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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?

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    $\begingroup$ Yes, they are correct $\endgroup$ Apr 9, 2019 at 19:08
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    $\begingroup$ Looks good to me. $\endgroup$
    – Andrei
    Apr 9, 2019 at 19:08

1 Answer 1

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Almost.

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.

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  • $\begingroup$ These are precisely the points I was going to bring up. +1 for you. $\endgroup$
    – MPW
    Apr 9, 2019 at 19:10
  • $\begingroup$ 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? $\endgroup$
    – Doesbaddel
    Apr 9, 2019 at 19:54
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    $\begingroup$ Yes, those will go between what you currently have as the 3rd and 4th lines. $\endgroup$
    – kccu
    Apr 9, 2019 at 20:02
  • $\begingroup$ Alright, thanks! $\endgroup$
    – Doesbaddel
    Apr 9, 2019 at 20:42
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    $\begingroup$ Yes, between the 2nd and 3rd lines. And my apologies, my previous comment had a typo - what you wrote above will also go between the 2nd and 3rd lines (not the 3rd and 4th lines) of the first proof. $\endgroup$
    – kccu
    Apr 10, 2019 at 13:32

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