Proving a transformation is not linear

[Bipolarity]

For a certain transformation T, it is known that $T(x+y) = T(x) + T(y)$

It is required to determine whether this transformation is linear. Obviously it is not, since it need not satisfy the degree-1 homogeneity property of all linear maps.

I'm just having trouble cooking up the counterexample. Any ideas? Thanks!

BiP

 

[micromass]

Is $T:\mathbb{R}\rightarrow \mathbb{R}$?
And are there other things known about $T$ such as continuity?

If $T:\mathbb{R}\rightarrow \mathbb{R}$ and it is indeed true that the only thing you know is $T(x + y) = T(x) + T(y)$ then there is a counterexample for linearity. This is not an easy counterexample however. It consist in looking at $\mathbb{R}$ as a $\mathbb{Q}$-vector space. As such, we have a basis $\{e_i\}_{i\in I}$ of $\mathbb{R}$. So any element $x\in \mathbb{R}$ can be written uniquely as

$$x = \sum_{i\in I}\alpha_i e_i$$

for some rational numbers $\alpha_i\in \mathbb{Q}$ such that only finitely many $\alpha_i$ are nonzero. Now take $j\in I$ fixed an consider

$$T(x) = \alpha_j$$

this map satisfies $T(x+y) = T(x) + T(y)$, but not $T(\lambda x) = \lambda T(x)$ for all $\lambda \in \mathbb{R}$ (it does satisfy it for $\lambda in \mathbb{Q}$ though).

 

[Bipolarity]

I see. What about the following transformation $T: ℂ → ℝ$ where $T(x) = Re(x)$. If the field is the complex field, then the scalar $i$ causes homogeneity to fail. Additivity is trivially shown.

Is this valid?

BiP

 

[micromass]

If you have a map $T:V\rightarrow W$ then it's assumed that both maps have the same scalar field (otherwise linear makes no sense). So I assume that both $\mathbb{C}$ and $\mathbb{R}$ are $\mathbb{R}$-vector spaces. But then the map you mention is linear.

 

[Bipolarity]

If you have a map $T:V\rightarrow W$ then it's assumed that both maps have the same scalar field (otherwise linear makes no sense). So I assume that both $\mathbb{C}$ and $\mathbb{R}$ are $\mathbb{R}$-vector spaces. But then the map you mention is linear.

I'm sorry, if I redefined the transformation as:
$T:ℂ\rightarrow ℂ$ then is my example valid?

BiP

 

[micromass]

I'm sorry, if I redefined the transformation as:
$T:ℂ\rightarrow ℂ$ then is my example valid?

BiP

Yes, that would be a valid example.