Serlo: EN: Vector space structure on matrices

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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 ${\mathrm{Hom}}_K(V,W)$ of linear maps from $V$ to $W$ is also a $K$-vector space. The representing matrix of a linear map $f\in {\mathrm{Hom}}_K(V,W)$ 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 ${\mathrm{Hom}}_K(V,W)$ 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 $f,g:V\to W$ 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 ${\mathrm{Hom}}_K(V,W)\to K^{m\times n};f\mapsto M_C^B(f)$ 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: Mathe für Nicht-Freaks: Vorlage:Satz

Vorlage:Anker 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: $A+A^{\prime }=M_C^B((M_C^B{)}^{-1}(A)+(M_C^B{)}^{-1}(A^{\prime }))$. Now let $A=(a_{ij}),A^{\prime }=(a{\text{'}}_{ij})\in K^{m\times n}$ be arbitrary and $f,g:V\to W$ be the linear maps associated with $A$ and $A^{\prime }$ with $M_C^B(f)=A,M_C^B(g)=A^{\prime }$. Then Vorlage:Einrücken

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 $f+g$, Vorlage:Einrücken Since the representation of $f(v_j)$ is unique with respect to $C$, it follows that $b_{ij}=a_{ij}+a{\text{'}}_{ij}$. 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 $(a{\text{'}}_{ij})=A^{\prime }=\lambda \cdot A$. We have that Vorlage:Einrücken Furthermore we have Vorlage:Einrücken Since $(\lambda \cdot f)(v_j)=\lambda f(v_j)$ we obtain Vorlage:Einrücken Thus, from the uniqueness of the representation it follows that $a{\text{'}}_{ij}=\lambda a_{ij}$. We see, the scalar multiplication induced from ${\mathrm{Hom}}_K(V,W)$ 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:

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

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

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:

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

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

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