# Prove that Distributive law holds: $(\lambda+\mu)u=\lambda u+\mu u$ and $\lambda(u+v)=\lambda u+\lambda v$

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?

• Yes, they are correct – Shubham Johri Apr 9 at 19:08
• Looks good to me. – Andrei Apr 9 at 19:08

You skipped a step in your first proof. Namely, why is: $$((\lambda+\mu)u_1,\dots,(\lambda+\mu)u_n)=\lambda u + \mu u?$$ Hint: First distribute each $$(\lambda+\mu)u_i$$ using the distributive property of real numbers. Then use the definition of vector addition to break the one vector into the sum of two vectors, and use the definition of multiplication by a scalar to pull out the $$\lambda$$ and $$\mu$$.
You are missing parentheses in the second line of your second proof. In addition, you have skipped a step. Why is: $$(\lambda u_1 + \lambda v_1, \dots,\lambda u_n + \lambda v_n) = \lambda u + \lambda v?$$ Hint: Use the definition of vector addition to break up the left-hand side, then the definition of multiplication by a scalar to pull out the $$\lambda$$'s.
• In the first proof I'll need to add $(\lambda u_1+\mu u_1,\lambda u_2+\mu u_2,\cdots,\lambda u_n+\mu u_n)=(\lambda u_1, \lambda u_2,\cdots, \lambda u_n)+(\mu u_1 + \mu u_2, \cdots,\mu u_n)$ right? – Doesbaddel Apr 9 at 19:54