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?

  • $\begingroup$ 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
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    $\begingroup$ 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

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 $T\left(\sum_{i=1}^n\lambda_iv_i\right)=\sum_{i=1}^n\lambda_iw_i$.

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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.

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  • $\begingroup$ 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

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