7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.