Matrices and Linear Maps

Coordinate Representation

We now examine the relationship between matrices and linear maps. This can be regarded 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 \(x_i\) of the \(e_i\) in this linear combination is called the coordinate representation with respect to \(\mathcal{B}\), denoted by \([x]_\mathcal{B}\). It is also worth noting that, using the coordinate form, we can write \(x_i\) simply as

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

without further ado. (§Dual Spaces, ⁋Definition 6)

Matrix Representation of Linear Maps

For 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 above situation, suppose an arbitrary \(A\)-linear map \(u:M \rightarrow N\) is given. Then the matrix representation of \(u\) is the matrix

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

We first note the following.

Proposition 2 The \(i\)-th column of the matrix representation \([u]_\mathcal{C}^\mathcal{B}\) of a linear map \(u:M \rightarrow N\) equals 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

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

The \(j\)-th component of this column vector is, by the preceding formula (1), precisely 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. Moreover, if \(\alpha\) lies 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

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

In summary, \(u\mapsto [u]_\mathcal{C}^\mathcal{B}\) 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, since \(u=0\) is equivalent to \(u(e_i)=0\) for all \(i\in I\). On the other hand, if \(J\) is finite, then for any element \((x_{ji})\) of \(\Mat_{J\times I}(A)\), we can define \(u\in\Hom_\lMod{A}\) by the 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; hence it is a \(Z(A)\)-isomorphism.

Product of Matrix Representations

We previously examined how the product of two matrices is defined. As in [Linear Algebra] §Fundamental Theorem of Linear Algebra, ⁋Theorem 5, this product 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 any \(x\in M\), the formula

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

holds.

Proof

We can verify that the right-hand side yields a column vector, and by formula (2) of §Matrices, §§Matrix Multiplication, its \(j\)-th component 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, since \(x=\sum_{i\in I}x_i e_i\), the \(j\)-th component of \([u(x)]_\mathcal{C}\) is

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

yielding the desired result.

Combining this with Proposition 2, we obtain the following.

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 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

The transpose of a matrix also has a corresponding notion for linear maps.

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

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

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 \(M\) with \(M^{\ast\ast}\), and then \(\mathcal{B}\) corresponds to the dual basis \(\mathcal{B}^{\ast\ast}\) of \(\mathcal{B}^\ast\). Now

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

so we obtain the desired result.

Matrix Representations and Trace

Previously we defined the trace of a linear map in §Hom and the Tensor Product, ⁋Definition 6. Now, for any \(n\times n\) matrix \(X\), we define the trace of \(X\) by

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

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 more general situation in which 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 the special case where all the \(M_i\) and \(N_j\) are equal to \(A\), this reduces to the situation above. Then any \(x\) can be written uniquely as

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

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 matrix

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

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

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

Even with this generalization, all the propositions examined above remain valid. In particular, the product of matrices is noteworthy: suppose \(I,J\) are both finite, and moreover each \(M_i\) and \(N_j\) has finite bases \(\mathcal{B}_i\), \(\mathcal{C}_j\). Then the unions of these bases form bases \(\mathcal{B},\mathcal{C}\) of \(M\) and \(N\) respectively. The matrix representing a linear map \(u:M \rightarrow N\) with respect to these bases can be verified to equal the matrix obtained by substituting, for each entry \(u_{ji}\) in the above \([u]_J^I\), its matrix representation with respect to the bases \(\mathcal{B}_i\), \(\mathcal{C}_j\); and this behaves meaningfully under 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 manner, the matrix representation of \(v\circ u\) with respect to the bases \(\mathcal{B}, \mathcal{D}\) is the matrix whose \((k,i)\)-entry is the matrix

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

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.