Question about lineal transformation of dimension finite.

Asked May 01 '24 at 10:56

Active May 01 '24 at 10:56

Viewed 19 times

I know that if we have a $T:V\to W$ is a lineal injective transformation and $V,W$ has finite dimension, then $T$ has to be an isomorphism.

But, if we consider $T:\Bbb R\to \Bbb R^2$ given by $T(x)=(x,x)$ that is a lineal injective transformation but is not surjective, so is not isomorphism. And, $\Bbb R,\Bbb R^2$ has finite dimension. So, why this is not fail with theorem?

I dont know if my question is dumb, but thanks for asking.

asked May 01 '24 at 10:56

James A.

1

You've just provided a counter-example to the aforementionned statement. – anonymous67 May 01 '24 at 11:01
2

You forgot that $V$ and $W$ need to have the same dimension. Otherwise, the statement is clearly false. – Mark May 01 '24 at 11:04
1

See this post for the correct statement, i.e., what Mark said. The duplicate gives the map $T(x)=(x,0)$, which obviously is not surjective. – Dietrich Burde May 01 '24 at 11:05

Question about lineal transformation of dimension finite.

0 Answers0