Proof of Theorem IFDVS

(⇒) This is just the statement proved in Theorem IVSED.

(⇐) This is the advertised converse of Theorem IVSED. We will assume $U$ and $V$ have equal dimension and discover that they are isomorphic vector spaces. Let $n$ be the common dimension of $U$ and $V$. Then by Theorem CFDVS there are isomorphisms $\ltdefn{T}{U}{\complex{n}}$ and $\ltdefn{S}{V}{\complex{n}}$.

$T$ is therefore an invertible linear transformation by Definition IVS. Similarly, $S$ is an invertible linear transformation, and so $\ltinverse{S}$ is an invertible linear transformation (Theorem IILT). The composition of invertible linear transformations is again invertible (Theorem CIVLT) so the composition of $\ltinverse{S}$ with $T$ is invertible. Then $\ltdefn{\left(\compose{\ltinverse{S}}{T}\right)}{U}{V}$ is an invertible linear transformation from $U$ to $V$ and Definition IVS says $U$ and $V$ are isomorphic.