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As almost everyone reading this probably knows, a function $T$ from one real vector space to another is defined to be a linear transformation if for all vectors $\mathbf{u}, \mathbf{v}$ in the domain and all real $a$, (i) $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$, and (ii) $T(a\mathbf{u}) = aT(\mathbf{u})$. Does anyone know of an example of two real vector spaces and a function $T$ from one to the other that satisfies one of the requirements but not both? I'm assuming there is one, or else it wouldn't be defined that way.

I can show that if $T$ satisfies (i), then (ii) holds for all rational $a$.

Is there such an example where the vector spaces are both finite-dimensional?

I am not interested in complex vector spaces or additional requirements such as norms.

I browsed the Similar Questions and couldn't find an answer. I apologize if this is a duplicate question.

Stefan (STack Exchange FAN)

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Related: math.stackexchange.com/questions/106785/… –  Jonas Meyer Jan 31 '13 at 2:06
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5 Answers

up vote 2 down vote accepted

For ii) but not i), I think $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ given by $f(x,y) = (x^3 + y^3 )^{1/3}$ is the simplest example. For i), examples actually exist for when the domain of $f$ is simply $\mathbb{R}$ - see here.

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Thank you. Yours is the first obviously correct answer that I spotted, so I am accepting it. I am embarrassed that there is such a simple example. I looked at the link you gave, and it appears that finding an example of (i) but not (ii) is much more difficult, and I didn't have time absorb it fully. In those examples, can the target also be finite-dimensional? Is the Axiom of Choice required? I'll try to read it myself tomorrow. To anyone else who gave a correct example, I apologize, but I can only accept one answer. –  Stefan Smith Jan 31 '13 at 2:43
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I'll give one example of a function satisfying (ii) but not (i) that I think is simple. Consider $T:\mathbb R^2 \to \mathbb R$ mapping $(r\cos\theta, r\sin\theta)$ to $r\theta$ for $\theta \in [0, \pi)$ and to $-r(\theta - \pi)$ for $\theta \in [\pi, 2\pi)$.

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You're right. Forgot that the scalar can be negative. I'll fix that now. –  Tunococ Jan 31 '13 at 2:21
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$r\cos^3\theta$ would also work. –  Jonas Meyer Jan 31 '13 at 2:30
    
@JonasMeyer That's right. Your other example is also simpler. (I think you have $r|\cos\theta|\cos\theta$ somewhere.) –  Tunococ Jan 31 '13 at 2:40
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An example that satisfies $(i)$ but not $(ii)$:

We can construct a function from $\mathbb{R}$ to itself; construct a basis for $\mathbb{R}$ over $\mathbb{Q}$ by taking the set $\{1, \sqrt{k}\}$ for some squarefree $k$, and extend this to a basis using the axiom of choice. Then define $f(1)=1$ and $f$ to be $0$ for all other basis elements. If we also define $f$ to be linear from this vector space of $\mathbb{R}$ over $\mathbb{Q}$, then $f$ is only $\mathbb{Q}$-linear and additive in $\mathbb{R}$, but it can be seen that it's nonhomogeneous, as $f(\sqrt{k}\cdot 1) = 0$ while $\sqrt{k}f(1)=\sqrt{k}$.

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Thanks. I mostly understand this. Is this basis what is often called a "Hamel basis"? Do you know if there are any examples satisfying (i) but not (ii) that don't require the Axiom of Choice, or can this be proven to be impossible? –  Stefan Smith Jan 31 '13 at 4:48
    
Yep, this is exactly a Hamel basis. It seems likely that there is such an example, but unfortunately I don't know any off the top of my head. –  bzprules Feb 1 '13 at 3:17
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In one dimension, the i) but not ii) half of this reduces to Cauchy's functional equation $$ f(x+y)=f(x)+f(y) $$ which has nonlinear solutions, but not nice ones.

Essentially, i) guarantees that your function is a linear transformation between rational vector spaces. But not every $\Bbb{Q}$-linear transformation is also $\Bbb{R}$-linear, even if its domain and range are $\Bbb{R}$-vector spaces...

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Can you give an example of a $\mathbb{Q}$-linear transformation that is not $\mathbb{R}$-linear? Does such a linear transformation exist between finite-dimensional vector spaces? Is there one that does not require the Axiom of Choice, or can it be proven that the Axiom of Choice or some weak form of it is necessary? I'll check out the link. –  Stefan Smith Jan 31 '13 at 4:53
    
@StefanSmith: Any such thing is non-Lebesgue-measurable, which as I understand it means you need some version the axiom of choice to prove it exists (though I'm no expert). I don't think the Wikipedia article mentions this, but here's a proof. Writing down such a function is either impossible or easy, depending on how explicit you want to be; it boils down to "pick a $\Bbb{Q}$-basis for $\Bbb{R}$, and then arbitrarily choose values for the function on these basis elements." –  Micah Jan 31 '13 at 5:05
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(ii) but not (i):

$f:\mathbb{R}^2\to\mathbb{R}^2$,

$f(r\cos\theta,r\sin\theta) = \theta(r\cos\theta,r\sin\theta)$

where $\theta\in[0,\pi), r\in\mathbb{R}$

This is well defined because every point in $\mathbb{R}^2$ can be written in terms of polar coordinates, $f(0,0)=(0,0)$ independently of $\theta$.

$f(ax,ay)=f(ar\cos\theta,ar\sin\theta)=a\theta(r\cos\theta,r\sin\theta)=af(x,y)$.

$f((0,1)+(1,0)) = f(1,1) = \frac{\pi}{4}(1,1)$

$f(1,0) + f(0,1) = 0 + \frac{\pi}{2}(0,1) \ne \frac{\pi}{4}(1,1)$

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A variant of this idea: $T:\mathbb R^2\to\mathbb R$, $T(x,y)=\dfrac{|x|x}{\sqrt{x^2+y^2}}$ if $(x,y)\neq (0,0)$, $T(0,0)=0$. –  Jonas Meyer Jan 31 '13 at 2:03
    
You're right, fixed polar coordinates so that it works. –  Alfonso Fernandez Jan 31 '13 at 2:50
    
I see, thank you. –  Jonas Meyer Jan 31 '13 at 2:53
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