linear independence in GF(3)

Question

I have a question regarding linear independence in finite fields, more specifically the GF(3)-field.

My textbook claims that

$ (1,2,0)^T,(2,1,0)^T,(0,0,1)^T, $

are linearly independent wrt. the field $\mathbb{R}$, but not wrt. GF(3). I've googled myself to death on the definition of finite fields, but I can't seem to find an answer. Obviously the three vectors are lienarly independent wrt. $\mathbb{R}$, but why this doesn't mean that they are wrt. GF(3) I don't understand, GF(3) is a subfield of $\mathbb{R}$? I guess the definition of inner products on GF(3) are different somehow, but I can't find anything sensible online.

Thank

Sorry it was supose to be the GF(3) field – 1233023 Aug 16 '17 at 14:11 — 1233023, Aug 16 '17 at 14:11

Dietrich Burde · Answer 1 · 2017-08-16T14:17:35.413

First you really need to google the field $GF(2)$ with two elements. It is sometimes defined by $\mathbb{Z}/2$, and then $(1,2,0)$ just denotes the class modulo $2$, i.e., the vector $(1,0,0)$. You may say, $2=0$ in $GF(2)$. Obviously we then just have the standard basis. If you also google the characteristic of a field, it will be clear that $GF(2)$ cannot be a subfield of the real numbers.

Edit: For the new question: also for $K=GF(3)$ you can use the result that the vectors $x_1,x_2,x_3$ are linearly independent over $K$ if and only if the determinant of the matrix $A=(x_1,x_2,x_3)$ with colums vectors has nonzero determinant over $K$. Now $\det(A)=-3=0$ over $K$, and hence this vectors do not form a basis.

Ethan Bolker · Answer 2 · 2017-08-16T14:18:43.567

1

GF($2$) is not a subfield of $\mathbb{R}$, it's the two element field $\{0,1\}$ with multiplication mod $2$. Arithmetic with $1$ and $0$ works as you'd expect except for the fact that $1+1 = 0$.

If you do all your scalar computations with that arithmetic you'll see the answer to your question.

Edit:

The OP changed the question, asking about GF($3$) instead of GF($3$). The answer above needs the appropriate generalization, which will work for any prime $p$. His comment below suggests that he now understands everything he needs to.

edited Aug 16 '17 at 14:18

answered Aug 16 '17 at 14:02

Ethan Bolker

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Thanks, just found the definition. Then everything is a lot clearer. – 1233023 Aug 16 '17 at 14:05
I used the modulo definition of wiki. with this definition of the product, I find $2(1,2,0)T+2(2,1,0)^T+2*(0,0,1)^T = 0$ – 1233023 Aug 16 '17 at 14:09
@DietrichBurde The joys of cut-and-paste and no proofreading. I won't fix it so your comment will stay relevant. – Ethan Bolker Aug 16 '17 at 14:23
Well, my comment is, I fear, not so relevant...I was just enjoying your nice answer. – Dietrich Burde Aug 16 '17 at 14:27

score 0 · Answer 3 · answered Aug 16 '17 at 14:42

It is obvious if you write the elements of $\mathbf F_3$ as $\{0,1, -1\}$. In this case the three vectors are $$\begin{bmatrix}\phantom-1\\-1\\\phantom-0\end{bmatrix},\begin{bmatrix}-1\\\phantom-1\\\phantom-0\end{bmatrix},\begin{bmatrix}0\\0\\1\end{bmatrix}.$$

linear independence in GF(3)

3 Answers3