Probability that arbitrary rank-$w$ matrix $mathbf{M} in mathbb{F}_2^{n times m}$ splits in two rank-$(w /...











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For simplicity, assume $n$ and $w$ are even.




Let $mathbf{M} in mathbb{F}_2^{n times m}$ be an arbitrary matrix with $operatorname{rank} mathbf{M} = w$. What is the probability that we can write
$$
mathbf{M} = begin{bmatrix} mathbf{M}_1 \ mathbf{M}_2 end{bmatrix}text,
$$

where $mathbf{M}_1, mathbf{M}_2 in mathbb{F}_2^{(n / 2) times m}$ and $operatorname{rank} mathbf{M}_1 = operatorname{rank} mathbf{M}_2 = w / 2$?




What follows is what I have worked on so far.





There are (see 1)



$$
C := binom{n}{w}_2 prod_{i = 0}^{w - 1} left(2^m - 2^iright)
$$



choices for $mathbf{M}$, where $binom{n}{w}_2$ denotes the $2$-binomial coefficient.



Given some $mathbf{M}_2$ with $operatorname{rank} mathbf{M}_2 = w / 2$, the number of ways we can "extend" it (as per problem statement) by an $mathbf{M}_1$ and get a rank-$w$ matrix is (see Introduction in 2)



$$
C_1 := 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^{w / 2 + i}right)text.
$$



Analogously to $mathbf{M}$, there are



$$
C_2 := binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^iright)
$$



choices for $mathbf{M}_2$.



Therefore, the wanted probability is



$$
frac{C_1 C_2}{C} = 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2^2 binom{n}{w}_2^{-1}text.
$$



(The products cancel out.) Any hopes to verify this? Are you aware of related work that might help?










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    up vote
    0
    down vote

    favorite












    For simplicity, assume $n$ and $w$ are even.




    Let $mathbf{M} in mathbb{F}_2^{n times m}$ be an arbitrary matrix with $operatorname{rank} mathbf{M} = w$. What is the probability that we can write
    $$
    mathbf{M} = begin{bmatrix} mathbf{M}_1 \ mathbf{M}_2 end{bmatrix}text,
    $$

    where $mathbf{M}_1, mathbf{M}_2 in mathbb{F}_2^{(n / 2) times m}$ and $operatorname{rank} mathbf{M}_1 = operatorname{rank} mathbf{M}_2 = w / 2$?




    What follows is what I have worked on so far.





    There are (see 1)



    $$
    C := binom{n}{w}_2 prod_{i = 0}^{w - 1} left(2^m - 2^iright)
    $$



    choices for $mathbf{M}$, where $binom{n}{w}_2$ denotes the $2$-binomial coefficient.



    Given some $mathbf{M}_2$ with $operatorname{rank} mathbf{M}_2 = w / 2$, the number of ways we can "extend" it (as per problem statement) by an $mathbf{M}_1$ and get a rank-$w$ matrix is (see Introduction in 2)



    $$
    C_1 := 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^{w / 2 + i}right)text.
    $$



    Analogously to $mathbf{M}$, there are



    $$
    C_2 := binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^iright)
    $$



    choices for $mathbf{M}_2$.



    Therefore, the wanted probability is



    $$
    frac{C_1 C_2}{C} = 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2^2 binom{n}{w}_2^{-1}text.
    $$



    (The products cancel out.) Any hopes to verify this? Are you aware of related work that might help?










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      For simplicity, assume $n$ and $w$ are even.




      Let $mathbf{M} in mathbb{F}_2^{n times m}$ be an arbitrary matrix with $operatorname{rank} mathbf{M} = w$. What is the probability that we can write
      $$
      mathbf{M} = begin{bmatrix} mathbf{M}_1 \ mathbf{M}_2 end{bmatrix}text,
      $$

      where $mathbf{M}_1, mathbf{M}_2 in mathbb{F}_2^{(n / 2) times m}$ and $operatorname{rank} mathbf{M}_1 = operatorname{rank} mathbf{M}_2 = w / 2$?




      What follows is what I have worked on so far.





      There are (see 1)



      $$
      C := binom{n}{w}_2 prod_{i = 0}^{w - 1} left(2^m - 2^iright)
      $$



      choices for $mathbf{M}$, where $binom{n}{w}_2$ denotes the $2$-binomial coefficient.



      Given some $mathbf{M}_2$ with $operatorname{rank} mathbf{M}_2 = w / 2$, the number of ways we can "extend" it (as per problem statement) by an $mathbf{M}_1$ and get a rank-$w$ matrix is (see Introduction in 2)



      $$
      C_1 := 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^{w / 2 + i}right)text.
      $$



      Analogously to $mathbf{M}$, there are



      $$
      C_2 := binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^iright)
      $$



      choices for $mathbf{M}_2$.



      Therefore, the wanted probability is



      $$
      frac{C_1 C_2}{C} = 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2^2 binom{n}{w}_2^{-1}text.
      $$



      (The products cancel out.) Any hopes to verify this? Are you aware of related work that might help?










      share|cite|improve this question















      For simplicity, assume $n$ and $w$ are even.




      Let $mathbf{M} in mathbb{F}_2^{n times m}$ be an arbitrary matrix with $operatorname{rank} mathbf{M} = w$. What is the probability that we can write
      $$
      mathbf{M} = begin{bmatrix} mathbf{M}_1 \ mathbf{M}_2 end{bmatrix}text,
      $$

      where $mathbf{M}_1, mathbf{M}_2 in mathbb{F}_2^{(n / 2) times m}$ and $operatorname{rank} mathbf{M}_1 = operatorname{rank} mathbf{M}_2 = w / 2$?




      What follows is what I have worked on so far.





      There are (see 1)



      $$
      C := binom{n}{w}_2 prod_{i = 0}^{w - 1} left(2^m - 2^iright)
      $$



      choices for $mathbf{M}$, where $binom{n}{w}_2$ denotes the $2$-binomial coefficient.



      Given some $mathbf{M}_2$ with $operatorname{rank} mathbf{M}_2 = w / 2$, the number of ways we can "extend" it (as per problem statement) by an $mathbf{M}_1$ and get a rank-$w$ matrix is (see Introduction in 2)



      $$
      C_1 := 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^{w / 2 + i}right)text.
      $$



      Analogously to $mathbf{M}$, there are



      $$
      C_2 := binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^iright)
      $$



      choices for $mathbf{M}_2$.



      Therefore, the wanted probability is



      $$
      frac{C_1 C_2}{C} = 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2^2 binom{n}{w}_2^{-1}text.
      $$



      (The products cancel out.) Any hopes to verify this? Are you aware of related work that might help?







      combinatorics matrices finite-fields






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      edited 7 hours ago

























      asked 11 hours ago









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