There appear to be a number of rational canonical forms. The best thing about standards is how many there are to choose from. However, the standard I choose seems to have a centralizer that is difficult to describe.



> 
> 1) Is there a reference that chooses a specific (hopefully pretty) rational canonical form, and computes its (hopefully pretty) centralizer?
> 
> 
> 


Alternatively,



> 
> 2) Is there a pretty description of the centralizer of my choice of canonical form?
> 
> 
> 


Matrices are over commutative (usually finite prime) fields.


Every matrix is similar to a direct sum of matrices whose minimal polynomial is of the form irr ^ pow, for some irreducible polynomial irr and some positive integer pow. While there is some disagreement on how to organize this, a common idea is to have canonical forms associated to pairs [ irr, pow ], (the other being to group coprime irr together: is it Z/2Z × Z/3Z or is it Z/6Z? we choose Z/2Z × Z/3Z).


So given a pair [ irr, pow ], I have seen two main ways to associate the canonical block: either take "the" companion matrix of irr^pow, or take a block diagonal matrix with pow blocks equal to the companion matrix of irr, and then fill in 1s on the sub/sup diagonal you used for the companion matrix. The former more clearly lays out an indecomposable direct decomposition, but it hides a composition series. The latter more subtly lays out the direct sum decomposition, but makes the composition series very clear. Both are quite pretty. In case irr has degree 1, then the latter is a Jordan block, and the former is not.


So I chose the latter. For instance, if irr = x2−x−1 is irreducible and pow = 3, then we get the block B:


`B = $\begin{bmatrix}
.&1&.&.&.&.\\%
1&1&1&.&.&.\\%
.&.&.&1&.&.\\%
.&.&1&1&1&.\\%
.&.&.&.&.&1\\%
.&.&.&.&1&1\\%
\end{bmatrix}$


If α is a root of x2−x−1, then I expected this guy to be the blow up of b:


`b = $\begin{bmatrix}
\alpha&1&.\\
.&\alpha&1\\
.&.&\alpha\\
\end{bmatrix}$


and so I expected any scalar matrices (blown up) to be in the centralizer of B, since they are in the centralizer of b. In other words, the matrix a:


`a = $\begin{bmatrix}
\alpha&.&.\\
.&\alpha&.\\
.&.&\alpha\\
\end{bmatrix}$


centralizes b, so I expected the matrix A:


`A = $\begin{bmatrix}
.&1&.&.&.&.\\%
1&1&.&.&.&.\\%
.&.&.&1&.&.\\%
.&.&1&1&.&.\\%
.&.&.&.&.&1\\%
.&.&.&.&1&1\\%
\end{bmatrix}$


to centralize B, but of course AB ≠ BA. For any particular B, one can just solve a bunch of equations, but at least to me the solutions do not seem easy to describe algorithmically. I worry that we might have wanted the slightly uglier canonical form that is actually the blowup of b:


$\tilde B = \begin{bmatrix}
.&1&1&.&.&.\\%
1&1&.&1&.&.\\%
.&.&.&1&1&.\\%
.&.&1&1&.&1\\%
.&.&.&.&.&1\\%
.&.&.&.&1&1\\%
\end{bmatrix}$


However, I have not found this form listed in any reference, and I'd like to have a reasonable reference to point to to justify my choices (especially if they choose uglier, less sparse matrices).



> 
> 1′) What reference uses $\tilde B$ as the rational canonical form of B?
> 
> 
> 


Or alternatively:



> 
> 2′) Where are the "scalars" in the centralizer of B?
> 
> 
> 


I'd much prefer to use B, but if I cannot even see the scalar matrices in this form, then it seems like a very poor form indeed.


For those curious the other leading canonical form of B is:


$\hat B = \begin{bmatrix}
.&1&.&.&.&.\\%
.&.&1&.&.&.\\%
.&.&.&1&.&.\\%
.&.&.&.&1&.\\%
.&.&.&.&.&1\\%
1&3&0&-5&0&3\\%
\end{bmatrix}$


It is pretty, but also non-obvious what its composition factors are. This form (along with the Z/6Z style grouping of irreducible factors) is used by GP/Pari. Transposes and alternate groupings of factors allow for a wide variety of canonical forms.


The best answer will address the ease of explicitly writing down generators of the centralizer of B over a finite prime field given just (irr,pow), or will address both writing down the canonical form from (irr,pow) and the centralizer. An inductive answer might prefer to allow non-prime fields, but added generality at the cost of clarity and explicit algorithms is not useful to me.


**Edit:** Both B and $\hat B$ have the nice property that given a generator w for the indecomposable module acted on by the original operator M, a basis realizing the matrix is easy to find. For B this is vi = w⋅Mj⋅irr(M)k where i−1 = k⋅deg(irr) + j and 0 ≤ j < deg(irr), for i = 1, 2, … deg(irr)⋅pow. For $\hat B$ this is vi = w⋅Mi−1, for i = 1, 2, … deg(irr)⋅pow.


In order to use $\tilde B$, I need a similar explicit understanding of the basis defining it.



> 
> 1″) How does one express a basis of k[x]/(irr^pow) in terms of the coset of 1 such that the operator corresponding to x has the form $\tilde B$, that is, block Toeplitz with the diagonal the companion matrix of irr, the super-diagonal an identity, and the other diagonals 0?
> 
> 
> 


I am of course still curious about 2′. I think an answer to 1″ would both provide an answer for 2′ and be enough, even without an explicit reference, to finish Robin Chapman's answer to 1.



Centralizers in GL(n,p)
For a start the accepted usage for "rational canonical form" in the literature
is for a diagonal sum $C(f\_1)\oplus C(f\_2)\oplus\cdots\oplus C(f\_k)$ where
$C(f\_i)$ is the companion matrix for a monic polynomial $f$ and
$f\_1\mid f\_2\mid\cdots\mid f\_k$. That said, if I needed to find a centralizer
explcitly it isn't the canonical form I would choose.


As ever we should think of $V=k^n$ as a $k[X]$-module where $X$ acts via $A$.
If the minimum polynomial of $A$ is $u\_1^{a\_1}\cdots u\_k^{a\_k}$ with the
$p\_i$ distinct irreducibles then $V$ splits *uniquely* into a direct sum of submodules
$M\_1\oplus\cdots\oplus M\_k$, where $M\_i$ is annihilated by a power of $u\_i$.
Both $A$ and ts centralizer fix each $M\_i$, so we may
reduce to the case where the minimum polynomial $u^a$ with $u$ irreducible.


Let $u$ have degree $r$ and let $C\in M\_r(k)$ be the companion matrix for $u$
or any conjugate for it. I claim that $A$ is conjugate to a diagonal sum of
``Jordan-style blocks'' like
$$J=\left(\begin{matrix}
C&I&O&O\\\
O&C&I&O\\\
O&O&C&I\\\
O&O&O&C
\end{matrix}\right).$$
This is just a question of checking this has the minimum polynomial $u^a$.
As we are working in $GL(n,p)$ then $C\ne0$ and over a finite field then the diagonal
sum of copies of $C$ is a polynomial in $A$ (indeed $A^{p^rs}$ for suitable $s$).
Thus each matrix in the centralizer of $A$ has a block decomposition into $r$-by-$r$
blocks which commute with $C$ and so are polynomials in $C$. So,
finding the centralizer of $A$ is equivalent to finding the centralizer of the matrix
$A'$ over $k'=k(\alpha)$ where $\alpha$ is a zero of $u$ and $A'$ is obtained by
replacing the above Jordan-style blocks by standard Jordan blocks
$$J'=\left(\begin{matrix}
\alpha&1&0&0\\\
0&\alpha&1&0\\\
0&0&\alpha&1\\\
0&0&0&\alpha
\end{matrix}\right).$$
The centralizer of $A'$ is the same as that of $A'-\alpha I$ which is
a very sparse matrix. At this stage I would just find the centralizer in the matrix
algebra by explicit calculation and extract the nonsingular elements as the centralizer
in the matrix group.


**Added** (6/6/2010) I claimed that $J$ had the minimum polynomial $u^a$.
Let $k'=k(\alpha)$. The matrix $J'$ gives the action of $x$ on a standard
$k'$-basis for the $k'[x]$-module $k'[x]/((x-\alpha)^a)$. Therefore $J$
gives the action of $x$ on a $k$-basis for $k'[x]/((x-\alpha)^a)$ considered as a
$k[x]$-module. To see that $J$ has minimum polynomial $u^a$ it suffices
to show that some element of $k'[x]/((x-\alpha)^a)$ has annihilator $u^a$
over $k[x]$. But the element $1$ has annilhator $(x-\alpha)^a k'[x]$
over $k'[x]$ and so has annihilator
$k[x]\cap (x-\alpha)^a k'[x]=u^a k[x]$ over $k[x]$,
**provided** the extension $k'/k$ is separable (so that $x-\alpha$ is not a repeated
factor of $u$).


So the assertion holds for finite fields, bt not necessarily for general
fields of prime characteristic. It would be interesting to work out the
details for inseparable irreducible polynomials.


