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:
(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

  • 1
    $\begingroup$ Perhaps you want $rL(u)+sL(v)$? $\endgroup$ Mar 21 '15 at 3:01
  • $\begingroup$ @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. $\endgroup$
    – Mike
    Mar 21 '15 at 3:03
  • $\begingroup$ 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. $\endgroup$
    – coffeemath
    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.

So the first pair of conditions implies the second single condition.

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.

Hence the second single condition implies the 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.]


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