Find the basis of $H = \{(x_1, x_2, x_3) \in \Bbb Z^3 | 6 \text{ divides } 2x_1+3x_2+4x_3\}$.

My attempt was to say that $$\{(3,0,0), (0,2,0), (0,0,3)\}$$ is the basis, since this automatically allows every element in $H$ to be divided by $6$. I do not know the answer, but apparently I'm wrong.

I would like some help on this problem, and a general method to solve these types of problems would be appreciated.

edit: Apparently, the answer (or one of the answers) is $\{(3,0,0), (-6,2,0), (-2,0,1)\}$. Still can't figure out why, or how to obtain the answer, though.

  • $\begingroup$ So your vector space is over $\mathbb{Z}$? $\endgroup$ – 雨が好きな人 May 5 at 10:28
  • $\begingroup$ Yes H would be the subgroup of the free abelian group $\Bbb Z^3$ $\endgroup$ – SKYejin May 5 at 10:30
  • $\begingroup$ One way to see that your attempt is incorrect: you can spot that $(1,0,1)\in H,$ but there is no way to express $(1,0,1)$ as a $\mathbb{Z}$-linear combination of your proposed basis elements, because $3$ is not a unit in $\mathbb{Z}.$ Also, note that the answer you have been told is correct is equivalent to $\{(3,0,0),(0,2,0),(1,0,1)\},$ since we are obviously allowed to add integer multiples of $(3,0,0).$ $\endgroup$ – Will R May 6 at 6:27

We have $2(x_1-x_3)+3x_2 = 6k$. Let $y = x_1-x_3$ and consider $2y+3x_2 = 6k$. It has a basis $(y,x_2) = (3,0),(0,2)$.

Thus we may consider $x_2$ as independent. We have the first vector $(x_1,x_2,x_3) = (0,2,0)$.

For $(y,x_2) = (3,0)$, we can split it to two vectors $(x_1,x_2,x_3) = (3,0,0),(1,0,1)$, by considering the problem $x_1-x_3 = 3m$.

Therefore, $(0,2,0),(3,0,0),(1,0,1)$ is a basis. You may check that your basis can be obtained by this basis, and vise versa.


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