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The question

Map closed under addition but not multiplication

asks for a map between two vector spaces where vector addition is preserved but also where scalar multiplication is not preserved.

A student of mine switched this, and I have not found an example.

Thus, what is an example of a map between two vector spaces where vector addition is not preserved but scalar multiplication is preserved?

Thank you.

Community
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Daniel Kiteck
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    How about ${0,1}^2\to{0,1}$ with the field $\Bbb F_2$ and usual addition and multiplication on $\Bbb F_2^n$ where $f(0,0)=0$, $f(1,0)=1, f(0,1)=1, f(1,1)=1$, i.e. $f$ is the "or" operation. Addition fails because $f(1,0)+f(0,1)\neq f(1,1)$, but multiplication succeeds since $0f(a,b)=0=f(0a,0b)$ and $1f(a,b)=f(a,b)=f(1a,1b)$ for all $a,b$ – JMoravitz Mar 31 '17 at 16:07

3 Answers3

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Unlike the reverse question, in this case the condition is weak enough that we can find examples even over $\mathbb R$. Essentially, this is telling us that the map acts like a linear map on every line through the origin, but isn't consistent about what it does to different lines. (So we'll need at least a two-dimensional vector space to find a counterexample.)

One possible example is the map $T : \mathbb R^2 \to \mathbb R^2$ defined in polar coordinates as $$T\begin{bmatrix}r \cos \theta \\ r \sin \theta\end{bmatrix} = \frac1{|\cos \theta| + |\sin \theta|} \begin{bmatrix} r \cos \theta \\ r \sin \theta \end{bmatrix}$$ which bends the unit circle into a diamond. In general, you can distort the unit circle however you like (that's symmetric with respect to reflection through the origin), and then make sure that scalar multiplication is respected. (This corresponds to multiplying by $f(\theta)$ for any function $f : [0, 2\pi] \to \mathbb R$ such that $f(\theta + \pi) = f(\theta)$.)

Misha Lavrov
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  • This is such a nice example getting to the heart of comparing vector addition versus scalar multiplication. Thank you. – Daniel Kiteck Mar 31 '17 at 19:36
  • Is the comment about distorting the unit circle in any way that's symmetric with respect to reflection through the origin correct? That would seem to imply that any linear function preserves the unit function, but in general a linear function will map the unit circle to a (possibly degenerate) non-circular ellipse. Is the condition for not being linear correspond more to sending the unit circle to something that isn't an ellipse? – Chill2Macht Feb 13 '20 at 18:47
  • Also which part corresponds to multiplying by $f(\theta)$ for an even function $f$? The distortion of the unit circle in a way that's symmetric with respect to reflection through the origin? Making sure that scalar multiplication is respected? Both? – Chill2Macht Feb 13 '20 at 18:50
  • In general, linear functions do preserve scalar multiplication, and so turning the unit circle into an ellipse is fine. I'm not sure what you mean by "preserving the unit function". – Misha Lavrov Feb 13 '20 at 19:04
  • Also also is the condition that $f$ is even? It seems, for example, that $\theta \mapsto \theta^2$ would not work. It seems like the condition on $f$, at least for guaranteeing symmetry about the origin, would be $f(\theta) = f(\theta + \pi)$ for any angle $\theta$, i.e. that $f$ is $\pi$-periodic. The function in the example you gave satisfies this condition, for example. – Chill2Macht Feb 13 '20 at 19:06
  • You are right; I should have said $\pi$-periodic rather than even. I was thinking that the angle opposite $\theta$ is $-\theta$, which is wrong. – Misha Lavrov Feb 13 '20 at 19:07
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How about $L:\mathbb R^2\to \mathbb R$ defined by:

$$ L(x,y) = \begin{cases} x, & \text{if $y=0$ } \\ y, & \text{if $y\neq 0$ } \end{cases}$$

lulu
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I here share another example:

$T: \mathcal A\to \mathbb R$ defined by:

where $\mathcal A=\{[x,y]\in{\mathbb R}^2\mid\neg (x=0~\cap~y=0)\}$

$T(x,y) = \begin{cases} \cfrac{xy^2}{x^2+y^2} \end{cases}$

Shih-Yu Yang
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