0
$\begingroup$

Suppose I want to prove that $P_2$ is isomorphic to $R^3$. Why is it not sufficient to prove dim($P_2$)=dim($R^3$) and apply Thm part (b)? Why do I have to show that the ker($T$)=$0$?

$\endgroup$
  • $\begingroup$ No, you don't have to. It's enough to prove $\dim(P_2)=\dim(R^3)$ to prove it's isomorphic. $\endgroup$ – Surb Apr 19 '19 at 15:44
  • $\begingroup$ The textbook says it. I am not able to attach the screenshot in the comments. $\endgroup$ – math Apr 19 '19 at 15:45
  • $\begingroup$ Do you want to show that $P_2$ is isomorphic to $R^2$? Or do you want to show that some specific $T\colon P_2\to R^2$ is an isomorphism? $\endgroup$ – Hagen von Eitzen Apr 19 '19 at 15:45
  • $\begingroup$ what is the difference between the two? $\endgroup$ – math Apr 19 '19 at 15:47
  • 1
    $\begingroup$ @math It's essentially the same as the difference between the statements "the crime was witnessed" and "Joseph Schmoe witnessed the crime". $\endgroup$ – Daniel Schepler Apr 19 '19 at 15:52
1
$\begingroup$

There is a difference between "two spaces are isomorphic" and "the linear transformation $T$ is an isomorphism between the two spaces".

If have two (finite dimensional) vector spaces $V$ and $W$, and you want to check that two are isomorphic, it's enough to compare dimensions. If you also have a linear transformation $T:V\to W$ ($V, W$ still finite dimensional), and you want to check that $T$ is an isomorphism, then you have to check that $V$ and $W$ are isomorphic (i.e. compare dimensions) and also that $T$ is injective (i.e. check the kernel).

The theorem you provided is all about the latter case, while it seems that your actual problem (comparing $P_2$ to $\Bbb R^2$) is about the former.

$\endgroup$
0
$\begingroup$

You need to prove $\ker(T) = \{0\}$ or $\text{im}(T) = W$, and then ($V$ and $W$ are finite dimensional spaces with the same dimension) the other is automatically true.

$\endgroup$
  • $\begingroup$ why can't I just use theorem part (b) and show that dimensions are the same and hence they are isomorphic? $\endgroup$ – math Apr 19 '19 at 15:48
  • 1
    $\begingroup$ If you just want to prove they are isomorphic (without specifying a particular isomorphism $T$), then you just need to show the dimensions are the same. I thought you wanted to prove that a particular $T$ is an isomorphism. $\endgroup$ – Robert Israel Apr 19 '19 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.