Why do we define a linear transformation to have the property that $f(A+B)=f(A)+f(B)$?

Question

My question is opposite of Why do we define a linear transformation to have the property that $f(cW)=c f(W)$?. I understand that there are cases where $f(cW)=cf(W)$ fails but $f(A+B)=f(A)+f(B)$ passes (e.g. Complex conjugate). But if for all $c$, $f(cW)=cf(W)$, then can we have a case where $f(A+B)=f(A)+f(B)$ fails?

Answer 1

Let $f:\mathbb{C}\to \mathbb{R}$ be “$\mathbb R$-multiplicative” so that $f(cx)=cf(x)$ for all $c\in\mathbb R, x\in\mathbb C$. Scaling a complex number by a real factor moves it radially, so that it goes from $re^{i\theta}$ to $cre^{i\theta}$. Notably, the arg is unchanged. So by saying that this property holds, we are saying that the value of $f$ at some non-zero point uniquely determines the value of $f$ at every point on the line through that point and the origin.

However, this tells us nothing about the relationship between $f(re^{i\theta})$ and $f(re^{i\varphi})$, so there is a ton of flexibility in how $f(x+y)=f(x)+f(y)$ can fail. In particular, the function is determined by its value on the unit circle, but even if you knew its values across all the unit circle except two diametrically opposite points, you wouldn’t be able to infer the value at those two points.

One example would be to define $f(e^{i\theta})=\sin(\theta)$ and extend $f$ to all of $c$ by using $f(cx)=cf(x)$. The resulting function can be easily described in polar coordinates, as $f(re^{i\theta})=r\sin(\theta)$, also known as the real part function, $\Re(z)$. More generally, if $g: [0,\pi/2)\to\mathbb R$ is a non-linear function, then defining $f(e^{i\theta})=g(\theta)$ and then extending in this way works.

I had to choose $\mathbb C$ as the domain for a reason: in $\mathbb R$ any function that satisfies $cf(x)=f(cx)$ is entirely determined by the value $f(x_0)=y_0$ (for any $x_0\neq 0$) and the formula is $f(x)=\frac{y_0}{x_0}x$. This function is linear.

Answer 2

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be defined by

$$f(x,y) = \sqrt[3]{x^3 + y^3}$$

Answer 3

Why do we define a linear transformation to have the property $f(a+b)=f(a)+f(b)$?

The best explanation from this is from category theory. This argues that mathematics ought to be organised into categories which essentially group together mathematical objects with similar structures: so we have $Vec$, the category of all vector spaces; $Grp$, the category of all groups, $Ab$, the category of all abelian groups and so on.

It further suggests that the morphisms between mathematical objects ought to be those maps that preserve all significant structure. So in $Vec$, the objects are vector spaces and their structure is the linear structure and so the structure preserving maps are the linear maps; similarly, in $Grp$ and $Ab$, the group homomorphisms.

Answer 4

Clearly, if $V$ and $W$ are two (possibly infinite-dimensiional) vector spaces and $f:V\to W$ is multiplicative but not additive, we must have $\dim V\ge2$ and $\dim W\ge1$.

Suppose this dimension constraint is indeed satisfied. Pick any two nonzero vectors $v\in V$ and $w\in W$ and define $f$ by $f(cv)=cw$ for any scalar $c$ and $f(u)=0$ for any $u\notin\operatorname{span}(v)$. Then $f$ is multiplicative by construction, but it is not additive. In fact, if $u\notin\operatorname{span}(v)$, we have $$ f\big((v-u)+u\big)=f(v)=w\ne0+0=f(v-u)+f(u). $$