Change of Basis

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.

Square Matrices

Definition 1 An \(I\times I\) matrix is called a square matrix. Their collection is denoted by \(\Mat_I(A)\).

In particular, when \(I\) is a finite set and \(A\) is commutative, \(\Mat_{n}(A)\) has special properties: this object is not only an \(A\)-module but also carries a multiplication defined on it. In other words, \(\Mat_{n}(A)\) is an \(A\)-algebra.

Proposition 2 In this situation, \(\Mat_n(A)\) is a unital associative algebra.

Proof

That \(\Mat_n(A)\) is an associative \(A\)-algebra follows immediately from §Matrices, §§Matrix Multiplication. One can verify that the identity element for multiplication in \(\Mat_n(A)\) is the following identity matrix

\[I_n=\begin{pmatrix}1&0&\cdots&0\\0&1&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&1\end{pmatrix}\]

\(M_n(A)\) has the canonical basis \((E_{ij})\), and considering the structure constants for these yields the formula

\[E_{ij}E_{jk}=\delta_{jh}E_{ik}\]

Definition 3 We denote by \(\GL_n(A)\) the collection of elements of \(\Mat_n(A)\) that have multiplicative inverses.

Fix a basis \(\mathcal{B}=(e_i)_{i\in I}\) of a free \(A\)-module \(M\), and let \(\lvert I\rvert=n\). Then for any \(u\in \End_{\lMod{A}}(M)\), we have \(\[u\]_{\mathcal{B}}^\mathcal{B}\in\Mat_n(A)\), and if \(u\) is an isomorphism then by §Matrices and Linear Maps, ⁋Corollary 4 we have \(\[u\]_{\mathcal{B}}^\mathcal{B}\in\GL_n(A)\). Then by §Dual Spaces, ⁋Proposition 5 and §Matrices and Linear Maps, ⁋Proposition 5, the formula

\[\bigl([u^{-1}]_{\mathcal{B}}^\mathcal{B}\bigr)^t=\bigl(\bigl[u^\ast\bigr]_{\mathcal{B}^\ast}^{\mathcal{B}^\ast}\bigr)^{-1}\]

holds.

Change of Basis

Proposition 4 Let \(M\) be an arbitrary \(A\)-module, and let \(\mathcal{B}=(e_i)_{i\in I}\) be a finite basis of \(M\). Then the condition that the formula

\[e_i'=\sum_{j=1}^n a_{ji}e_i,\qquad 1\leq i\leq n\]

gives a basis of \(M\) is equivalent to the square matrix \((a_{ji})\) having an inverse.

Proof

The given matrix \((a_{ji})\) is simply the matrix representation \(\[u\]_{\mathcal{B}}^\mathcal{B}\in\Mat_n(A)\) of the linear map \(u\in\End_{\lMod{A}}(M)\) defined by

\[u:e_i\mapsto e_i'=\sum_{j=1}^n a_{ji}e_i\]

Now this matrix having an inverse is equivalent to \(u\) being an isomorphism, which in turn is equivalent to \((u(e_i))_{i\in I}\) being a basis of \(M\).

Conversely, we may also regard the matrix \((a_{ji})\) as the matrix representation of the identity map \(\id_M:M \rightarrow M\) with respect to different bases. We write the basis \((e_i')\) as \(\mathcal{B}'\). Then since

\[\id_M(e_i')=\sum_{j=1}^n a_{ji}e_i\]

we have

\[([\id_M]^{\mathcal{B}'}_\mathcal{B})=(\langle \id_M(e_i'), e_j^\ast\rangle)_{(j,i)\in J\times I}=(a_{ji})_{(j,i)\in J\times I}\]

From this perspective, this matrix is also called the change-of-basis matrix from \(\mathcal{B}'\) to \(\mathcal{B}\).

More generally, the following holds.

Proposition 5 Let \(M\) and \(N\) be two \(A\)-modules, and let \(\mathcal{B}=(e_i)_{i\in I}\) and \(\mathcal{C}=(f_j)_{j\in J}\) be finite bases of \(M\) and \(N\), respectively. Then for other bases \(\mathcal{B}'=(e_i')_{i\in I}\) of \(M\) and \(\mathcal{C}'=(f_j')_{j\in J}\) of \(N\), the formula

\[[u]_{\mathcal{C}'}^{\mathcal{B}'}=[\id_N]^\mathcal{C}_{\mathcal{C}'}[u]^\mathcal{B}_\mathcal{C}[\id_M]^{\mathcal{B}'}_{\mathcal{B}}\]

holds.

Similar Matrices

Definition 6 Two \(m\times n\) matrices \(X, X'\) are called equivalent if there exist square matrices \(P\in\GL_m(A)\) and \(Q\in\GL_n(A)\) such that \(X'=PXQ\).

In the same context as the discussion preceding [Basic Linear Algebra] §Change of Basis, ⁋Definition 2, it is better to consider the following finer equivalence relation than that of equivalent matrices.

Definition 7 Two \(n\times n\) matrices \(X, X'\) are called similar if there exists a square matrix \(P\in\GL_n(A)\) such that \(X'=PXP^{-1}\).

Then setting \(M=N\), \(\mathcal{B}=\mathcal{C}\), and \(\mathcal{B}'=\mathcal{C}'\) in Proposition 5 above, we see that the matrix representations of an element \(u\) of \(\End_\rMod{A}(M)\) obtained through different bases are similar to each other.