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Let $a,b,c $ be a fix elements of a field $F$ Show that $$W= \big((x,y,z)| \ \ ax+by+cz=0 ; x,y,z\in F\big) $$ is a subspace of $V_3(F)$

The necessary and sufficient condition for non empty subset $W$ of a vector space $V(F)$ to be a subspace of V is that $\alpha u+\beta v\in W $ for all $\alpha ,\beta\in F$ and $u,v\in W$

Now in this question I used above thought and proceeded in following manner.

Let $u=(x_1,y_1,z_1)$

$v=(x_2,y_2,z_2)$

$\alpha u+\beta v = \alpha(x_1,y_1,z_1)+ \beta(x_2,y_2,z_2)$

$\implies(\alpha x_1 +\beta x_2,\alpha y_1+\beta y_2,\alpha z_1+\beta z_2)$

Now can I say $(\alpha x_1 +\beta x_2,\alpha y_1+\beta y_2,\alpha z_1+\beta z_2)\in W \implies \alpha u+\beta v \in W $ ??

Did I do it correct?

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You are supposed to verify that $\alpha u + \beta v \in W$ rather than just claiming that it is in $W$.

Verify that $a(\alpha x_1+\beta x_2) + b (\alpha y_1 + \beta y_2)+ c( \alpha z_1 +\beta z_2)$ is equal to zero.

Also remember to check that $W$ is non-empty and closed under scalar multiplication.

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  • $\begingroup$ Why do we have to show it is equal to zero ? Because of this condition $ax+by+cz=0$? $\endgroup$ – Daman Mar 29 '18 at 5:30
  • $\begingroup$ yup, that is how you verify an element is in $W$. and of course we have $\alpha x_1 + \beta x_2 \in F$, similarly for $y$ and $z$. $\endgroup$ – Siong Thye Goh Mar 29 '18 at 5:31
  • $\begingroup$ And how to check W is non empty? $\endgroup$ – Daman Mar 29 '18 at 5:33
  • $\begingroup$ $(0,0,0) \in W $ making it non empty set right? $\endgroup$ – Daman Mar 29 '18 at 5:35
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    $\begingroup$ congratulations. $\endgroup$ – Siong Thye Goh Mar 29 '18 at 5:35
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Let $(l,m,n)=(\alpha x_1 +\beta x_2,\alpha y_1+\beta y_2,\alpha z_1+\beta z_2)$. It is easy to verify that $al+bm+cn=0$. Thus, $(l,m,n)\in W$ which was what was required to be shown.

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