Equivalence of Matrices

The last example (Page ) touched upon one of the core objectives of this chapter and we surely don't need to explain why we spend so much time to talk about the seemingly unrelated subjects in the last two sections. Let us go back to the main track: Classifying matrices into equivalence classes.

Given ${\cal D}$-homomorphism $\alpha : {\cal D}^{n} \longrightarrow {\cal D}^{m}$, finding suitable bases ${\cal B}_{1}$, ${\cal B}_{2}$ for ${\cal D}^{n}$ and ${\cal D}^{m}$ respectively, we know how to represent $\alpha$ as a matrix $M_{m \times n}$ with entries in ${\cal D}$. We note here that the dimension $n$ of ${\cal D}^{n}$ is no longer assumed to be less than or equal to the dimension $m$ of ${\cal D}^{m}$ and $\alpha$ may not be one-to-one. The sequence ${\cal D}^{n} \stackrel{\textstyle{\alpha}}{\longrightarrow} {\cal D}^{m} \longr... ...c{\textstyle{{\cal D}^{m}}}{\textstyle{{\alpha(\cal D}^{n})}} \longrightarrow 0$ is exact, but not necessary short exact, where $\frac{\textstyle{{\cal D}^{m}}}{\textstyle{{\alpha(\cal D}^{n})}}$ is easily seen to be finitely generated.

Given matrices $M, N \in {\cal M}\!\!\!\!\!\!\!{\cal M}_{m \times n}({\cal D})$ over a P.I.D. ${\cal D}$, by Definition 65.1, $M, N$ are equivalent if and only if there are invertible matrices $P \in {\cal M}\!\!\!\!\!\!\!{\cal M}_{m \times m}({\cal D})$, $Q \in {\cal M}\!\!\!\!\!\!\!{\cal M}_{n \times n}({\cal D})$ so that $N = P \cdot M \cdot Q$. As usual, matrices $P$ and $Q$ can be interpreted as "changing bases" operations on the ${\cal D}$-modules ${\cal D}^{m}$ and ${\cal D}^{n}$. In this section, our focal point is to analyze the matrix equivalence relation in terms of the elementary and secondary (Definitions 65.2 and 65.3) row and column operations. In other words, we will show that each invertible square matrix can be decomposed as a product of a sequence of elementary and secondary matrices. In Section 7.1.2, we have seen that the effects of multiplying these types of matrices on the left or right to a matrix $M$ are equivalent to doing the correspondent elementary and secondary row, column operations on the matrix $M$.

Exercise: Given free ${\cal D}$-module ${\cal D}^{m}$,

1. Let ${\cal B}_{1} = \{u_{1}, \ldots, u_{m}\}, {\cal B}_{2} = \{v_{1}, \ldots, v_{m}\}$ be bases for ${\cal D}^{m}$, $P$ be the coefficient matrix obtained by expressing each $v_{i}$ as a linear combination of elements in ${\cal B}_{1}$, $1 \leq i \leq m$. Show that $P$ is invertible.
2. Let $\{u_{1}, \ldots, u_{m}\}$ be a basis for ${\cal D}^{m}$, $P = \left[\!\!\begin{array}{ccc} a_{11}& \cdots & a_{1m} \\ & \ddots & \\ ... ...d{array}\!\!\right]\! \in {\cal M}\!\!\!\!\!\!\!{\cal M}_{m \times m}({\cal D})$ be an invertible matrix. Show that $\left\{ w_{j} \mid w_{j} = \sum_{i=1}^{m} a_{ij} u_{i}, 1 \leq j \leq m\right\}$ is again a basis for ${\cal D}^{m}$.

Hint: Can be done easily by mimicking the proof for Theorem 35.

Let $M = \left[\!\!\begin{array}{ccc} a_{11}& \cdots & a_{1n} \\ & \ddots & \\ ... ...d{array}\!\!\right]\! \in {\cal M}\!\!\!\!\!\!\!{\cal M}_{m \times n}({\cal D})$ be a matrix over the P.I.D. ${\cal D}$, $\{\varepsilon_{1}, \ldots, \varepsilon_{n}\}$, $\{\epsilon_{1}, \ldots, \epsilon_{m}\}$ be the standard bases for the free ${\cal D}$-modules ${\cal D}^{n}$ and ${\cal D}^{m}$, respectively. Clearly, this matrix $M$ induces a ${\cal D}$-homomorphism $\alpha_{M} : {\cal D}^{n} \longrightarrow {\cal D}^{m}$ which is the extension of $\varepsilon_{1} \mapsto \raisebox{.0in}{${\mathrel{\mathop{\sum}\limits^{m}\limits_{i=1}}}$} a_{i1} \cdot \epsilon_{i}$, $\cdots$, $\varepsilon_{n} \mapsto \raisebox{.0in}{${\mathrel{\mathop{\sum}\limits^{m}\limits_{i=1}}}$} a_{in} \cdot \epsilon_{i}$. The image of $\alpha_{M}$ is a free submodule of ${\cal D}^{m}$. By Theorems 83 and 84, we may find a new basis $\{v_{1}, \ldots, v_{m}\}$ for ${\cal D}^{m}$ and a list of scalars $\{a_{1}, \ldots, a_{k}\} \subset {\cal D}$ with $a_{i} \,\,\rule[-.05in]{0.005in}{.2in}\,\, a_{i + 1}$ for $1 \leq i \leq k -1$ so that $\{a_{1} \cdot v_{1}, \ldots, a_{k} \cdot v_{k}\}$ is a basis for $\alpha_{M}({\cal D}^{n})$. Our second focal point in this section is to fulfill the promise we made in Section 7.1.2 by proving that, for each ${\bf0} \not= M_{m \times n} \in {\cal M}\!\!\!\!\!\!\!{\cal M}_{m \times n}({\cal D})$ over a P.I.D. ${\cal D}$, $M_{m \times n}$ is equivalent to a matrix $N_{m \times n}$ of the following format:

Exercise: Clearly, $\frac{\textstyle{\cal D}^{m}}{\textstyle{\alpha_{M}({\cal D}^{n})}}$ is a finitely generated ${\cal D}$-module. Show that $\{a_{l + 1}, \ldots, a_{k}, 0, \ldots, 0\}$ is the invariant factors for $\frac{\textstyle{\cal D}^{m}}{\textstyle{\alpha_{M}({\cal D}^{n})}}$. Note: All the cyclic submodules of $\frac{\textstyle{\cal D}^{m}}{\textstyle{\alpha_{M}({\cal D}^{n})}}$ induced by the $I$ submatrix have order 1, hence, are trivial submodules. The $m-k$ copies of $0$ added to the list $\{a_{l + 1}, \ldots, a_{k}\}$ induce the torsion free summand of $\frac{\textstyle{\cal D}^{m}}{\textstyle{\alpha_{M}({\cal D}^{n})}}$.

We will devote the rest of this section to the following three tasks:

1. Find a practical way to calculate these invariant factors.
2. Show that each matrix $M_{m \times n} \in {\cal M}\!\!\!\!\!\!\!{\cal M}_{m \times n}({\cal D})$ is equivalent to an extended diagonal matrix (Definition 78). As a byproduct, we show that a square matrix over a P.I.D. ${\cal D}$ is invertible if and only if it is a product of a sequence of elementary and secondary matrices. From this, it is then easy to deduce that two matrices $M$ and $N$ are equivalent if and only if $M$ may be transfigured to $N$ by a sequence of elementary and secondary operations.
3. Show that the extended diagonal matrix which a matrix is equivalent to is unique. And the computed invariant factors are the diagonal entries of the diagonal submatrix which is the basic building block of the extended diagonal matrix.