I Questions about linear transformations

[Mayan Fung]

We learned that the condition of a linear transformation is
1. T(v+w) = T(v)+T(w)
2. T(kv) = kT(v)

I am wondering if there is any transformation which only fulfil either one and fails another condition. As obviously, 1 implies 2 for rational number k.

Could anyone give an example of each case? (Fulfilling 1 but 2 and 2 but 1)

Thanks!

 

[mathman]

If you don't have continuity, it is possible, using a Hamel basis to get (2) for rational k only..

http://mathworld.wolfram.com/HamelBasis.html

 

[andrewkirk]

Consider the real numbers $\mathbb R$ as a vector space over the rationals, and the operator $T:\mathbb R\to \mathbb R$ that is the identity on the rationals and maps to zero on the irrationals. Then T satisfies the second axiom, since it is a linear operator on $\mathbb Q$ considered as a subspace of $\mathbb R$ and, for $q\in\mathbb Q-\{0\},x\in\mathbb R$, $qx$ is in $\mathbb Q\cup\{0\}=\ker\ T$ iff $x$ is.

But the addition axiom does not hold, because $T(1+(\sqrt2-1))=T(\sqrt 2)=0$ but $T(1)+T(\sqrt2-1)=1\neq 0$.

 

[Stephen Tashi]

Define $M((x,y))$ by
if $x \ne y$ then $M((x,y)) = (x,y)$
if $x = y$ then $M((x,y)) = (2x, 2y)$
$M$ satisfies 2, but not 1

 

[Mayan Fung]

This really gives me new insight into linear transformation. thanks all!

 

[Mayan Fung]

If you don't have continuity, it is possible, using a Hamel basis to get (2) for rational k only..

http://mathworld.wolfram.com/HamelBasis.html

I am sorry but I don't quite understand. How can we construct a transformation like that?

 

[mathman]

I am sorry but I don't quite understand. How can we construct a transformation like that?

Unfortunately Hamel basis exists, but it is not constructable - existence is equivalent to axiom of choice.

 

[andrewkirk]

I'm still trying to think of a scenario with a map that satisfies 1 (additivity) but not 2 (scalar mult). Can anybody think of one?

All the examples I come up with either end up satisfying neither 1 nor 2, or satisfying 2 but not 1.

I assume there must be one, otherwise some texts would specify 1 as the sole requirement and derive 2 as a consequence of 1.

 

[Mayan Fung]

Andrew, I am thinking that as we have to apply the transformation to a vector space, and vectors in vector space obeys kv is also in the space. As we can have k be any real number, it seems that it somehow implies axiom2. The transformation only satisfy axiom1 must be of a very weird form.

 

[Stephen Tashi]

On the "talk" page for the current Wikipedia article on "Linear transformation", I found:

The complex numbers is a vector space over itself, so take f:C->C to be complex conjugation. Then f(a+b)=f(a)+f(b), but -1 = f(i*i) != i*f(i) = 1. JackSchmidt (talk) 16:47, 7 March 2008 (UTC)

 

[Mayan Fung]

That's a clear and direct example!