linear algebra and subspaces question?

please help:

find the number of 1-dimensional subspaces in the field Z_p^n, which consists of all possible n length vectors, composed of values from the set {0,1,2 ...p-1 } where p = a prime number. (hint: any nonzero vector generates a 1-dimensional subspace, but some vectors generate the same space)

thanks so much

2 Answers

  • JB
    Lv 7
    1 decade ago
    Favorite Answer

    Let V be the vector space whose vectors are n-tuples from the field with p elements, Z/pZ. The scalars that you can multiply these vectors by also come from the field Z/pZ.

    V has p^n elements since each vector has n positions that can independently filled in p ways.

    There are p^n-1 non-zero vectors in V, pick one and call it v. The subspace generated by v consists of {0, v, 2v, 3v, ..., (p-1)v }. Each non-zero vector in this subspace generates the same subspace. This uses up p-1 of the p^n-1 non-zero vectors in v. Each time we pick a new non-zero vector not in any of the subspaces generated so far we can generate a new subspace containing p-1 non-zero elements. Repeat, until no non-zero elements remain. We have partitioned the non-zero elements of V into pairwise disjoint subsets each containing p-1 elements, each being the non-zero elements of a subspace. So, to count the subspaces we only need to divide p^n - 1 by p-1:

    number of one dimensional subspaces of V = (p^n-1)/(p-1).

    Comments: (p^n-1)/(p-1) = 1 + p + p^2 + ... + p^(n-1) since this is the formula for the sum of a finite geometric series. I don't know if there is a counting argument that gives the right side of this equation directly. Does anyone? In any case this is comforting since the right side is clearly an integer and that is not ab initio obvious of the left side.

    Also, it is important to realize that the ring Z/p^nZ is not a field since it has zero divisors, so that is not what this problem is about in any way.

  • Anonymous
    4 years ago

    a=a million provides vector [3,3,13] = u, a=-a million provides vector [-3,3,-13] = v u+v = [0,6,0], which you may properly be in U we want 0=3a and 6=3a^4 and 0=13a for the comparable quantity a. not a danger. So U isn't closed under vector addition, and for this reason isn't a subspace.

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