(I took courses on linear algebra, but I don't know anything about $R$-modules or such things.)

  1. How do you define the rank of a matrix whose entries are polynomials in $K[X]$?
  2. If you assign some element of $K$ in the entries of such a matrix, what is the rank of the produced matrix (in $M_{mn}(K)$)? Is it larger, equal, or smaller than that of the original matrix?

EDIT: Here $K$ denotes an arbitrary field, but mostly I'm interested in $\mathbb{R}$ and $\mathbb{F}_p$.

  • $\begingroup$ Given $R = K[X]$ and $A \in R^{n\times n}$, what do you mean by "assign" some element of $K$ in the entries of $A$? $\endgroup$ – user2468 Jan 27 '12 at 15:19
  • $\begingroup$ I mean, take some $(b_{ij})\in M_{n}(K)$, and get a matrix $(a_{ij}(b_{ij}))$. $\endgroup$ – Pteromys Jan 28 '12 at 0:15
  1. You could use the Smith normal form, a generalization of Gaussian elimination for PID.

  2. You can expect the rank to be at most the rank of the original matrix.

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  • $\begingroup$ FWIW the Smith normal form is defined over PIRs are well: Irving Kaplansky. Elementary divisors and modules. Transactions of the American Mathematical Society, 66(2):464–491, 1949. Also the book Linear Algebra Over Commutative Rings by Bernard R. McDonald contains some material on Smith normal forms. $\endgroup$ – user2468 Jan 27 '12 at 14:09
  • $\begingroup$ I understand that the rank of a matrix in $M_{mn}(K[X])$ is the number of ones in the SNF. How do you prove the rank of the assigned matrix is at most the rank of the original matrix? $\endgroup$ – Pteromys Jan 27 '12 at 14:35
  • $\begingroup$ @Pteromys because by setting $x=a$ you may turn some columns into zeros. Consider for instance the diagonal matrix with entries $x$ and then set $x=0$. $\endgroup$ – lhf Jan 27 '12 at 15:22
  • $\begingroup$ @lhf is this argument valid? Let $R = K[x], A \in R^{n\times n}$. Let $f \in (K[x])[y]$ be minpoly of $A$. $rank(A)$ is the $y$-degree of $f$. Then setting $x=a$ will not increase the $y$-degree of $f$? $\endgroup$ – user2468 Jan 27 '12 at 16:07
  • $\begingroup$ @J.D., I did not say it will decrease but that it may decrease. $\endgroup$ – lhf Jan 27 '12 at 18:14

I assume that you are taking $K$ to be an arbitrary field here? Generally, the best way to view an m by n matrix with entries in a ring $R$ is as a linear map from a finite-dimensional $R$-module to another.

$R$-modules are nothing to be afraid of; they're like a normal vector space but your scalars don't have to be from a field, they can be from a general ring (like $K[X]$ in this case).

The rank is then the dimension of the image of this linear map as an $R$-module. Hence it is the smallest number of elements $x_1,...,x_k$ such that everything in the image of the linear map can be written as $a_1x_1+\cdots+a_kx_k$ where $a_i$ are in $K[X]$.

EDIT: To make things simpler, you may as well assume that the coefficients are in the field of rational functions $K(X)$ (where the elements are quotients of regular polynomials). Then the image genuinely is a vector space, and all is well.

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  • $\begingroup$ Yes, by $K$ I mean an arbitrary field. How are modules different from vector spaces, when linear independence and bases are concerned? $\endgroup$ – Pteromys Jan 27 '12 at 13:35
  • 1
    $\begingroup$ Maximal linearly independent subsets are not necessarily spanning sets. $\;$ For all rings $R$ and non-negative integers $n$, the range of an $n$-by-$n$ matrix with entries in $R$ will be a finitely generated $R$-submodule of the torsion-free $R$-module $R^n$. $\;\;$ If $\: R = K[x] \:$ then $R$ is a Principal Ideal Domain, in which case there will exist a unique non-negative integer $m$ such that the submodule is isomorphic to the $R$-module $R^m$, and that $m$ will satisfy $\: m\leq n \:$. $\;\;\;$ $\endgroup$ – user57159 Jan 28 '12 at 0:44

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