# Why do we have to check ker(T)=0 for isomorphism when I know that dim (V)=dim(W)?

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($$T$$)=$$0$$?

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

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.
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.
• 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. – Robert Israel Apr 19 '19 at 15:52