Proof of Linear Maps Form a Vector Space and a Monoid under Composition

Routine facts about the sum, the scalar multiple and the composite of linear maps as operations, and about the values of the inverse of a bijection, are used without comment (Toolkit for AL2).

Two functions with the same domain and codomain and the same values are equal (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (i)), so every identity between linear maps below is checked on values. We first show that sums and scalar multiples of linear maps are linear. Let $S,T\in\mathcal{L}(V,W)$, $\lambda,\mu\in K$ and $u,v\in V$. By Operations on Linear Maps (ii) and the linearity of $S$ and $T$ (Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (i)),

$$(S+T)(u+v)=(Su+Sv)+(Tu+Tv)=(Su+Tu)+(Sv+Tv)=(S+T)u+(S+T)v,$$

where the four terms are rearranged by Vector Space over a Field (i) and Vector Space over a Field (ii) in $W$; and $(S+T)(\mu v)=\mu Sv+\mu Tv=\mu(S+T)v$ by Vector Space over a Field (vii). Likewise, by Operations on Linear Maps (iii), $(\lambda T)(u+v)=\lambda(Tu+Tv)=\lambda Tu+\lambda Tv$ by Vector Space over a Field (vii), and

$$(\lambda T)(\mu v)=\lambda(\mu Tv)=(\lambda\mu)Tv=(\mu\lambda)Tv=\mu(\lambda Tv)$$

by Vector Space over a Field (v) and the commutativity of the multiplication of $K$. So $S+T$ and $\lambda T$ are linear.

Consequently the sum is a binary operation on the set $\mathcal{L}(V,W)$ of Operations on Linear Maps (i), and the scalar multiplication is a map from $K\times\mathcal{L}(V,W)$ to $\mathcal{L}(V,W)$. The conditions Vector Space over a Field (i), Vector Space over a Field (ii), Vector Space over a Field (v), Vector Space over a Field (vi), Vector Space over a Field (vii) and Vector Space over a Field (viii) hold for these operations because, by Operations on Linear Maps (ii) and Operations on Linear Maps (iii), each reduces at every $v\in V$ to the same condition in $W$; for instance, by Vector Space over a Field (viii) in $W$,

$$\bigl((\lambda+\mu)T\bigr)v=(\lambda+\mu)Tv=\lambda Tv+\mu Tv=(\lambda T+\mu T)v .$$

The conditions Vector Space over a Field (iii) and Vector Space over a Field (iv) are checked next. The zero map $z:v\mapsto0$ is linear, since $z(u+v)=0=0+0$ by The Zero Vector, Negatives and Differences in a Vector Space (i) and $z(\mu v)=0=\mu0$ by Elementary Identities in a Vector Space (iii), and it is a zero of $\mathcal{L}(V,W)$, as $(T+z)v=Tv+0=Tv$. It is the only one: any zero $z'$ satisfies $z=z+z'=z'+z=z'$. For the negatives condition it therefore suffices to find, for each $T$, a map whose sum with $T$ is $z$, and $(-1)T$ does this: its values are $(-1)Tv=-Tv$ by Elementary Identities in a Vector Space (iv), so $(T+(-1)T)v=Tv+(-Tv)=0$ (The Zero Vector, Negatives and Differences in a Vector Space (ii)). Thus $\mathcal{L}(V,W)$ is a vector space over $K$, and its zero vector is $z$ by The Zero Vector, Negatives and Differences in a Vector Space (i). This proves (i).

Next, composition. Let $R$, $S$, $S_{1}$, $S_{2}$, $T$, $T_{1}$, $T_{2}$ and $\lambda$ be as in (ii), and let $u,u'\in U$ and $\mu\in K$. By Operations on Linear Maps (iv), $(ST)u=S(Tu)$, so the linearity of $T$ and of $S$ (Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (i)) gives

$$(ST)(u+u')=S(Tu+Tu')=STu+STu',\qquad(ST)(\mu u)=S(\mu Tu)=\mu\,STu,$$

and $ST$ is linear. The values of $R(ST)$ and $(RS)T$ at $u$ are both $R(S(Tu))$, so the two maps are equal. The distributive laws hold on values: $S(T_{1}u+T_{2}u)=ST_{1}u+ST_{2}u$ and $S(\lambda Tu)=\lambda STu$ by the linearity of $S$, which gives $S(T_{1}+T_{2})=ST_{1}+ST_{2}$ and $S(\lambda T)=\lambda(ST)$; while $(S_{1}+S_{2})T=S_{1}T+S_{2}T$ and $(\lambda S)T=\lambda(ST)$ follow from Operations on Linear Maps (ii) and Operations on Linear Maps (iii) alone. This proves (ii).

By Operations on Linear Maps (v), $I_{V}v=v$ for $v\in V$, so $I_{V}(u+v)=I_{V}u+I_{V}v$ and $I_{V}(\mu v)=\mu I_{V}v$, and $I_{V}\in\mathcal{L}(V)$. For $T\in\mathcal{L}(V,W)$ we have $TI_{V}=T=I_{W}T$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (iv). This proves (iii).

Taking $U=V=W$ in (ii), the composite of two operators on $V$ is again one, so $(S,T)\mapsto ST$ is a binary operation on $\mathcal{L}(V)$ (a map by Maps from Functional Predicative Relations, and Families of Sections (ii)). It is associative, since $R(ST)=(RS)T$, so $\mathcal{L}(V)$ with it is a semigroup; and $I_{V}$ is a neutral element by (iii) with $W=V$. Hence $\mathcal{L}(V)$ is a monoid with identity $I_{V}$ by Monoids (i), which is (iv).

Now let $T\in\mathcal{L}(V,W)$ be an isomorphism of $V$ onto $W$. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vi), $T^{-1}$ is a bijective map from $W$ to $V$ with

$$T(T^{-1}w)=w\quad(w\in W),\qquad T^{-1}(Tv)=v\quad(v\in V).$$

For $w,w'\in W$ and $\mu\in K$, the linearity of $T$ gives

$$T(T^{-1}w+T^{-1}w')=w+w',\qquad T(\mu T^{-1}w)=\mu w,$$

and applying $T^{-1}$ to both sides of these equalities yields $T^{-1}(w+w')=T^{-1}w+T^{-1}w'$ and $T^{-1}(\mu w)=\mu T^{-1}w$. So $T^{-1}$ is linear, and as it is bijective it is an isomorphism of $W$ onto $V$ (Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (iii)). This proves (v).

Finally, let $T\in\mathcal{L}(V)$. If $T$ is a unit, with inverse $S\in\mathcal{L}(V)$ (Units of a Monoid (i)), then $TS=ST=I_{V}$, so $T$ is bijective by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vii). Conversely, if $T$ is bijective, it is an isomorphism of $V$ onto itself, so $T^{-1}\in\mathcal{L}(V)$ by (v), and $TT^{-1}=T^{-1}T=I_{V}$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vi); thus $T$ is a unit. That the set $\mathrm{GL}(V)$ of units, with the restriction of composition, is a group with identity $I_{V}$ is Units of a Monoid: Unique Inverses, Products, and the Group of Units (v), applied to the monoid of (iv). This proves (vi).