# 2-tuple Vector Space: GF(2)?

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.

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

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.
• @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. – user1551 Apr 8 '13 at 22:07
• 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? – Trancot Apr 8 '13 at 22:17
• 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$\bf{s}$"? – Trancot Apr 8 '13 at 22:21
• @Trancot $(a_1,a_2)\in V$ is a zero element if $(a_1,a_2)+(b_1,b_2)=(b_1,b_2)$ $\color{red}{\mathbf{\text{for all}}}$ $(b_1,b_2)\in V$. – user1551 Apr 8 '13 at 22:34