3
$\begingroup$

I have proved (=>) but showing (<=) is hard for me.

I think it should start with letting {$v_1, ... ,v_n$} be a basis of V and {$w_1, ... ,w_n$} be a basis of W then suppose there exists a linear transformation T such that $T:V->W$ defined by $T(v_i)=w_i$ . Then I have a feeling that it's an isomorphism. How do I show that?

$\endgroup$
  • $\begingroup$ Hint: a linear map between two vector spaces is completely defined by the images of the basis vectors. $\endgroup$ – B. Pasternak Jun 4 '16 at 11:36
  • $\begingroup$ completely defined by the images of the basis vectors? Could you please explain in detail? $\endgroup$ – Kim Jun 4 '16 at 13:44
  • $\begingroup$ Since any element in $V$ is a linear combination of the basis vectors $\{v_1,\ldots,v_n\}$, and any element in $W$ is a linear combination of the basis vectors $\{w_1,\ldots,w_n\}$, if you know where each basis vector is send to, you know where each element is send to, since the map is linear. Mathematically, for any $v\in V$ we can write $v=\lambda^iv_i$, then $T(\lambda^iv_i)=\lambda^iT(v_i)=\lambda^iw_i$, so you know where any element in $V$ is send to, hence the map is defined completely by the images of the basis vectors. $\endgroup$ – B. Pasternak Jun 4 '16 at 13:48
  • $\begingroup$ Aha... so the map is 1-1 because: if $T(v_i)=T(v_j)$ , then $v_i=v_j$ and is onto because: suppose $w_i$ belongs to W, then there exists $v_i$ such that $T(v_i)=w_i$ ? I looked up injective and bijective in some set theory books. $\endgroup$ – Kim Jun 4 '16 at 15:10
  • 1
    $\begingroup$ Do note that it should be stipulated that both vector spaced are over the same field. For example, the two dimensional vector spaces over $\Bbb C$ and $\Bbb R$ are not isomorphic, while $\Bbb R^2$ and the set of polynomials of degree at most $1$ with real coefficients are. $\endgroup$ – pjs36 Jun 4 '16 at 17:13
2
$\begingroup$

Just to be sure that there are no misunderstandings, I will add an answer. Suppose that $\{v_1,\ldots,v_n\}$ is a basis for $V$, and that $\{w_1,\ldots,w_n\}$ is a basis for $W$. Define a map $T:V\to W$ by the linear extension of the assignment $v_i\to w_i$ for all $i\in\{1,\ldots,n\}$. To prove that $T$ is injective, it suffices to show that the $\ker(T)=\{0\}$. So suppose that $v=\sum_{i=1}\lambda_iv_i\in\ker(T)$, then \begin{equation} \begin{split} 0&=T\left(\sum_{i=1}^n\lambda_iv_i\right) \\ &=\sum_{i=1}^n\lambda_iT(v_i) \\ &=\sum_{i=1}^n\lambda_iw_i, \end{split} \end{equation} which implies that $\lambda_i=0$ for all $i\in\{1,\ldots,n\}$, since $\{w_1,\ldots,w_n\}$ is a basis. Surjectivitiy of $T$ is of course trivial, for if $w=\sum_{i=1}^n\lambda_iw_i\in W$ is given, then $\sum_{i=1}^n\lambda_iv_i\in V$ is such that $T\left(\sum_{i=1}^n\lambda_iv_i\right)=\sum_{i=1}^n\lambda_iw_i$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Let $V$ and $W$ be vector spaces with $\dim V=\dim W$, say $n$. Then $V$ and $W$ have bases $(v_1,\ldots,v_n)$ and $(w_1,\ldots,w_n)$, and the linear map $f$ defined by $f(v_i)=w_i$ is an isomorphism.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ and my question is that why is f an isomorphism? what am I missing...? :( $\endgroup$ – Kim Jun 4 '16 at 13:43
  • $\begingroup$ It is linear by definition (check that this does indeed define a linear map), it is injective, and it is surjective. Which one of these facts troubles you? $\endgroup$ – Servaes Jun 4 '16 at 13:46
  • $\begingroup$ @Servaes Most probably everyone of them. For Kim, you should go back to your definition of injective and surjective if you are confused by all of this. $\endgroup$ – B. Pasternak Jun 4 '16 at 13:49
  • $\begingroup$ @Servaes I appriciate your comment, too. I was not sure how it is injective and bijective, but B. Pasternak explained it above. Thanks though! $\endgroup$ – Kim Jun 4 '16 at 15:14
  • $\begingroup$ Anyone care to explain the downvote? $\endgroup$ – Servaes Jun 4 '16 at 17:18

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.