2024-11-21

This following is a very brief summary of what happened in class on 2024-11-21.

We spent nearly the entire class period working through the matrix in this exercise, namely the matrix

A = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ - 2 & - 2 & 0 & 1 \\ - 2 & 0 & - 1 & - 2 \end{matrix}] .

We briefly talked about how the standard basis for $V = Q^{4}$ corresponds to a surjective $Q [x]$ -module morphism $π : F_{Q [x]} ({z_{1}, z_{2}, z_{3}, z_{4}}) \to V$ , where the first module is the free $Q [x]$ -module on four generators. This surjection establishes the $Q [x]$ -module isomorphism

V ≃ F_{Q [x]} ({z_{1}, z_{2}, z_{3}, z_{4}}) / \ker (π) .

Our main structure theorem for free $Q [x]$ -modules then tells us we should be able to find a new set of generators ${z_{1}^{'}, z_{2}^{'}, z_{3}^{'}, z_{4}^{'}}$ for that free module, together with $a_{1}, a_{2}, a_{3}, a_{4} \in Q [x]$ such that ${a_{1} z_{1}^{'}, \dots, a_{4} z_{4}^{'}}$ generate $\ker (π)$ . How do we find those generators?

First we find generators for $\ker (π)$ and express those generators in terms of the original generators ${z_{1}, \dots, z_{4}}$ for the free module. We can encode this information in a relations matrix. We noted that there are several very easy elements one can see in $\ker (π)$ . For example, from the matrix $A$ we know that

T (e_{1}) = e_{1} - 2 e_{3} - 2 e_{4},

where $T : V \to V$ is the linear transformation given by $A$ with respect to the standard basis. Since the action of $x$ on $V$ is given by $T$ , another way to express the above equation is

x \cdot e_{1} = e_{1} - 2 e_{3} - 2 e_{4},

or equivalently,

(x - 1) \cdot e_{1} + 2 e_{3} + 2 e_{4} = 0 .

By the construction of $π$ , this means that the element

y_{1} = (x - 1) z_{1} + 2 z_{3} + 2 z_{4}

is in the kernel of $π$ . By the same reasoning, the following additional three elements are in the kernel of $π$ :

\begin{aligned} y_{2} & = (x - 1) z_{2} + 2 z_{3} \\ y_{3} & = x z_{3} + z_{4} \\ y_{4} & = - z_{3} + (x + 2) z_{4} . \end{aligned}

One can prove that the set ${y_{1}, \dots, y_{4}}$ generates $\ker (π)$ . By construction, the relations matrix for these four generators is the matrix

[\begin{matrix} x - 1 & 0 & 2 & 2 \\ 0 & x - 1 & 2 & 0 \\ 0 & 0 & x & 1 \\ 0 & 0 & - 1 & x + 2 \end{matrix}] = x I_{4} - A^{⊤}

Note that this is exactly the transpose of the matrix $x I_{4} - A$ , which for unfortunate historical reasons is the matrix everyone uses to compute the Smith normal form. This means that row (resp., column) operations on $x I_{4} - A$ correspond to column (resp., row) operations on the relations matrix above. BOO.

In any case, we then detailed how modifying our generators for $\ker (π)$ correspond to row operations on the relations matrix, while modifying our generators for the free module correspond to column operations (but with a little twist) for the matrix above.

We then listed the row and column operations that transformed $x I_{4} - A$ into the matrix

[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & x - 1 & 0 \\ 0 & 0 & 0 & (x - 1) (x + 1)^{2} \end{matrix}] .

This corresponding to a new choice of generators ${z_{1}^{'}, z_{2}^{'}, z_{3}^{'}, z_{4}^{'}}$ for the free module such that ${z_{1}^{'}, z_{2}^{'}, (x - 1) z_{3}^{'}, (x - 1) (x + 1)^{2} z_{4}^{'}}$ is a set of generators for $\ker (π)$ . This allows us to conclude that

V ≃ F_{Q [x]} ({z_{1}, z_{2}, z_{3}, z_{4}}) / \ker (π) ≃ Q [x] / ⟨ x - 1 ⟩ \oplus Q [x] / ⟨ (x - 1) (x + 1)^{2} ⟩ .

This allowed us to conclude that the invariant factors of $A$ are $a_{1} (x) = x - 1$ and $a_{2} (x) = (x - 1) (x + 1)^{2} = x^{3} + x^{2} - x - 1$ .

We then discussed how to write down the rational canonical form, $R$ , of $A$ , as well as how to compute the change-of-basis matrix, $P$ , such that $R = P^{- 1} A P$ . That latter step involved first tracking the changes to the generators of the free module to produce an "auxiliary matrix"

P^{'} = [\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]

The third and four columns in that matrix told us $Q [x]$ -module generators for two invariant summands in $V$ , from which we were able to determine the corresponding $Q$ -basis (by acting on those generators by $x$ ).

We ended by briefly previewing the idea of the Jordan canonical form. We'll cover that form in detail in the next class.

Concepts

References

Dummit & Foote: Section 12.2