Let $u,v,w$ n-tuples (vectors) and $\lambda,\mu$ any number.
Prove that Distributive law holds: $(\lambda+\mu)u=\lambda u+\mu u$
We defined in our lecture, that
Let $u=\left(u_1,u_2,\cdots,u_n\right) \text{, } v=\left(v_1,v_2,\cdots,v_n\right)\text{ be two }n\text{-tuples and }\lambda \text{ be any number.}$
We define:
$u+v:=(u_1+v_1,u_2+v_2,\cdots,u_n+v_n)$
$\lambda u:=(\lambda u_1,\lambda u_2,\cdots,\lambda u_n)$
Begin Proof: \begin{align} (\lambda+\mu)u&=(\lambda + \mu)\cdot(u_1,u_2,\cdots,u_n)\\ &=\left(\left(\lambda+\mu\right)u_1,\left(\lambda+\mu\right)u_2,\cdots,\left(\lambda+\mu\right)u_n\right)\\ &=\lambda(u_1,u_2,\cdots,u_n)+\mu(u_1,u_2,\cdots,u_n)\\ &=\lambda u + \mu u \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad _\blacksquare \end{align}
Let $u,v,w$ n-tuples (vectors) and $\lambda,\mu$ any number.
Prove that $\lambda(u+v)=\lambda u+\lambda v$
Begin Proof: \begin{align} \lambda(u+v)&=\lambda(u_1+v_1,u_2+v_2,\cdots,u_n+v_n)\\ &=(\lambda u_1+\lambda v_1,\lambda u_2+\lambda v_2,\cdots,\lambda u_n+\lambda v_n)\\ &=\lambda u + \lambda v \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad _\blacksquare \end{align}
Are those proofs correct?