Proof for Moore-Penrose inverse of transposed matrix

Question

I would like to ask how to prove that Moore-Penrose inverse of $A^T$ is $(A^+)^T$. I know that I can do it by proving all 4 properties but I am stuck at proving that $(A^+A)^T=A^+A$ and $(AA^+)^T=AA^+$

Answer 1

We work over $\Bbb R$. So the transpose and the hermitean conjugation coincide.

We start with $A^+$ satisfying the defining properties of a pseudoinverse:

• $(1)$ $AA^+A = A$,
• $(2)$ $A^+AA^+ = A^+$,
• $(3)$ $(A^+A)^T = A^+A$, i.e. $(3')$ $A^T(A^+)^T=A^+A$,
• $(4)$ $(AA^+)^T = AA^+$, i.e. $(4')$ $(A^+)^TA^T=AA^+$.

Let $B=A^T$. We denote by $X$ the matrix $(A^+)^T$, $$X=(A^+)^T\ .$$ We apply the transposition in the above relations $(1)$, $(2)$, and use $(3')$, $(4')$ in terms of the new notations.

• $(1)^T$ $BXB = B$,
• $(2)^T$ $XBX = X$,
• $(3')$ $BX=X^TB^T$,
• $(4')$ $XB=B^TX^T$.

Above, we can rewrite $X^TB^T=(BX)^T$, $B^TX^T=(XB)^T$.

By definition $X$ is $B^+$.

This is explicitly $(A^+)^T=(A^T)^+$.

$\square$

Note: I hope this also explains my comment.

Answer 2

I am stuck at proving that $(A^+A)^T=A^+A$ and $(AA^+)^T=AA^+$.

You cannot prove that because it isn't true in general (unless $A$ is real).

What you should prove is that both $A^T(A^+)^T$ and $(A^+)^TA^T$ are Hermitian, but this is obviously true because they are the transposes of $A^+A$ and $AA^+$, and the latter two products are Hermitian by the definition of $A^+$.