Matrices, the Matrix of a Linear Map, and Change of Basis

For $i<m$ and $\ell<q$, expansion by [L1] gives $((AB)C)_{i\ell}=\sum_{k<p}\sum_{j<n}(a_{ij}b_{jk})c_{k\ell}$ and $(A(BC))_{i\ell}=\sum_{j<n}\sum_{k<p}a_{ij}(b_{jk}c_{k\ell})$.

Associativity and distributivity in $F$, followed by [L2], identify the two sums in step 1.1, so $(AB)C=A(BC)$.

The entry formulas also give $(I_mA)_{ik}=a_{ik}=(AI_n)_{ik}$; splitting sums gives both distributive laws; pulling out $\lambda$ gives scalar compatibility; and a sum of zero summands gives each zero law. Thus all stated identities hold entrywise.

If any indexing dimension is zero, every asserted matrix equality either has no entries to compare or reduces to an empty sum, whose value is $0$. Hence the same calculations cover all zero-sized shapes, including $I_0$.

The published pointwise vector-space structure on $M_n(F)$ makes entrywise addition an abelian group operation, with the zero matrix and entrywise negatives.

By [L1], matrix multiplication is associative, has identity $I_n$, and satisfies both distributive laws over that addition.

These are exactly the ring axioms. If $n=0$, there is one empty matrix, so its zero and identity coincide; the ring convention permits $0=1$, making $M_0(F)$ the zero ring.

For $r<m$ and $s<p$, the product entry is $(E_{ij}E_{k\ell})_{rs}=\sum_{t<n}\delta_{ri}\delta_{tj}\delta_{tk}\delta_{s\ell}$.

If $j\ne k$, every summand is zero. If $j=k$, only $t=j$ can contribute and the value is $\delta_{ri}\delta_{s\ell}$.

The two cases combine as $(E_{ij}E_{k\ell})_{rs}=\delta_{jk}(E_{i\ell})_{rs}$ for every entry, proving the matrix identity.

By [L1], $E_{01}E_{10}=E_{00}$ while $E_{10}E_{01}=E_{11}$; these products differ at entry $(0,0)$ because $1\ne0$ in a field.

Thus two elements of $M_n(F)$ fail to commute, so the matrix ring is noncommutative.

Swapping indices in the entrywise sum and scalar product gives the two linearity identities, and swapping twice gives $(A^{\mathsf T})^{\mathsf T}=A$.

For conformable $A\in M_{m\times n}(F)$ and $C\in M_{n\times p}(F)$, one has $((AC)^{\mathsf T})_{ki}=(AC)_{ik}=\sum_{j<n}a_{ij}c_{jk}$.

Commutativity in $F$ rewrites the sum in step 2.1 as $\sum_{j<n}(C^{\mathsf T})_{kj}(A^{\mathsf T})_{ji}=(C^{\mathsf T}A^{\mathsf T})_{ki}$, proving the product law entrywise.

If $S,T$ are linear, then $(S+T)(au+bv)=a(S+T)(u)+b(S+T)(v)$, and if $\lambda\in F$, then $(\lambda T)(au+bv)=a(\lambda T)(u)+b(\lambda T)(v)$; hence the pointwise operations remain inside $\mathcal L(V,W)$.

Evaluating at an arbitrary $v\in V$ reduces associativity, commutativity, both distributive laws, scalar associativity, and the scalar identity law to the corresponding vector-space laws in $W$.

The zero function is the additive identity and $(-T)(v):=-T(v)$ is the additive inverse of $T$. The same formulas cover the zero domain and zero codomain, so all vector-space axioms hold.

The identity satisfies $\operatorname{id}_V(av+bw)=av+bw=a\operatorname{id}_V(v)+b\operatorname{id}_V(w)$, so it is linear.

Linearity of $T$ and then of $S$ gives $S(T(au+bv))=S(aT(u)+bT(v))=aS(T(u))+bS(T(v))$.

The equality in step 2.1 is the defining linearity equation for $S\circ T$, so the composite is linear.

Write $[v]_{\mathcal B}=(x_j)_{j<n}$, so [L1] gives $v=\sum_{j<n}x_jb_j$.

By linearity, $T(v)=\sum_{j<n}x_jT(b_j)$; writing $T(b_j)=\sum_{i<m}t_{ij}c_i$ gives $T(v)=\sum_{i<m}(\sum_{j<n}t_{ij}x_j)c_i$.

The inner sum is the $i$-th row-by-column entry of $[T]_{\mathcal B}^{\mathcal C}[v]_{\mathcal B}$, and uniqueness of $\mathcal C$-coordinates identifies this column with $[T(v)]_{\mathcal C}$.

For every basis vector $b_j$, coordinate uniqueness in [L2] gives $[(S+T)(b_j)]_{\mathcal C}=[S(b_j)]_{\mathcal C}+[T(b_j)]_{\mathcal C}$ and $[(\lambda T)(b_j)]_{\mathcal C}=\lambda[T(b_j)]_{\mathcal C}$, so $\Phi$ is linear column by column.

If $\Phi(S)=\Phi(T)$, then [L2] gives $S(b_j)=T(b_j)$ for every $j$; linearity and the unique expansion of every vector in $\mathcal B$ give $S=T$, so $\Phi$ is injective.

Given $A=(a_{ij})\in M_{m\times n}(F)$, prescribe $T(b_j):=\sum_{i<m}a_{ij}c_i$ and, for the unique expansion $v=\sum_{j<n}x_jb_j$ from [L2], define $T(v):=\sum_{j<n}x_jT(b_j)$. This is well defined and linear, and the $j$-th matrix column is the $j$-th column of $A$; hence $\Phi(T)=A$. Together with steps 1.1 and 2.1, $\Phi$ is a linear bijection. Its set-theoretic inverse is linear: if $A=\Phi(S)$ and $B=\Phi(T)$, then injectivity and linearity give $\Phi^{-1}(A+B)=S+T$ and $\Phi^{-1}(\lambda A)=\lambda S$. Thus $\Phi$ is a linear isomorphism by [L3].

If $n=0$, then $V$ is the zero space and both sides contain only their zero element; if $m=0$, then $W$ and $M_{m\times n}(F)$ are zero spaces and the only map is the zero map. Thus the construction also proves the isomorphism in every zero-dimensional case.

Every matrix $A=(a_{ij})$ has the expansion $A=\sum_{(i,j)\in m\times n}a_{ij}E_{ij}$, and a linear relation among the $E_{ij}$ has each coefficient zero when its corresponding entry is read. Thus the matrix units form a basis.

By [L2], the index set $m\times n$ has cardinality $mn$, so this basis has $mn$ elements; if either dimension is zero, it is the empty basis of the zero matrix space. Hence $\dim_FM_{m\times n}(F)=mn$.

The isomorphism in [L3] transports a basis and preserves dimension; substituting $n=\dim_FV$ and $m=\dim_FW$ gives the second formula.

Apply [L2] to $T$ and then to $S$: $[S(T(u))]_{\mathcal D}=[S]_{\mathcal C}^{\mathcal D}[T(u)]_{\mathcal C}=[S]_{\mathcal C}^{\mathcal D}[T]_{\mathcal B}^{\mathcal C}[u]_{\mathcal B}$.

Associativity from [L3] rewrites step 1.1 as $[S(T(u))]_{\mathcal D}=([S]_{\mathcal C}^{\mathcal D}[T]_{\mathcal B}^{\mathcal C})[u]_{\mathcal B}$ for every $u$.

Evaluating at each vector of $\mathcal B$ makes $[u]_{\mathcal B}$ a standard coordinate column, so the columns of $[S\circ T]_{\mathcal B}^{\mathcal D}$ equal those of the displayed product. Therefore the matrices are equal.

Composition of endomorphisms is associative, has $\operatorname{id}_V$ as identity, and distributes over pointwise addition; together with the additive group from [L1], this makes $\operatorname{End}_F(V)$ a ring.

By [L2], matrix representation is a bijective linear map, so it preserves addition and zero.

It preserves products by the composition formula in [L2], and $[\operatorname{id}_V]_{\mathcal B}^{\mathcal B}=I_n$ by coordinate action. Thus it is a bijective unital ring homomorphism and hence a ring isomorphism.

By [L3], $L_A$ and $L_B$ are linear. If $AB=BA=I_n$, then $L_AL_B=L_{AB}=\operatorname{id}$ and $L_BL_A=L_{BA}=\operatorname{id}$, so $L_A$ is a linear isomorphism with inverse $L_B$. This includes $n=0$, where the unique empty matrix and the unique zero-space map are their own inverses.

Conversely, if $L_A$ has a linear inverse $S$, [L4] represents $S$ by a matrix $B$ in the standard coordinate basis. The two identity composites and [L2] give $AB=I_n=BA$, so $A$ is invertible.

For a general linear isomorphism $T$, the list $T(\mathcal B)$ is an ordered basis of $W$: $T^{-1}$ transfers both linear independence and spanning back to $\mathcal B$. Hence [L5] shows that $\mathcal B$ and $\mathcal C$ have the same length, so the two displayed representation matrices are square. Representing $T^{-1}\circ T$ and $T\circ T^{-1}$ in the chosen bases, [L2] gives both inverse equations for $[T]_{\mathcal B}^{\mathcal C}$ and $[T^{-1}]_{\mathcal C}^{\mathcal B}$.

If $T:V\to W$ is an isomorphism and $(b_j)_{j<n}$ is a basis of $V$, then $(T(b_j))_{j<n}$ is independent because applying $T^{-1}$ to a vanishing linear combination makes every coefficient zero, and it spans because every $w$ equals $T(v)$ and $v$ expands in the $b_j$. Thus it is a basis of $W$, so the dimensions agree.

Conversely, if the dimensions agree, choose ordered bases $(b_j)_{j<n}$ of $V$ and $(c_j)_{j<n}$ of $W$ and define $T(\sum_jx_jb_j)=\sum_jx_jc_j$. Unique coordinates make this a linear map with $T(b_j)=c_j$.

Defining $S(\sum_jy_jc_j)=\sum_jy_jb_j$ gives a linear inverse to $T$. For $n=0$, both bases are empty and both spaces are zero, so the same formulas give the unique isomorphism.

By [L1], $\operatorname{GL}_n(F)$ is exactly the unit set of the ring $M_n(F)$.

Applying [L2] gives closure, associativity inherited from the ring, identity $I_n$, and inverse $A^{-1}$ for every element, so $\operatorname{GL}_n(F)$ is a group.

If $n=0$, $M_0(F)$ has one element, the empty matrix $I_0$, which is its own inverse; hence its unit group is the trivial group.

Applying coordinate action from [L2] to $\operatorname{id}_V$ and using [L1] gives $[v]_{\mathcal C}=P_{\mathcal C\leftarrow\mathcal B}[v]_{\mathcal B}$.

Represent the identity composition $\operatorname{id}_V\circ\operatorname{id}_V$ first from $\mathcal B$ through $\mathcal C$ back to $\mathcal B$, and then from $\mathcal C$ through $\mathcal B$ back to $\mathcal C$.

The composite-matrix formula in [L2] gives $P_{\mathcal B\leftarrow\mathcal C}P_{\mathcal C\leftarrow\mathcal B}=I$ and $P_{\mathcal C\leftarrow\mathcal B}P_{\mathcal B\leftarrow\mathcal C}=I$, so the two matrices are inverses. Empty ordered bases give the same two equations in $M_0(F)$.