How do I combine additivity and homogeneity into one question? (proof)

Prove that the set of conditions:
L(u+v) = L(u)+L(v)
L(cv) = cL(v)
(valid for all vectors u, v, and any scalar c)

Is equivalent to the single condition:
L(ru+sv)=rL(u)+sL(v)
(For all vectors u, v and any scalars r, s)

I understand obviously that additivity and homogeneity conditions are being combined into one condition.

I understand that I have to do a biconditional proof going from 1=>2 and then proving 2=>1 but I'm having trouble with the mechanics.

Just for fun : what is this new condition called? I heard superposition and/or convulsion. Please enlighten me

• Perhaps you want $rL(u)+sL(v)$? Mar 21 '15 at 3:01
• @Michael Please check again the formatting was off. I don't think that's what I need. I'm looking for a mechanical biconditional proof.
– Mike
Mar 21 '15 at 3:03
• Michael Burr is right, you need the scalars on the outside on the right side of the single condition. Otherwise from that condition one could not derive $L(cv)=cL(v)$ of the first "two rule" condition. Mar 21 '15 at 3:08

The second "single condition" should be $L(ru+sv)=rL(u)+sL(v).$
Assume the first "pair of conditions (additive/scalar multiple)", then $$L(ru+sv)=L(ru)+L(sv)=rL(u)+sL(v),$$ by using the additive rule and then scalar multiple rules twice.
On the other hand, assume the second single condition, and let $r=s=1$ to get the additive rule of the pair of conditions, while let $r=c,s=0$ to get the scalar multiple rule of the first pair of conditions.
About a name: The second single condition could be expressed, provided one has already defined a linear combination of vectors $u,v$ as a sum $ru+sv$ with scalars $r,s$, by saying that a map is linear if and only if it preserves linear combinations. [Naturally this extends to linear combinations of more than two vectors.]