Csdrhrt

I am reading a linear algebra textbook and it mentions that, to be linear, a function $f : mathbb{R}^n rightarrow mathbb{R}^m$ must fulfill $f(x+y) = f(x) + f(y)$ and $f(cx) = c(fx)$.

I'm trying to understand why the second condition is necessary, since taking $y = (c-1)x$ and recursively expanding the sum into $f(x)$ terms seems to be a way to arrive at the same conclusion from only the first condition.

A problem with the proposed way of defining $f(cx)$,

$f(cx) = f(x + (c - 1)x) = f(x) + f((c - 1)x), ; text{and so forth}, tag 1$

is that, unless $c in Bbb N$, the recursive process won't terminate. What happens is $c = sqrt 2$ or $c = pi$, for example? Or even if $0 > c in Bbb Z$? Of course here one may take

$f(cx) = f(-x + (c + 1)x) = f(-x) + f((c + 1)x), tag 2$

since we have $f(-x) = -f(x)$ from

$f(x) + f(-x) = f(x + (-x)) = f(0) = 0. tag 3$

But with $c notin Z$, we will never arrive at a result for $f(cx)$.

If $c in Bbb Q$, one can make some progress in this direction via the observation that with

$c = dfrac{p}{q}, ; p, q in Bbb Z, tag 4$

we can write

$pf(x) = f(px) = f left (q dfrac{p}{q}x right ) = qf left (dfrac{p}{q}x right ), tag 5$

whence

$f left (dfrac{p}{q}x right ) = dfrac{p}{q}f(x). tag 6$

We can handle $f(cx) = cf(x)$, $c in Bbb Z$, via $f(x + y) = f(x) + f(y)$ without axiomatizing $f(cx) = cf(x)$ since the addition axiom essenially posits an abelian group homomporphism between $Bbb R^m$ and $Bbb R^n$, and such homomorphisms extend in a natural way to $Bbb Z$-module homomorphisms; as we have seen, rational $c$ then obey $f(cx) = cf(x)$; but for $c$ irrational we are faced with a non-terminating process . . .

Typically the assumption that $f(x)$ is continuous may be invoked to address the case of irrational $c$. Then if $c_n to c$ with $c_n in Bbb Q$, we have

$c_n f(x) = f(c_n x) to f(cx) tag 7$

by the continuity of $f(x)$.

You are correct that induction combined with $f(x+y)=f(x)+f(y)$ yields $f(nx)=nf(x)$ for $nin mathbb{N}$. However, for $c=pi$ (say) it isn't obvious how one would show that it follows that $f(pi x)=pi f(x)$.

Of course, we want our maps to respect scalar multiplication, so it necessitates making the definition $f(lambda x)=lambda f(x)$ for all $lambda in mathbb{R}$.