Talk:Condition for Linear Transformation

Isn't it equivalent to show $\forall x, y \in G: \forall \lambda \in R: \phi \left({\lambda x + y}\right) = \lambda \phi \left({x}\right) + \phi \left({y}\right)$? And isn't this slightly easier to use in proving things are linear transformations? --Cynic 22:45, 7 November 2008 (UTC)

Possibly. Talk me through it. (Glad someone's paying attention - respect.) --Matt Westwood 22:56, 7 November 2008 (UTC)

• Any linear transformation clearly satisfies the condition.
• Let $\phi$ be such that the condition is satisfied.

Let $\lambda = 1_R$. Then $\phi \left({x + y}\right) = \phi \left({x}\right) + \phi \left({y}\right)$.
Now let $y = 0$. Then $\phi \left({\lambda x}\right) = \lambda \phi \left({x}\right)$.
Thus by $R$-algebraic structure homomorphism the conditions are fulfilled for $\phi$ to be a homomorphism, that is, a linear transformation.
--Cynic 01:17, 8 November 2008 (UTC)

Yes of course you're dead right. However, the theorem as it stands is more directly useful in the form given, otherwise when you do find yourself with a structure that fulfils the condition with $\mu$ in it you still have to declare the truth of that. Bear with me on this one, we may come back to revisit it but I have an Agenda. --Matt Westwood 06:30, 8 November 2008 (UTC)

I don't, if it fulfills the condition with $\mu$, then simply let $\mu = 1$ and it trivially fulfills the other shorter condition. But whatever, it doesn't really matter. --Cynic 22:41, 8 November 2008 (UTC)