Moore-Penrose pseudoinverse by rank decomposition

Question

The wikipedia page on Moore Penrose pseudo inverse shows a method using rank decomposition to compute the pseudo inverse : $$A^+=C^+B^+=C^*(CC^*)^{-1}(B^*B)^{-1}B^* $$

But why $(CC^*)$ is inversible ? In particular I just realised that this is not true in some finite field, so the hypothesis that $\mathbb K$ is $\mathbb R$ or $\mathbb C$ seems important.

Answer 1

Theorem: If $C$ has full row-rank in a field of characteristic zero, then $CC^*$ is invertible.

Proof: Note that for $x \in \Bbb F^{n} = \Bbb F^{n \times 1}$, we have $$ CC^*x = 0 \implies\\ x^*CC^*x = 0 \implies\\ (C^*x)^*(C^*x) = 0 \overset{\text{char}(\Bbb F) = 0}{\implies}\\ C^*x \overset{\text{full row rank}}\implies\\ x = 0 $$ Since $C^*Cx = 0$ has only the trivial solution, $C^*C$ is invertible.