$A{A}^{+}$ and $A^{+}A$ are the orthogonal projections onto $\mathrm{im}A$ and $\mathrm{im}{A}^{\ast }$

By [L4], write $A=U\mathrm{\Sigma }{V}^{\ast }$. Define $B:=V{\mathrm{\Sigma }}^{+}{U}^{\ast }$ by reciprocating the positive singular values and leaving the zero block fixed. Direct diagonal multiplication verifies the four Penrose equations, so uniqueness in [L1] gives $B=A^{+}$. Hence $A{A}^{+}=U(\mathrm{\Sigma }{\mathrm{\Sigma }}^{+})U^{\ast },A^{+}A=V({\mathrm{\Sigma }}^{+}\mathrm{\Sigma })V^{\ast }\mathrm{.}$

The products $\mathrm{\Sigma }{\mathrm{\Sigma }}^{+}\in M_m(\mathbb{F})$ and ${\mathrm{\Sigma }}^{+}\mathrm{\Sigma }\in M_n(\mathbb{F})$ are the respective rank-$r$ diagonal projections, each with $1$ in its nonzero singular directions and $0$ elsewhere, where $r$ is the number of nonzero singular values from [L2]. Hence $A{A}^{+}$ and $A^{+}A$ are self-adjoint idempotents.

By [L3], $A{A}^{+}$ and $A^{+}A$ are orthogonal projections onto their image spaces. In the singular basis, $\mathrm{im}(A{A}^{+})$ is the span of the left singular vectors corresponding to the nonzero singular values, which is exactly $\mathrm{im}A$.

The same computation shows that $\mathrm{im}(A^{+}A)$ is the span of the right singular vectors corresponding to the nonzero singular values, which is exactly $\mathrm{im}{A}^{\ast }$.

Therefore $A{A}^{+}$ and $A^{+}A$ are the orthogonal projections onto $\mathrm{im}A$ and $\mathrm{im}{A}^{\ast }$ respectively.