Matrices and Linear Maps

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Coordinate Representation

Now we examine the relationship between matrices and linear maps. This can be thought of as a generalization of [Linear Algebra] §Fundamental Theorem of Linear Algebra, ⁋Theorem 5. For convenience,

let a free \(A\)-module \(M\) be given, and fix a basis \(\mathcal{B}=(e_i)_{i\in I}\) of \(M\). Then any \(x\in M\) can be written as

\[x=\sum_{i\in I} x_i e_i,\qquad x_i\in A\]

and the column vector \((x_{i0})_{(i,0)\in I\times\{0\}}\) consisting of the coefficients \(x_i\) corresponding to the \(e_i\) when expressed as such a linear combination is called the coordinate representation with respect to \(\mathcal{B}\) and is denoted by \([x]_\mathcal{B}\). On the other hand, it is also worth noting that using the coordinate form, we can write \(x_i\) simply as the formula

\[x_i=\langle x,e_i^\ast\rangle\tag{1}\]

without complicated explanation. (§Dual Spaces, ⁋Definition 6)

Matrix Representation of Linear Maps

In the remainder of this post, we assume that two free \(A\)-modules \(M,N\) are given, and fix their bases \(\mathcal{B}=(e_i)_{i\in I}\), \(\mathcal{C}=(f_j)_{j\in J}\).

Definition 1 In the situation above, suppose any \(A\)-linear map \(u:M \rightarrow N\) is given. Then the matrix representation of \(u\) is the following matrix

\([u]_\mathcal{C}^\mathcal{B}=(f_j^\ast(u(e_i)))_{(j,i)\in J\times I}=(\langle u(e_i), f_j^\ast\rangle)_{(j,i)\in J\times I}\).

Then the following holds first.

Proposition 2 The \(i\)-th column of the matrix representation \([u]_\mathcal{C}^\mathcal{B}\) of a linear map \(u:M \rightarrow N\) is equal to the coordinate representation \([u(e_i)]_\mathcal{C}\) of \(u(e_i)\) with respect to \(\mathcal{C}\).

Proof

By definition, the \(i\)-th column of \([u]_\mathcal{C}^\mathcal{B}\) is given by the following formula

\((f_j^\ast(u(e_i)))_{j\in J}=(\langle u(e_i), f_j^\ast\rangle)_{j\in J}\).

Now the \(j\)-th component of this column vector is, by the preceding formula (1), exactly the coefficient of \(f_j\) when \(u(e_i)\) is expressed as a linear combination with respect to the basis \(\mathcal{C}\).

If another \(A\)-linear map \(v:M \rightarrow N\) is given, we can verify that

\[[u+v]_\mathcal{C}^\mathcal{B}=[u]_\mathcal{C}^\mathcal{B}+[v]_\mathcal{C}^\mathcal{B}\]

holds. Also, if \(\alpha\) is contained in the center of \(A\), then \(\alpha u\) is an \(A\)-linear map, and the matrix representation of this \(A\)-linear map is given by

\([\alpha u]_\mathcal{C}^\mathcal{B}=\alpha[u]_\mathcal{C}^\mathcal{B}\).

To summarize, \(u\mapsto [u]_\mathcal{C}^\mathcal{B}\) is a \(Z(A)\)-module homomorphism from \(\Hom_{\lMod{A}}(M,N)\) to \(\Mat_{J\times I}(A)\). This \(Z(A)\)-linear map is injective, because \(u=0\) and \(u(e_i)=0\) for all \(i\in I\) are equivalent. On the other hand, if \(J\) is a finite set, then for any element \((x_{ji})\) of \(\Mat_{J\times I}(A)\), we can define \(u\in\Hom_\lMod{A}\) by the following formula

\[u(e_i)=\sum_{j\in J} \langle u(e_i),f_j^\ast\rangle f_j\]

to construct the inverse of the above \(Z(A)\)-linear map, so it becomes a \(Z(A)\)-isomorphism.

Product of Matrix Representations

We previously examined how to define the product of two matrices. As with [Linear Algebra] §Fundamental Theorem of Linear Algebra, ⁋Theorem 5, the product of these matrices corresponds to the composition of linear maps. Let us first prove the following proposition.

Proposition 3 If \(I,J\) are finite sets, then for any linear map \(u:M \rightarrow N\) and \(x\in M\), the following formula

\[[u(x)]_\mathcal{C}=[u]_\mathcal{C}^\mathcal{B}[x]_\mathcal{B}\]

holds.

Proof

We can verify that the expression on the right-hand side yields a column vector, and then by formula (2) of §Matrices, §§Matrix Multiplication, the \(j\)-th component of the right-hand side expression is

\(\left([u]_\mathcal{C}^\mathcal{B}[x]_\mathcal{B}\right)_{j0}=\sum_{i\in I}\left([u]_\mathcal{C}^\mathcal{B}\right)_{ji}\left([x]_\mathcal{B}\right)_{i0}=\sum_{i\in I}\left\langle u(e_i),f_j^\ast\right\rangle \left\langle x,e_i^\ast\right\rangle\).

On the other hand, examining the left-hand side, since \(x=\sum_{i\in I}x_i e_i\), the \(j\)-th component of \([u(x)]_\mathcal{C}\) becomes

\[\langle u(x),f_j^\ast\rangle=\left\langle u\left(\sum_{i\in I} x_i e_i\right), f_j^\ast\right\rangle=\left\langle \sum_{i\in I} x_i u(e_i), f_j^\ast\right\rangle=\sum_{i\in I}x_i\langle u(e_i),f_j^\ast\rangle=\sum_{i\in I}\left\langle u(e_i),f_j^\ast\right\rangle \left\langle x,e_i^\ast\right\rangle\]

yielding the desired result.

Combining this with Proposition 2, we obtain the following result.

Corollary 4 Suppose three \(A\)-modules \(M,N,L\) are given, and fix finite bases \(\mathcal{B}=(e_i)_{i\in I},\mathcal{C}=(f_j)_{j\in J},\mathcal{D}=(g_k)_{k\in K}\). Then for any linear maps \(u:M \rightarrow N\), \(v:N \rightarrow L\), the following formula

\[[v \circ u]_\mathcal{D}^\mathcal{B}=[v]_\mathcal{D}^\mathcal{C}[u]_\mathcal{C}^\mathcal{B}\]

holds.

Proof

For any \(x\in M\),

\[[v \circ u]_\mathcal{D}^\mathcal{B}[x]_\mathcal{B}=[(v \circ u)(x)]_\mathcal{D}=[(v(u(x))]_\mathcal{D}=[v]_\mathcal{D}^\mathcal{C}[u(x)]_\mathcal{C}=[v]_\mathcal{D}^\mathcal{C}[u]_\mathcal{C}^\mathcal{B}[x]_\mathcal{B}\]

so from the \(Z(A)\)-isomorphism \(\Mat_{K\times I}(A)\cong\Hom_\lMod{A}(M,L)\) we obtain the desired result.

Transpose of Matrix Representations

On the other hand, the transpose of a matrix also has a corresponding concept in linear maps.

Proposition 5 If \(I,J\) are finite sets, then for any linear map \(u:M \rightarrow N\), the following formula

\[\left([u]_\mathcal{C}^\mathcal{B}\right)^t=\left[u^\ast\right]_{\mathcal{B}^\ast}^{\mathcal{C}^\ast}\]

holds. Here \(\mathcal{B}^\ast\) and \(\mathcal{C}^\ast\) are the dual bases of \(\mathcal{B},\mathcal{C}\) respectively.

Proof

By §Dual Spaces, ⁋Proposition 8, we may identify \(M\) and \(M^{\ast\ast}\), and then \(\mathcal{B}\) corresponds to the dual basis \(\mathcal{B}^{\ast\ast}\) of \(\mathcal{B}^\ast\). Now

\[\left(\left[u^\ast\right]_{\mathcal{B}^\ast}^{\mathcal{C}^\ast}\right)_{ji}=\langle u^\ast(f_j^\ast), e_i^{\ast\ast}\rangle=\langle e_i, u^\ast(f^\ast)\rangle=\langle u(e_i), f_j^\ast\rangle=\left([u]_\mathcal{C}^\mathcal{B}\right)_{ij}=\left(\left([u]_\mathcal{C}^\mathcal{B}\right)^t\right)_{ji}\]

so we obtain the desired result.

Matrix Representations and Trace

Previously we defined the trace of a linear map in §Hom and Tensor Products, ⁋Definition 4. This time, for any \(n\times n\) matrix \(X\), let us define the trace of \(X\) by the following formula

\(\tr(X)=\sum_{i=1}^n x_{ii}\).

Then for any \(u\in\End_\rMod{A}(M)\), fixing a basis \(\mathcal{B}=(e_i)_{1\leq i\leq n}\) and considering \([u]_\mathcal{B}^\mathcal{B}\), we have

\(\tr([u]_\mathcal{B}^\mathcal{B})=\sum_{i=1}^n ([u]_\mathcal{B}^\mathcal{B})_{ii}=\sum_{i=1}^n\langle u(e_i), e_i^\ast\rangle\).

On the other hand, for obvious reasons

\[u(x)=\sum_{i=1}^n \langle x, e_i^\ast\rangle u(e_i)\]

so we obtain \(\tr(u)=\tr([u]_\mathcal{B}^\mathcal{B})\). From this, for \(X\in\Mat_{m\times n}(A)\), \(Y\in\Mat_{n\times m}(A)\)

\[\tr(XY)=\tr(YX)\]

holds.

Block Matrices

Finally, we consider the case where two free \(A\)-modules \(M,N\) are each written as a direct sum of submodules

\(M=\bigoplus_{i\in I}M_i,\qquad N=\bigoplus_{j\in J} N_j\).

In particular, if all the \(M_i\), \(N_j\) are equal to \(A\), this is the same situation as above. Then any \(x\) can be written uniquely in the form

\[x=\sum_{i\in I} x_i,\qquad x_i\in M_i\]

and we define the coordinate representation of \(x\) with respect to this decomposition as

\([x]_I=(x_{i0})_{i\in I}\).

Also, for any \(A\)-linear map \(u: M \rightarrow N\), writing

\[u(x_i)=\sum_{j\in J} u_{ji}(x_i),\qquad u_{ji}(x_i)\in N_j\]

the following matrix

\[[u]^I_J=(u_{ji})_{(j,i)\in J\times I}\]

is well-defined. This matrix can be thought of as a \(J\times I\) matrix defined over the following ring

\(H=\bigoplus_{(j,i)\in J\times I}\Hom_{\lMod{A}}(M_i,N_j)\).

Even with this generalization, we can verify that all the propositions examined above still hold as they are. In particular, the product of matrices is worth noting: suppose \(I,J\) are both finite, and furthermore each \(M_i\) and \(N_j\) has finite bases \(\mathcal{B}_i\), \(\mathcal{C}_j\). Then the collections of all these bases form bases \(\mathcal{B},\mathcal{C}\) of \(M\) and \(N\) respectively. Then representing a linear map \(u:M \rightarrow N\) as a matrix with respect to this basis can be verified to be equal to the matrix obtained by substituting, for each component \(u_{ji}\) in the above \([u]_J^I\), its matrix representation with respect to the bases \(\mathcal{B}_i\), \(\mathcal{C}_j\), and this behaves meaningfully with respect to matrix multiplication. That is, for another direct sum \(L=\bigoplus_{k\in K} L_k\) and basis \(\mathcal{D}=\bigcup \mathcal{D}_k\), writing \(v:N \rightarrow L\) in the same way, the matrix representation of \(v\circ u\) with respect to the bases \(\mathcal{B}, \mathcal{D}\) becomes the matrix whose \((k,i)\)-component is the matrix

\(\sum_{j\in J}[v_{kj}]_{\mathcal{D}_k}^{\mathcal{C}_j}[u_{ji}]_{\mathcal{C}_j}^{\mathcal{B}_i}\).