 Solution

A linear map preserves linear combinations: $\psi (av+bw)=a\psi (v)+b\psi (w)$ for all vectors and real scalars. The rank-nullity theorem gives $\dim V=\dim \ker \psi +\dim \mathrm{im}\psi$. For an endomorphism of the same finite-dimensional vector space, surjectivity means rank $\dim V$, equivalent to zero kernel and hence injectivity.

If $\psi \varphi =I$, then every $v=\psi (\varphi v)$ lies in the image, so $\psi$ is surjective and consequently injective. Now $\psi (\varphi \psi v)=\psi v$; injectivity implies $\varphi \psi v=v$ for every $v$. Thus the right inverse is also a left inverse: $\boxed{\varphi \psi =I}$. This is the one-sided inverses of finite-dimensional endomorphisms property.

Apply it to the linear maps on ${\mathbb{R}}^n$ represented by $A$ and $B$: $AB=I$ implies $\boxed{BA=I}$. Equal finite dimensions are essential to the argument.