1
$\begingroup$

Let $V = F^2$ for a field $F$. For $(a_1,a_2) \in V$ and $c \in F$, define $(a_1,a_2)+(b_1,b_2)=(a_1+2b_1,a_2+3b_2)$ and $c(a_1,a_2)=(ca_1,ca_2)$. Is $V$ a vector space over $F$ given these operations?

My approach was to attempt to demonstrate that there does not exist a unique zero, and so this is not a vector space:

$(a_1,a_2)+(b_1,b_2)=(a_1+2b_1,a_2+3b_2)$

$b_1=-a_1$, $b_2=-a_2$

$(a_1,a_2)+(-a_1,-a_2)=(a_1+2(-a_1), a_2+3(-a_2))=\bf{(-a_1,-2a_2)}$

$a_1=-b_1$, $a_2=-b_2$

$(-b_1,-b_2)+(b_1,b_2)=(-b_1+2b_1,-b2+3b_2)=\bf{(b_1,2b_2)}$

$F^2$ isn't $GF(2)$ according to my professor's notation. He uses $\mathbb{F}_2$ to denote the finite field of two elements. $F^2$ is just $F\times F$, a 2-tuple.

Thank you, @user1551, for your help with this problem.

$\endgroup$
  • 1
    $\begingroup$ Do the vectors form an abelian group under addition? $\endgroup$ – Jyrki Lahtonen Apr 8 '13 at 18:15
1
$\begingroup$

Hint: Actually $V$ has no zero elements. Show that there does not exist $(a_1,a_2)\in V$ such that $(a_1,a_2)+(b_1,b_2)=(b_1,b_2)$ for all $(b_1,b_2)\in V$. Alternatively, by definition, the addition operation in a vector space must be commutative. Show that the given one is not.

$\endgroup$
  • $\begingroup$ How is my approach invalid? Just curious... $\endgroup$ – Trancot Apr 8 '13 at 21:54
  • $\begingroup$ @Trancot I haven't said that your approach is invalid, but so far, you have only shown that if $u=(a_1,a_2)$ and $v=(b_1,b_2)=(-a_1,-a_2)$, you can write $u+v$ in terms of the $a_i$s and $b_i$s in two equivalent expressions, and I don't see how this is related to the question of whether $V$ is a vector space. $\endgroup$ – user1551 Apr 8 '13 at 22:07
  • $\begingroup$ Are there values $a_1$ and $a_2$ such that $(a_1,a_2)+(b_1,b_2)=(b_1,b_2)$? Let us see. Because the LHS of this equation is defined as $(a_1+2b_1,a_2+3b_2)$, then we must find a appropriate values for the following equations: \begin{eqnarray} a_1+2b_1 & = & b_1 \\ a_2+3b_2 & = & b_2, \end{eqnarray} correct? $\endgroup$ – Trancot Apr 8 '13 at 22:17
  • $\begingroup$ Are not $a_1 = -b_1$ and $a_2 = -2b_2$ such values for the defined condition? What do you mean by "$\bf{no}$ zero element$\bf{s}$"? $\endgroup$ – Trancot Apr 8 '13 at 22:21
  • $\begingroup$ @Trancot $(a_1,a_2)\in V$ is a zero element if $(a_1,a_2)+(b_1,b_2)=(b_1,b_2)$ $\color{red}{\mathbf{\text{for all}}}$ $(b_1,b_2)\in V$. $\endgroup$ – user1551 Apr 8 '13 at 22:34

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.