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.
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.
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.