Hi,

i have the following question:

If a function satisfies:

for all

then

for all and ?

i think that this is true, but i only have a proof for rational numbers

thanks

Printable View

- Sep 17th 2009, 02:43 PMAndres PerezLinear transformation
Hi,

i have the following question:

If a function satisfies:

for all

then

for all and ?

i think that this is true, but i only have a proof for rational numbers

thanks - Sep 17th 2009, 04:35 PMJameson
I don't think that the first statement implies the second. When both are true then f is a linear transformation, but it must satisfy both. I am not 100% and can't think of an example, but I am sure that if this were true then this would be indicated in the definition of a linear transformation.

If you assume that f is linear then proving the second statement is easy. Is that what you want to do? - Sep 17th 2009, 04:51 PMAndres Perez
in this case i'm not assuming that is linear (this is because i use the word "function"). It's true that in general vector spaces the first statement doesn't implies the second (i agree with you that it would be indicated on the definition), but i'm thinking in the special case when .

Fortunately, i found a negative answer to that implication but the construction of such a function requires the axiom of choice, so we don't have an explicit function.

(Evilgrin) - Sep 17th 2009, 05:02 PMJameson
- Sep 17th 2009, 05:06 PMJose27
- Sep 17th 2009, 08:38 PMNonCommAlg