# Vector spaces of the same finite dimension are isomorphic

Suppose that V and W are vector spaces with the same dimension. We wish to show that V is isomorphic to W, i.e. show that there exists a bijective linear function, mapping from V to W.

I understand that it will suffice to find a linear function that maps a basis of V to a basis of W. This is because any element of a vector space can be written as a unique linear combination of its basis elements.

However I'm not sure how to show that such a map exists.

Just write it down. That is, given a basis $\{ e_i \ | i = 1..n \}$ of $V$ and $\{ f_i \ | i = 1..n \}$ of $W$ define $T : V \rightarrow W$ first on the basis vectors,

$$T(e_i) = f_i \ \ \ \ \text{ for each } i = 1, 2, ..., n$$

Now how would you extend $T$ to all of $V$?

• Don't you need scalar fields to be same ? – Aaqib Iqbal Jun 15 '19 at 5:05
• Yes. If that isn’t implied by the context of the question then it should be explicitly stated. I guessed in this case it was. – Simon S Jun 17 '19 at 1:30
• Yeah, vector space is so nice because it is over field. For general module, it may happen that two modules of different dimensions are isomorphic. – user5280911 Jun 1 at 15:23

When you have two vector spaces $V$ and $W$, if you are given a basis $\{v_1,v_2,\dots,v_n\}$ of $V$ and any set $\{w_1,w_2,\dots,w_n\}$, there is a unique linear map $f\colon V\to W$ such that $$f(v_i)=w_i\quad (i=1,2,\dots,n)$$

How's it defined? Since every vector $v\in V$ can be written in one and only one way as $$v=\alpha_1v_1+\dots+\alpha_nv_n,$$ we define $$f(v)=\alpha_1w_1+\dots+\alpha_nw_n.$$ The verification that this is a linear map are just tedious, but basically easy.

If moreover $\{w_1,w_2,\dots,w_n\}$ is a basis of $W$, the map $f$ so defined maps a basis to a basis, so it's an isomorphism.

• To check that it is well defined linear function, does one need only show that every element of V maps to an element of W and that the element it maps to in W is not ambiguous ? And of course that it is linear? – Bysshed Nov 17 '14 at 16:58
• The “non ambiguity” is clear, because the coefficients $\alpha_i$ are unique. The tedious part is proving linearity. – egreg Nov 17 '14 at 17:10
• Why is the verification of the linearity tedious? Let $v=\sum \alpha_i v_i$ and $u=\sum \beta_i v_i$ and $\lambda$ a scalar, then $f(v+\lambda u) = f(\sum (\alpha_i + \lambda \beta_i) v_i) = \sum (\alpha_i + \lambda \beta_i) w_i = \sum \alpha_i w_i + \lambda \sum \beta_i w_i = f(v) + \lambda f(u)$ Is that correct? I expected worse. – philmcole May 2 '18 at 20:41
• @philmcole Isn't that tedious? ;-) – egreg May 2 '18 at 20:51

Denote the dimension by $n$ and let $(v_1,\ldots,v_n)$ a basis for $V$ and $(w_1,\ldots,w_n)$ a basis for $W$ and let the linear transformation $f:V\to W$ defined by

$$f(v_i)=w_i$$ then $f$ transforms a basis to a basis then $f$ is an isomorphism of vector spaces.

List your bases $\{v_1, \ldots , v_n \}$ for $V$ and $\{w_1, \ldots, w_n\}$ for $W$. I'm assuming these are both vector spaces over the same field $F$. Suppose that $\sum_{i = 1}^n a_i v_i \in V$ for some $a_i \in F$. Then define $\phi : V \to W$ by $$\phi \big(\sum_{i = 1}^n a_i v_i\big) = \sum_{i = 1}^n a_i w_i$$

Not too hard to see it's a well defined vector space isomorphism.