Do either of the linear transformation properties imply the other?

Question

I'm new to linear algebra and curious about the two properties of a linear transformation:

1) $f(x_i) + f(x_j) = f(x_i+ x_j)$

2) $cf(x) = f(cx)$ where $c$ is some scalar constant

Do either of these two properties entail the other one? I was thinking about it abstractly, and it seems like the linear transformation means that any step you take in the domain is identical in the co-domain.

Or, I can stretch and shift in the domain and that movement will appear exactly the same in the co-domain.

Or, the mapping preserves addition (subtraction) and multiplication (division).

I had two ways to try and answer this question, but got stuck.

First I asked myself if the addition of any two arbitrary reals a, b could be reached by multiplication of any two other arbitrary reals, c, d. In some vague sense I thought this would help me see if there was a link between (1) and (2).

It occurred to me that you can set c = (a+b) and d = 1 to achieve c * d = a+b for all reals a, b. It occurred to me that even if you prohibit using the identity you can simply set c = 1/(a+b) and d = (a+b)² and this works so long as we avoid (a+b)=0. To me this felt like an additive "shift" (a+b) could yield the same location as a "stretch" (c*d). However, this did not feel rigorous.

---

For my second attempt I tried to find a counter-example, where one of the two linear transformation properties was preserved but the other was not, but came up empty handed.

Help appreciated, thank you!

EDIT: I assumed domains and co-domains in the reals. Not sure how this changes things.

EDIT 2: It has been claimed that these conditions are equivalent only over the reals. Does anyone have a proof of their equivalence over the reals?

Answer 1

See vector spaces are not just about reals, If any of the second conditions were not there things would have been different in other vector spaces other than that of the reals, in reals the two conditions are equivalent.

Answer 2

Consider two vector spaces V, V' and a map $T: V\rightarrow V'$. If for any $v_1,v_2\in V, T(v_1+v_2)=Tv_1+Tv_2$, then $\forall n \in \Bbb Z, T(nv_1)=Tv_1+...+Tv_1=nTv_1$, and hence $\forall \alpha \in \Bbb Q, v \in V, T(\alpha v)=\alpha Tv$.

Given any irrational number $0\lt r\lt 1,$ we can write $r=\sum_{k=1}^{\infty}10^{-k}r_k.$ That is, $\forall \epsilon\gt 0, \exists N\in \Bbb N$ such that every positive integer $n\ge N$ gives $0\lt r-\sum_{k=1}^{n}10^{-k}r_k\lt \epsilon$. In this case, $\forall v\in V,rTv\gt T(\sum_{k=1}^{n}10^{-k}r_kv)=Tv\sum_{k=1}^{n}10^{-k}r_k\gt (r-\epsilon)Tv$. This implies $T(rv)=rT(v) \;\forall r\in \Bbb R$.

For the other direction, $\forall v_1,v_2\in V, T(v_1+v_2)=T(v_1(1+\dfrac{v_2}{v_1}))=(1+\dfrac{v_2}{v_1})Tv_1=Tv_1+Tv_2$, provided $v_1,v_2$ are scalar multiples of each other or V is 1 dimensional.

Note that given $T(1,0,0)$ and $T(0,0,1)$, it is impossible to find $T(1,0,1)$ as $(1,0,1)$ is not a scalar multiple of either $(1,0,0)$ or $(0,0,1)$.