I have a question about a particular step of the invariant factor decomposition algorithm in Dummit and Foote on page 480. My question is about step $2$. Here is the general set up and the first three steps (paraphrased):
Let $V$ be an $F[x]$-module with vector space basis $E=\{e_1,...,e_n\}$. Let $T$ be a linear transformation of $V$ to itself defined by $x$ and let $A$ be the $n \times n$ matrix associated to $T$ and this choice of basis for $V$.
Step $1$: Use the following three elementary row and column operations to diagonalize the matrix $xI-A$ over $F[x]$.
a) Interchange two rows or columns.
b) Add a multiple (in $F[x]$) of one row or column to another.
c)Multiply any row or column by a unit in $F[x]$
Step $2$: Beginning with the generators $\{e_1,...,e_n\}$, for each row operation used in $1$, change the set of generators by the following rules:
a) If the $i$th row is interchanged with the $j$th row then interchange the $i$th and $j$th generators.
b) If $p(x)$ times the $j$th row is added to the $i$th row, then subtract $p(x)$ times the $i$th generator from the $j$th generator.
c) If the $i$th row is multiplied by the unit $u \in F$ then divide the $i$th generator by $u$.
At this point we have obtained a modification of the basis $E$. Namely, we have a set of $F[x]$-linear combinations of $\{e_1,...,e_n\}$.
In step $3$ we use $T$ to convert these $F[x]$-linear combinations into $F$-linear combinations of $\{e_1,...,e_n\}$, which turn out to be generators for the cyclic factors of the decompositions of $V$. So it is easy enough to understand why step $2$ was done, but what exactly was going on in step $2$? The book says nothing.
The "then" parts in step two correspond to places where if you had done this to the generators, then some substep of step 1 would not have been needed. For example, consider 2a: Say as one substep you had to exchange rows 3 and 7. If $e_3$ and $e_7$ had been exchanged in the generator set, then in the diagonalization in step 1 you would not have needed to exchange rows 3 and 7.
So if you do all the things prescribed for step 2, you end up with a basis in which you could have done the entire step 1 iwithout doing any operations; in other words, in this basis $xI-A$ was already diagonal.