$Count number of matrices over finite field with certain rank and zeros on the diagonal$

Question

$Count number of matrices over finite field with certain rank and zeros on the diagonal$

Count the number of $m \times n$ (where $m \leq n$) matrices over the finite field $\mathbb{F}_q$ with rank $i$ and all zeros in the main diagonal. I think it might be simple, but there may be some tedious calculations involved.

Linear Algebra Algebra

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M F H

+1

Yes it will be tedious mainly because of the "rank i" requirement. Already each of i=0, i=1 and i=m require (IMO) a separate, distinct and nontrivial treatment. And for a generic 1 < i < m it will get even more messy ...
- M F H
  
  +1
  
  PS: OK, the case rank i=0 is trivial, it must be the all zero matrix so there's only one! :-)
Kr1Staps

+2

Actually the rank i condition is the easy part. See Proposition 2 of Two Remarks on Graded Nilpotent Classes by Zelevinsky; the formula for the number of rank i m-by-n matrices over Fq is just a special case of this result.
- M F H
  
  +1
  
  That sounds interesting, didn't know that reference. But it defines "M(V) the set of matrices with nonnegative entries..." and I don't see how that could relate to matrices over finite fields.
Kr1Staps

+1

M(V) is just a set of parameters classifying the graded nilpotent orbits. That is, the actual space of matrices is what he calls E(V) in the beginning, and the Aut(V)-orbits on E(V) are written as X_M for in M(V). When V =V_0 + V_1 has only two non-zero graded pieces, E(V) is the space of mxn matrices, and the orbits X_M are stratified by rank. Thus when you unpack Corollary 2 in this case, it counts the the number of mxn matrices of a given rank over Fq.
- Ribs
  
  0
  
  Thank you for your reply! Would you mind elaborating a bit: Why the vector space (let's say (F_q)^n) has only two graded pieces, and why is E(V) just mxn matrices in this case? Also I assume the action of Aut(V) is the following? for each linear functional f in E(V), g in Aut(V) acts by taking f(x) \mapsto f(g(x)), for each x in V? Thanks in advance!
Kr1Staps

0

This might have to be spread out over several comments due to the length restriction. Letting V = V_0 + V_1, then E(V) is the space of all linear operators N : V --> V such that N(V_0) is contained in V_1 and N(V_1) is the 0-vector space. Choosing a basis e_1, ..., e_n for V_0 and f_1, ..., f_m for V_1 gives us a basis for V. Since N in E(V) has to map V_0 into V_1, it must be that N(e_i) =0 for all i. If you think about what this means for the matrix of N in the basis e_1, ..., f_m,
Kr1Staps

0

SORRY! I actually got that backwards, rather that means N(e_i) is a linear combination of the f_j and so the (i, j) entries of the matrix of N when 1 <= i <= n and 1<= j <= m are all 0. When 1 <= i <=n and m+1 < = j <= m+n, then you can have non-zero entries (and any choice of non-zero entries works). It's the condition that N(V_1) = 0 that forces the *last* m columns to be 0. In other words, looking at N in the basis e_1, ..., f_m, one obtains an (m+n)-by-(m+n) matrix
Kr1Staps

0

Where the only non-zero entries are in the lower left m-by-n corner. (And all choices of non-zero entries in the lower m-by-n corner defines an element of E(V)). Therefore, E(V) is isomorphic to the space of all m-by-n matrices where m = dim (V_1) and n = dim(V_0). Note that by Aut(V) Zelevinsky means GL(V_0)x GL(V_1) (viewed as embedded in GL(V)). So an element of Aut(V) realized as a matrix wrt. to the basis e_1, ..., f_m if going to be a block-diagonal matrix [[A, 0], [0, B]] where
Kr1Staps

0

A in GL(V_0) and B in GL(V_1). The action is given by conjugation on N, which recall that by the above, we can write N = [[0, 0], [X, 0]] for some X an m-by-n matrix, and thus when you conjugate, you find that the action of [[A, 0], [0, B]] on N is [[0, 0], [AXB^{-1}, 0]]. One can then prove (or unpack Zelevinsky's formulas to see that) the orbits are parameterized by the rank of X, and then apply the point-counting proposition.
- Ribs
  
  0
  
  Thank you so much! Feel free to claim the bounty from the other question, I can even give you more if you want!
M F H

0

The page had disappeared, I could not access it any more, I don't know why :-(

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M F H

+1

Yes it will be tedious mainly because of the "rank i" requirement. Already each of i=0, i=1 and i=m require (IMO) a separate, distinct and nontrivial treatment. And for a generic 1 < i < m it will get even more messy ...

M F H

+1

PS: OK, the case rank i=0 is trivial, it must be the all zero matrix so there's only one! :-)
M F H

+1

PS: OK, the case rank i=0 is trivial, it must be the all zero matrix so there's only one! :-)
Kr1Staps

+2

Actually the rank i condition is the easy part. See Proposition 2 of Two Remarks on Graded Nilpotent Classes by Zelevinsky; the formula for the number of rank i m-by-n matrices over Fq is just a special case of this result.

M F H

+1

That sounds interesting, didn't know that reference. But it defines "M(V) the set of matrices with nonnegative entries..." and I don't see how that could relate to matrices over finite fields.
M F H

+1

That sounds interesting, didn't know that reference. But it defines "M(V) the set of matrices with nonnegative entries..." and I don't see how that could relate to matrices over finite fields.
Kr1Staps

+1

M(V) is just a set of parameters classifying the graded nilpotent orbits. That is, the actual space of matrices is what he calls E(V) in the beginning, and the Aut(V)-orbits on E(V) are written as X_M for in M(V). When V =V_0 + V_1 has only two non-zero graded pieces, E(V) is the space of mxn matrices, and the orbits X_M are stratified by rank. Thus when you unpack Corollary 2 in this case, it counts the the number of mxn matrices of a given rank over Fq.

Ribs

0

Thank you for your reply! Would you mind elaborating a bit: Why the vector space (let's say (F_q)^n) has only two graded pieces, and why is E(V) just mxn matrices in this case? Also I assume the action of Aut(V) is the following? for each linear functional f in E(V), g in Aut(V) acts by taking f(x) \mapsto f(g(x)), for each x in V? Thanks in advance!
Ribs

0

Thank you for your reply! Would you mind elaborating a bit: Why the vector space (let's say (F_q)^n) has only two graded pieces, and why is E(V) just mxn matrices in this case? Also I assume the action of Aut(V) is the following? for each linear functional f in E(V), g in Aut(V) acts by taking f(x) \mapsto f(g(x)), for each x in V? Thanks in advance!
Kr1Staps

0

This might have to be spread out over several comments due to the length restriction. Letting V = V_0 + V_1, then E(V) is the space of all linear operators N : V --> V such that N(V_0) is contained in V_1 and N(V_1) is the 0-vector space. Choosing a basis e_1, ..., e_n for V_0 and f_1, ..., f_m for V_1 gives us a basis for V. Since N in E(V) has to map V_0 into V_1, it must be that N(e_i) =0 for all i. If you think about what this means for the matrix of N in the basis e_1, ..., f_m,
Kr1Staps

0

SORRY! I actually got that backwards, rather that means N(e_i) is a linear combination of the f_j and so the (i, j) entries of the matrix of N when 1 <= i <= n and 1<= j <= m are all 0. When 1 <= i <=n and m+1 < = j <= m+n, then you can have non-zero entries (and any choice of non-zero entries works). It's the condition that N(V_1) = 0 that forces the *last* m columns to be 0. In other words, looking at N in the basis e_1, ..., f_m, one obtains an (m+n)-by-(m+n) matrix
Kr1Staps

0

Where the only non-zero entries are in the lower left m-by-n corner. (And all choices of non-zero entries in the lower m-by-n corner defines an element of E(V)). Therefore, E(V) is isomorphic to the space of all m-by-n matrices where m = dim (V_1) and n = dim(V_0). Note that by Aut(V) Zelevinsky means GL(V_0)x GL(V_1) (viewed as embedded in GL(V)). So an element of Aut(V) realized as a matrix wrt. to the basis e_1, ..., f_m if going to be a block-diagonal matrix [[A, 0], [0, B]] where
Kr1Staps

0

A in GL(V_0) and B in GL(V_1). The action is given by conjugation on N, which recall that by the above, we can write N = [[0, 0], [X, 0]] for some X an m-by-n matrix, and thus when you conjugate, you find that the action of [[A, 0], [0, B]] on N is [[0, 0], [AXB^{-1}, 0]]. One can then prove (or unpack Zelevinsky's formulas to see that) the orbits are parameterized by the rank of X, and then apply the point-counting proposition.

Ribs

0

Thank you so much! Feel free to claim the bounty from the other question, I can even give you more if you want!
Ribs

0

Thank you so much! Feel free to claim the bounty from the other question, I can even give you more if you want!
M F H

0

The page had disappeared, I could not access it any more, I don't know why :-(

$Count number of matrices over finite field with certain rank and zeros on the diagonal$

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