Given the following problem

Let $U$ and $V$ be finite dimensional vector spaces, such that $\dim(U) = \dim(V)$. Prove that $U$ and $V$ are isomorphic.

I was just wondering if somebody could critique and validate my proof.


Two vector spaces $U$ and $V$ are isomorphic if and only if there exists a bijection $\phi: U \to V$. That is, an invertible linear map between the two. Let $\dim(U) = \dim(V) = n$ and let $\left\{u_1, \dots, u_n\right\}$ and $\left\{v_1, \dots, v_n\right\}$ be bases for $U$ and $V$, respectively. It follows that any vector $u \in U$ can be written as

$$ u = c_1u_1 + \cdots + c_nu_n $$

We define $\phi: U \to V$ as

$$ \phi(c_1u_1 + \cdots + c_nu_n) = c_1v_1 + \cdots + c_nv_n $$

Observe that $\phi$ is invertible. That is, given a vector

$$ v = d_1v_1 + \cdots + d_nv_n $$

in $V$, we define $\phi^{-1}: V \to U$ as

$$ \phi^{-1}(d_1v_1 + \cdots + d_nv_n) = d_1u_1 + \cdots + d_nu_n $$

Since there exists a bijection between $U$ and $V$, they are isomorphic.

  • $\begingroup$ your proof is pretty good except that at the end you should say "since there is a linear bijection". it is pretty obvious that your map is linear, as you mention. however the isomorphism concept is about preserving structure, and so in any discussions of isomorphism the relevant structure should always be kept to the forefront of attention $\endgroup$ – David Holden Dec 2 '13 at 1:29
  • $\begingroup$ Our professor always just uses the term "bijection" since it is a linear algebra course and linear bijections are the only type of bijections we study. However, I will change it to "linear bijection" to be more rigorous. Thank you very much! $\endgroup$ – David Bradwell Dec 2 '13 at 1:31
  • $\begingroup$ all the best with your studies! i love math, but sometimes suffer from short-circuit of the brain which leads me to make bizarre assertions! i'm trying to learn to be more careful and thorough. $\endgroup$ – David Holden Dec 2 '13 at 1:43
  • $\begingroup$ You might want to write “can be written uniquely as”, otherwise $\phi$ would not be well-defined. $\endgroup$ – Carsten S Dec 14 '13 at 14:58
  • $\begingroup$ You need to mention (or show a little) bit why the map $\varphi$ that you defined is linear. And then you need to mention again that by the same argumen the inverse map that you defined is linear as well. $\endgroup$ – user9077 Dec 11 '18 at 17:29

This is a common theorem that qualifies as
• Theorem 6.25 on P512 of Linear Algebra by David Poole,
Theorem 3.18 on P55 of Linear Algebra Done Right by Sheldon Axler,
• and can be found here at Google Books.

Also, in your first line, I think that you should specify $\phi$ to be a bijective linear transformation, and not just a bijection. I hope that this helps.


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