cdot )$ is linear and hence $f:(V,+,\cdot )\to (W,\oplus ,\odot )$ is also linear.

Proof step: Uniqueness

Uniqueness: Suppose we have a vector space structure $(\oplus ,\odot )$ such that $f:(V,+,\cdot )\to (W,\oplus ,\odot )$ is linear. Then $f^{-1}:(W,\oplus ,\odot )\to (V,+,\cdot )$ is the inverse function of a bijective linear function and hence also linear. Therefore we have that

$w\oplus w'=f(f^{-1}(w\oplus w'))=f(f^{-1}(w)+f^{-1}(w'))$,

$\lambda \odot w=f(f^{-1}(\lambda \odot w))=f(\lambda \cdot f^{-1}(w))$.

That is, any vector space structure on $W$ with respect to which $f$ is linear must be our previously defined vector space structure.

We would now like to explicitly determine the vector space structure of $K^{m\times n}$. Let $B=\{v_{1},\dots ,v_{n}\}$ be a basis of $V$, and $C=\{w_{1},\dots ,w_{m}\}$ a basis of $W$. We define the addition induced by $M_{C}^{B}$ on the space of matrices as in the last theorem: $A+A'=M_{C}^{B}((M_{C}^{B})^{-1}(A)+(M_{C}^{B})^{-1}(A'))$. Now let $A=(a_{ij}),A'=(a'_{ij})\in K^{m\times n}$ be arbitrary and $f,g:V\to W$ be the linear maps associated with $A$ and $A'$ with $M_{C}^{B}(f)=A,M_{C}^{B}(g)=A'$. Then

$(b_{ij})=A+A'=M_{C}^{B}((M_{C}^{B})^{-1}(A)+(M_{C}^{B})^{-1}(A'))=M_{C}^{B}(f+g).$

We now calculate this $b_{ij}$: In the $j$-th column, $(f+g)(v_{j})=\sum _{i=1}^{m}b_{ij}w_{i}$ must hold. However, by definition of $f+g$,

$(f+g)(v_{j})=f(v_{j})+g(v_{j})=\sum _{i=1}^{m}a_{ij}w_{i}+\sum _{i=1}^{m}a'_{ij}w_{i}=\sum _{i=1}^{m}(a_{ij}+a'_{ij})w_{j}.$

Since the representation of $f(v_{j})$ is unique with respect to $C$, it follows that $b_{ij}=a_{ij}+a'_{ij}$. That is, the addition induced by $M_{C}^{B}$ on $K^{m\times n}$ is a component-wise addition.

Let us now examine the scalar multiplication $\lambda \cdot A=M_{C}^{B}(\lambda (M_{C}^{B})^{-1}(A))$ induced by $M_{C}^{B}$. Let again $f=(M_{C}^{B})^{-1}(A)$ and consider $(a'_{ij})=A'=\lambda \cdot A$. We have that

$(\lambda \cdot f)(v_{j})=\sum _{i=1}^{m}a'_{ij}w_{i}.$

Furthermore we have

$\lambda f(v_{j})=\lambda \sum _{i=1}^{m}a_{ij}w_{i}=\sum _{i=1}^{m}\lambda a_{ij}w_{i}.$

Since $(\lambda \cdot f)(v_{j})=\lambda f(v_{j})$ we obtain

$\sum _{i=1}^{m}a'_{ij}w_{i}=\sum _{i=1}^{m}\lambda a_{ij}w_{i}.$

Thus, from the uniqueness of the representation it follows that $a'_{ij}=\lambda a_{ij}$. We see, the scalar multiplication induced from $\operatorname {Hom} _{K}(V,W)$ by $M_{C}^{B}$ on $K^{m\times n}$ is the component-wise scalar multiplication.

We also see here that the induced vector space structure is independent of our choice of $V,W,B$ and $C$.

Definition

We have just seen: To define a meaningful vector space structure on the matrices, we need to perform the operations component-wise. So we define addition and scalar multiplication as follows:

Definition (Addition of matrices)

Let $K$ be a field and let $A=(a_{ij})_{i=1,...,m;\,j=1,...,n}$ and $B=(b_{ij})_{i=1,...,m;\,j=1,...,n}$ be matrices of the same type $(m\times n)$ over $K$. Then

$A+B:=(a_{ij}+b_{ij})_{i=1,...,m;\,j=1,...,n}$

Written out explicitly in terms of matrices, this definition looks as follows:

$A+B={\begin{pmatrix}a_{11}&\ldots &a_{1n}\\\vdots &\ddots &\vdots \\a_{m1}&\ldots &a_{mn}\end{pmatrix}}+{\begin{pmatrix}b_{11}&\ldots &b_{1n}\\\vdots &\ddots &\vdots \\b_{m1}&\ldots &b_{mn}\end{pmatrix}}={\begin{pmatrix}a_{11}+b_{11}&\ldots &a_{1n}+b_{1n}\\\vdots &\ddots &\vdots \\a_{m1}+b_{m1}&\ldots &a_{mn}+b_{mn}\end{pmatrix}}$

Definition (Scalar multiplication of matrices)

Let $K$be a field and let $A=(a_{ij})$ be a matrix over $K$. Then, for $\lambda \in K$ we have

$\lambda \cdot A=\lambda \cdot (a_{ij}):=(\lambda a_{ij})$

Written out explicitly in terms of matrices, this definition looks as follows:

$\lambda \cdot A=\lambda \cdot {\begin{pmatrix}a_{11}&\ldots &a_{1n}\\\vdots &\ddots &\vdots \\a_{m1}&\ldots &a_{mn}\end{pmatrix}}={\begin{pmatrix}\lambda a_{11}&\ldots &\lambda a_{1n}\\\vdots &\ddots &\vdots \\\lambda a_{m1}&\ldots &\lambda a_{mn}\end{pmatrix}}$

Example (Addition of matrices)

We are in $\mathbb {R} ^{2\times 3}$.

${\begin{pmatrix}2&4&6\\1&3&5\end{pmatrix}}+{\begin{pmatrix}-1&2&-2\\3&2&1\end{pmatrix}}={\begin{pmatrix}2+(-1)&4+2&6+(-2)\\1+3&3+2&5+1\end{pmatrix}}={\begin{pmatrix}1&6&4\\4&5&6\end{pmatrix}}$

Example (Multiplication by a field element)

As an example we take the matrix ${\begin{pmatrix}1&-2&0\\2&-3&-1\end{pmatrix}}\in \mathbb {R} ^{2\times 3}$ and as field element the real number $(-3)$. Then

$(-3)\cdot {\begin{pmatrix}1&-2&0\\2&-3&-1\end{pmatrix}}={\begin{pmatrix}(-3)\cdot 1&(-3)\cdot (-2)&(-3)\cdot 0\\(-3)\cdot 2&(-3)\cdot (-3)&(-3)\cdot (-1)\end{pmatrix}}={\begin{pmatrix}-3&6&0\\-6&9&3\end{pmatrix}}$

Theorem (Matrices form a vector space)

The set of $(m\times n)$-matrices $K^{m\times n}$ forms a $K$-vector space with the addition and scalar multiplication defined above. The neutral element of addition of this vector space is the zero matrix $(0)\in K^{m\times n}$ and the additive inverse of a matrix $A=(a_{ij})\in K^{m\times n}$ is $-A:=(-a_{ij})$.

Proof (Matrices form a vector space)

Proof step: Component-wise addition and scalar multiplication form a vector space structure on $K^{m\times n}$

Let $B$ be a basis of $K^{n}$ and $C$ a basis of $K^{m}$. For example, we can choose the standard bases. Using the above theorem, we see that the bijective mapping $M_{C}^{B}:\operatorname {Hom} _{K}(K^{n},K^{m})\to K^{m\times n}$ induces a vector space structure on the space of matrices. We have already considered at the end of the derivation that in this vector space structure is given by component-wise addition and scalar multiplication. So, component-wise addition and scalar multiplication generate a vector space structure on $K^{m\times n}$.

Proof step: $(0)\in K^{m\times n}$ is the neutral element of the addition

We need to show that $A+(0)=A$ holds for any matrix $A\in K^{m\times n}$. So let $A=(a_{ij})\in K^{m\times n}$ be arbitrary. By definition of addition of matrices, $(a_{ij})+(0)=(a_{ij}+0)=(a_{ij})$ holds, where we have exploited that $0\in K$ is the neutral element of addition in $K$.

Proof step: Every matrix $A=(a_{ij})\in K^{m\times n}$ has additive inverse $-A:=(-a_{ij})$

We have to show that $A+(-A)=(0)$ holds for any matrix $A\in K^{m\times n}$. So let $A=(a_{ij})\in K^{m\times n}$ be arbitrary. Then by definition of $-A=(-a_{ij})$ and by the definition of addition of matrices, we have $A+(-A)=(a_{ij})+(-a_{ij})=(a_{ij}-a_{ij})=(0)$. In the last equality we used that $-a_{ij}$ is the additive inverse of $a_{ij}$ in $K$.

If we consider matrices just as tables of numbers (without considering them as mapping matrices), we see the following: Matrices are nothing more than a special way of writing elements of $K^{m\cdot n}$, since matrices have $m\cdot n$ entries. Just as in $K^{m\cdot n}$, the vector space structure for matrices is defined component-wise. So we get alternatively the following significantly shorter proof:

Alternative proof (Matrices form a vector space)

We can simply use the proof that coordinate spaces are vector spaces, since $K^{m\times n}$ is a certain coordinate space. As an example, we show the associativity of addition: If $A=(a_{ij}),B=(b_{ij}),C=(c_{ij})$ are three $(n\times m)$-matrices, then $(A+B)+C=((a_{ij}+b_{ij})+c_{ij})=(a_{ij}+(b_{ij}+c_{ij}))=A+(B+C)$. In the second step we used associativity in the field $K$.

Dimension of $K^{m\times n}$

By the above identification of $K^{m\times n}$ with $K^{m\cdot n}$ we obtain a canonical basis of $K^{m\times n}$: Let $B_{ij}$ be for $i\in \{1,...,m\},j\in \{1,...,n\}$ the matrix $B_{ij}=(b_{kl})$ with

$b_{kl}={\begin{cases}1&{\text{if }}k=i{\text{ and }}l=j\\0&{\text{else.}}\end{cases}}$

Example

In $K^{2\times 3}$, the basis elements are given by

${\begin{aligned}B_{11}={\begin{pmatrix}1&0&0\\0&0&0\end{pmatrix}},\quad B_{12}={\begin{pmatrix}0&1&0\\0&0&0\end{pmatrix}},\quad B_{13}={\begin{pmatrix}0&0&1\\0&0&0\end{pmatrix}},\\B_{21}={\begin{pmatrix}0&0&0\\1&0&0\end{pmatrix}},\quad B_{22}={\begin{pmatrix}0&0&0\\0&1&0\end{pmatrix}},\quad B_{23}={\begin{pmatrix}0&0&0\\0&0&1\end{pmatrix}}\end{aligned}}$

Thus, $K^{m\times n}$ is a $(m\cdot n)$-dimensional $K$-vector space. We constructed the vector space structure on $K^{m\cdot n}$ such that for $n$- and $m$-dimensional vector spaces $V$ and $W$ with bases $B$ and $C$, respectively, we have that the map

$M_{C}^{B}:\operatorname {Hom} _{K}(V,W)\to K^{m\times n},\quad f\mapsto M_{C}^{B}(f)$

is a linear isomorphism. So $\operatorname {Hom} _{K}(V,W)$ is a $(m\cdot n)$-dimensional $K$-vector space. This result can also be found in the article vector space of a linear map.