linear algebra - Prove transpose of pseudoinverse commutes

How can I show that $(A^T)^+=(A^+)^T$, where
$A^+$ is Moore-Penrose Inverse?

I know there are 4 properties of the Moore-Penrose Generalized inverse, for example:
$$AA^+A=A^+.$$

To prove it, could I take the transpose of the above
$$(AA^+A)^T=(A^+)^T$$
and somehow simplify the LHS so it looks like the LHS of the original statement? Is this on the right track at all? I can get the LHS to be:
$$A^+A(A^+)^T$$ but this does not seem to help me.

Any hints please?

Answer

You are on the right track; we just need to show both matrices satisfy the same 4 properties.

Let $Z=A^{+}$, so $AZA=A, \;ZAZ=Z, \;(AZ)^{T}=AZ, \;(ZA)^{T}=ZA$.

Let $Y=(A^{+})^{T}=Z^{T}$. Taking transposes in the 4 properties above gives

$\;\;\;A^{T}YA^{T}=A^{T},\;YA^{T}Y=Y, \;AY^{T}=YA^{T},\;Y^{T}A=A^{T}Y$.

Now show that if $X=(A^{T})^{+}$, so $A^{T}XA^{T}=A^{T}, \;XA^{T}X=X, \;(A^{T}X)^{T}=A^{T}X,\;(XA^{T})^{T}=XA^{T}$,

then $Y=X$ since $Y$ satisfies these same 4 properties of the Moore-Penrose inverse.