2025-11-17

This following is a summary of what happened in class on 2025-11-17.

We spent most of the class period adding additional details and insights to our previous explicit example of computing the Smith normal form (and hence invariant factors) for a given $3 \times 3$ matrix, $A$ . This time, we explained how the standard basis ${e_{1}, e_{2}, e_{3}}$ for $Q^{3}$ in our example corresponded to a surjection

π : F_{Q [x]} ({x_{1}, x_{2}, x_{3}}) \to Q^{3}

from the free $Q [x]$ -module on the three-element set ${x_{1}, x_{2}, x_{3}}$ , namely the map that sent $x_{i} \mapsto e_{i}$ . The kernel of this surjection consisted of all formal $Q [x]$ -linear combinations of $x_{1}, x_{2}, x_{3}$ that $π$ sent to $0$ , i.e.,

\ker (π) = {p_{1} (x) x_{1} \oplus p_{2} (x) x_{2} \oplus p_{3} (x) x_{3} ∣ p_{1} (x) e_{1} + p_{2} (x) e_{2} + p_{3} (x) e_{3} = 0} .

We noted that it is easy to write down some elements in $\ker (π)$ , as follows. We first noted that $x \cdot e_{1} = 2 e_{1}$ , since $x$ acted as multiplication by $A$ when we wrote our vectors with respect to the standard basis. It followed that $(x - 2) e_{1} = 0$ , and so one element in $\ker (π)$ was the element

y_{1} := (x - 2) x_{1} .

Similarly, we found two more elements in $\ker (π)$ , namely the elements

\begin{aligned} y_{2} & := 2 x_{1} \oplus (x - 3) x_{2} \\ y_{3} & := - 14 x_{1} \oplus 7 x_{2} \oplus (x - 2) x_{3} . \end{aligned}

It turned out that these three elements were a basis for $\ker (π)$ . It followed that the $Q [x]$ -module inclusion of $\ker (π)$ into $F_{[x]} ({x_{1}, x_{2}, x_{3}})$ could be described as multiplication by the matrix

[\begin{matrix} x - 2 & 2 & - 14 \\ 0 & x - 3 & 7 \\ 0 & 0 & x - 2 \end{matrix}] = x I_{3} - A,

with respect the bases ${y_{1}, y_{2}, y_{3}}$ for $\ker (π)$ and ${x_{1}, x_{2}, x_{3}}$ for $F_{[x]} ({x_{1}, x_{2}, x_{3}})$ . Indeed, this matrix simply gave the image of each basis element $y_{i}$ under its inclusion into the free module, written in terms of the basis ${x_{1}, x_{2}, x_{3}}$ for that free module.

Now, column operations on the above matrix corresponded in the obvious way to changing the basis ${y_{1}, y_{2}, y_{3}}$ for $\ker (π)$ . For example, swapping Column 1 and Column 2 in the above matrix was equivalent to swapping the basis elements $y_{1}$ and $y_{2}$ . (Note that bases should really be viewed as ordered lists.)

Row operations on the above matrix were related to changing the basis ${x_{1}, x_{2}, x_{3}}$ , but in a slightly more convoluted way. For example, scaling Row 1 by $\frac{1}{2}$ in the above matrix corresponded to replacing $x_{1}$ with $x_{1}^{'} := 2 x_{3}$ in our basis. (The exact details are described in our online notes.)

In any case, this correspondence explained why the Smith Normal form of the above matrix amounted to finding new bases ${y_{1}^{'}, y_{2}^{'}, y_{3}^{'}}$ and ${x_{1}^{'}, x_{2}^{'}, x_{3}^{'}}$ for $\ker (π)$ and the free module, respectively, such that $y_{i}^{'} = a_{i} (x) x_{i}^{'}$ with $a_{1} (x), a_{2} (x), a_{3} (x)$ monic polynomials satisfying $a_{1} (x) | a_{2} (x) | a_{3} (x)$ . This was exactly what we needed to conclude that

V ≃ F_{Q [x]} ({x_{1}, x_{2}, x_{3}}) / \ker (π) ≃ Q [x] / ⟨ a_{1} (x) ⟩ \oplus Q [x] / ⟨ a_{2} (x) ⟩ \oplus Q [x] / ⟨ a_{3} (x) ⟩ .

In our specific example, we found $a_{1} (x) = 1$ , $a_{2} (x) = x - 2$ and $a_{3} (x) = (x - 2) (x - 3) = x^{2} - 5 x + 6$ , and so

V ≃ Q [x] / ⟨ x - 2 ⟩ \oplus Q [x] / ⟨ x^{2} - 5 x + 6 ⟩ .

We explained how one can also obtain the change-of-basis matrix that will conjugate $A$ to $R$ . See our online notes for details.

Next class: the Jordan canonical form!

Concepts

Rational Canonical Form III - Computation

References

Dummit & Foote, Abstract Algebra: Section 12.2