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Why we define a lin tranfs to have the property that $f(cW)=c f(W)$ ?

let $V,T$ be any two vector spaces and

let $f:V\rightarrow T$ be a linear transformation between $V $and $T $

why do we assume this condition ?

i think you would say , to preserve the structure of the spaces ! and this is right

but i find it unnecessary , why ?

we know that $f(A+B)=f(A)+f(B)$ by the first condition of transformation so we can , using induction , prove that

$f(A_1 + A_2 +... +A_n ) = f(A_1) + f(A_2) + ... + f(A_n)$

putting , $A_1 = A_2 = ... =A_n = A$

we conclude that , $f(cA)=cf(A)$

so , why mathematicians assume the second condition although it's a followed from the first ?!

isn't this a repeat ? i think mathematicians don't like repeating things !

FNH
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    A vector space is an $R$- module where $R$ is a field. Recall that a (left) $R$ module is an abelian group $M$ with a map $R \times M \to M$ that satisfies certain properties. An $R$-module homomorphism $M \to N$ is a map that preserves the group structure and the action of $R$. i.e $f(a + b) = f(a) + f(b)$ and $f(r \cdot a) = r \cdot f(a)$. Linear maps are just $F$-module maps. – Ink Jun 19 '13 at 20:01

4 Answers4

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Hint: you are assuming in your demonstration that $c \in \mathbb Z$. The underlying field may not be the set of integers. So we have a problem when $c\notin \mathbb Z$, and we'd need to extend this to cover, e.g., rationals. But then what do we do, e.g, if $c$ is Irrational? Imaginary?

So in the general scenario, we need that a linear transformation $f$ on $W$ satisfies the constraint that $f(cW) = cf(W)$.

I am quite confident that the "extra baggage" of this requirement would have fallen by the wayside if it were not needed to characterize what conditions, exactly, must be satisfied by any and every linear transformation.

amWhy
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  • extra baggage like what does this requirement add up to the structure except the preserving of the structure of the spaces ? – FNH Jun 19 '13 at 20:06
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    I mean that if it wasn't needed to preserve the structure, it would have long ago been dispensed with as a condition. – amWhy Jun 19 '13 at 20:18
  • @amWhy: Sorry, but this gets extra baggage with another TU! :-) – Amzoti Jun 20 '13 at 00:34
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Your process can give you at most $f(qv)=qf(v)$, $q\in\mathbb Q$. Without some additional hypothesis on $f$ and the vector space (topology, continuity, completeness... sometimes we have neither of the above), you can't deduce the same for all $q\in\mathbb R$. Moreover, you should think about multiplication by $i=\sqrt{-1}$.

TZakrevskiy
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Not really, what you have is $f(nx)=nf(x)$, where $n$ is an integer. You can push this further to $f(rx)=rf(x)$ where $Q\subseteq F$ is the isomorphic copy of $\Bbb Q$ in $F$ (assuming $\operatorname{char}F=0$); $F$ the underlying field. That is far from the general $f(\lambda x)=\lambda f(x)$ for any scalar $\lambda\in F$. For example, in the case of real numbers, you can show this implies $f(rx)=rf(x)$ for $r$ rational, but no more without some further assumptions. See this for example.

Community
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Pedro
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Just in addition to the above: The additional requirement is also so that the definition of linearity is consistent with differentiation.

Take for example the conjugate complex operation (assume $\mathbb{C}$ as a vector space over $\mathbb{C}$), i.e.

$f: \mathbb{C}^1 \to \mathbb{C}^1: z \to f(z) = z^*$

Clearly, $f$ fulfils $f(A+B) = f(A) + f(B)$ but not $c f(W) = f(cW)$ for relevant elements, so it is not a linear transform.

Also recall that the conjugate complex is not differentiable over $\mathbb{C}$ so this is consistent with complex conjugation not being a linear transform.

So if we would not demand $c f(W) = f(cW)$ for linear transforms, then we could have linear transforms which might not have derivatives (in situations where these can be defined).

Situation is of course different when $\mathbb{C}$ is regarded as a $\mathbb{R}$ vector space (in which case conjugate complex is a linear transform and also differentiable).

Andreas H.
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