# Vector space structure on matrices – Serlo

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## Derivation

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

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

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

On ${\displaystyle K^{m\times n}}$, is there perhaps even a vector space structure, such that for all finite dimensional vector spaces ${\displaystyle V}$ and ${\displaystyle W}$ and all ordered bases ${\displaystyle B}$ of ${\displaystyle V}$ and ${\displaystyle C}$ of ${\displaystyle W}$, the mapping ${\displaystyle \operatorname {Hom} _{K}(V,W)\to K^{m\times n};f\mapsto M_{C}^{B}(f)}$ 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 ${\displaystyle V}$ be a vector space with addition ${\displaystyle +}$ and scalar multiplication ${\displaystyle \cdot }$ and ${\displaystyle W}$ be a set. Let ${\displaystyle f:V\to W}$ be a bijective mapping. Then there exists exactly one vector space structure ${\displaystyle (\oplus ,\odot )}$, on ${\displaystyle W}$, such that ${\displaystyle f}$ is linear.

Proof (Bijective maps induce vector space structures)

Proof step: Existence

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

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

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

Associativity of the addition on ${\displaystyle W}$ also follows from associativity of the addition on ${\displaystyle V}$:

{\displaystyle {\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 ${\displaystyle W}$. Let us now show that ${\displaystyle f}$ is linear with respect to ${\displaystyle (\oplus ,\odot )}$. Since ${\displaystyle f}$ is bijective, it suffices to show that the inverse map with respect to ${\displaystyle f}$ is linear (see isomorphism ). We have ${\displaystyle f^{-1}(w\oplus w')=f^{-1}(f(f^{-1}(w)+f^{-1}(w')))=f^{-1}(w)+f^{-1}(w')}$ and ${\displaystyle f^{-1}(\lambda \odot w)=f^{1}(f(\lambda \cdot f^{-1}(w')))=\lambda \cdot f^{-1}(w')}$. Thus ${\displaystyle f^{-1}:(W,\oplus ,\odot )\to (V,+,\cdot )}$ is linear and hence ${\displaystyle f:(V,+,\cdot )\to (W,\oplus ,\odot )}$ is also linear.

Proof step: Uniqueness

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

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

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

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

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

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

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

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

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

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

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

Furthermore we have

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

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

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

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

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

## 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:

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

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

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

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

Definition (Scalar multiplication of matrices)

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

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

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

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

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

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

Example (Multiplication by a field element)

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

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

Theorem (Matrices form a vector space)

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

Proof (Matrices form a vector space)

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

Let ${\displaystyle B}$ be a basis of ${\displaystyle K^{n}}$ and ${\displaystyle C}$ a basis of ${\displaystyle K^{m}}$. For example, we can choose the standard bases. Using the above theorem, we see that the bijective mapping ${\displaystyle M_{C}^{B}:\operatorname {Hom} _{K}(K^{n},K^{m})\to K^{m\times n}}$ 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 ${\displaystyle K^{m\times n}}$.

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

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

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

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

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 ${\displaystyle K^{m\cdot n}}$, since matrices have ${\displaystyle m\cdot n}$ entries. Just as in ${\displaystyle K^{m\cdot n}}$, 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 ${\displaystyle K^{m\times n}}$ is a certain coordinate space. As an example, we show the associativity of addition: If ${\displaystyle A=(a_{ij}),B=(b_{ij}),C=(c_{ij})}$ are three ${\displaystyle (n\times m)}$-matrices, then ${\displaystyle (A+B)+C=((a_{ij}+b_{ij})+c_{ij})=(a_{ij}+(b_{ij}+c_{ij}))=A+(B+C)}$. In the second step we used associativity in the field ${\displaystyle K}$.

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

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

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

Example

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

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

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

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