Linear transformation

**Andres Perez** · Sep 17th 2009, 02:43 PM

Hi,

i have the following question:

If a function $T:\mathbb{R}^n\to \mathbb{R}^m$ satisfies:

$T(u+v)=T(u)+T(v)$ for all $u,v\in \mathbb{R}^n$

then

$T(\alpha u)=\alpha T(u)$ for all $\alpha\in \mathbb{R}$ and $u\in \mathbb{R}^n$ ?

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

thanks

**Jameson** · Sep 17th 2009, 04:35 PM

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?

**Andres Perez** · Sep 17th 2009, 04:51 PM

in this case i'm not assuming that $T$ 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 $T:\mathbb{R}^n\to\mathbb{R}^m$ .

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.

**Jameson** · Sep 17th 2009, 05:02 PM

Originally Posted by Andres Perez

in this case i'm not assuming that $T$ 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 $T:\mathbb{R}^n\to\mathbb{R}^m$ .

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.

Interesting! I also seem to miss important details. I should have seen you were talking about reals. Would you share your solution?

**Jose27** · Sep 17th 2009, 05:06 PM

Originally Posted by Jameson

Would you share your solution?

Please do, and note that if f is assumed continous then it is valid (actually in any normed space).

**NonCommAlg** · Sep 17th 2009, 08:38 PM

Originally Posted by Andres Perez

Hi,

i have the following question:

If a function $T:\mathbb{R}^n\to \mathbb{R}^m$ satisfies:

$T(u+v)=T(u)+T(v)$ for all $u,v\in \mathbb{R}^n$

then

$T(\alpha u)=\alpha T(u)$ for all $\alpha\in \mathbb{R}$ and $u\in \mathbb{R}^n$ ?

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

thanks

a counter-example for $n = 1$ : let $B=\{x_i \}_{i \in I}$ be a basis for $\mathbb{R},$ as a vector space over $\mathbb{Q}.$ we may assume that $1 \in B.$ define $f(1)=1$ and $f(x_i)=0, \ x_i \neq 1,$ and extened $f$ linearly to $\mathbb{R}.$

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