I'm doing some review for linear algebra. I have some ideas about some of the questions but I'm not sure if I'm going about it in the wrong direction. The question is as follows:

Let F be the field of the integers modulo 5. Consider the vector space V defined as

V = {$\begin{bmatrix}a\\b\\c\end{bmatrix} : a,b,c\in F $}

(a) How many vectors are there in V?

(b) How many vectors are there in subspaces of dimension 2?

(c) List all of the vectors spanned by $([1], [2], [3])$

(d)What vectors (a, b, c) $\in$ V are solutions to $[1]a+[2]b+4[c] = [0]$?

(e) Show that every two distinct subspaces of dimension 2 intersect in a subspace of dimension 1

My Thoughts

(a) I'm not sure if the question is asking for the total number of possible vectors or the number of vectors in the basis. For total possible vectors, as a, b, c can only be an integer between 0 - 4, there are 5x5x5 = 125 vectors. For vectors in the basis, there are 3, they are the unit vectors.

(b) I'm having trouble understanding what the basis for subspaces of dimension 2 would look like for this example. I understand that there are two vectors in the basis for each subspace but that's as far as I get.

(c) The vectors spanned by $([1], [2], [3])$ should be the multiples of it from 1 - 5. Ex. $(4[1], 4[2], 4[3]) = ([4], [3], [2])$ This would be done for multiples of 1 - 5.

(d) I'm assuming that the sum will be moduled. Ex. By setting a = 1, b = 2 and c = 0, $[1]a + [2]b + [4]c = [1]1 + [2]2 + [4]0 = 5 = 0$ due to being integers modulo 5. Please tell me if this is wrong.

(e) I have no idea how to begin. Any help will be greatly appreciated!


1 Answer 1


For a), they're asking for number of vectors, which you've already found is $125$.

For b), when you have two linearly independent vectors $v_1, v_2$ as a basis, you can take any combination $av_1 + bv_2$ for $a, b \in \{ 0, 1, 2, 3, 4 \}$, and they will all be distinct - if any two are the same, you get some nonzero linear combination of $v_1$ and $v_2$ being $0$, contradicting linear independence. So we conclude there are $25$ elements in this subspace.

Your answer for c) looks good.

For d), we’re looking for elements in the plane $a + 2b + 4c =0$. This is a subspace (check this yourself as an exercise) so we all we need to do is find two basis vectors, which will let you generate all elements.

For e), we first claim that any two subspaces of dimension $2$ have to intersect. Indeed, we can easily find a basis of $V$ with three elements, so $V$ has dimension $3$; this means that any two subspaces of dimension $2$ must intersect. Suppose they intersect in vector $a$. Then $\{0, a, 2a, 3a, 4a\}$ is a subspace of dimension $1$ in both subspaces, as required.

  • $\begingroup$ Thank you for the clarification! For (d) I assumed that since a,b,c are elements of F, they should be integers. Is that wrong? That being said, what you stated makes a lot more sense in general. $\endgroup$
    – monkey1009
    Jun 12, 2019 at 1:59
  • $\begingroup$ @monkey1009 Ah, I misread the question. I’ve edited my answer, my bad. $\endgroup$
    – auscrypt
    Jun 12, 2019 at 2:25

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