Problem 72: V is the set of all $(x, y, z)$ such that $x + y + z = 0$. Show that V is a subspace of $\R^3$

Solution:
$S = \{(x,y,z) \in \R):x+y+z=0}$.
zero check:
$\mathbf(0) = (0,0,0) \in V$ since $0+0+0=0$
Closure under Addition:
Let $u=(u_1,u_2,u_3) \in V \Rightarrow u_1+u_2+u_3 =0$.
and $v=(v_1,v_2,v_3) \in V \Rightarrow v_1+v_2+v_3 =0$.
$u+v = (u_1,u_2,u_3) + (v_1,v_2,v_3) = (u_1+v_1,u_2+v_2,u_3+v_3)$
$(u_1+v_1)+(u_2+v_2)+(u_3+v_3) = (u_1+u_2+u_3)+(v_1+v_2+v_3) = 0+0 = 0$
So, $u+v in V$
Hence,S is closed under addition.
Closure under Scalar Multiplication:
Let $k in \R$ and $u=(u_1,u_2,u_3) \in V$
Then, $ku = (ku_1,ku_2,ku_3)$
$(ku_1)+(ku_2)+(ku_3) = k (u_1+u_2+u_3) = k0 = 0 $
which shows that $ku \in V$
Therefore S is a subspace of $\R^3$

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