# Moore-Penrose inverse of Moore-Penrose inverse

If Y is the pseudoinverse of matrix X, then X will be the pseudoinverse of Y. This is a trivial consequence once the Moore-Penrose conditions are written:

$$Y = X^+$$ implies

\begin{aligned} XYX&=X\\ YXY&=Y\\ (XY)^T&=XY\\ (YX)^T&=YX \end{aligned}

Let $$Z=Y^+$$. This would mean

\begin{aligned} YZY&=Y\\ ZYZ&=Z\\ (YZ)^T&=YZ\\ (ZY)^T&=ZY \end{aligned}

Substituting $$Z=X$$ results in the second set of conditions being identical to the first.

However, when I try to simplify the expression for the Moore-Penrose inverse, the resulting equations are a mess, and I don't see how to move forward.

\begin{align} Y &= (X^TX)^{-1}X^T\\ \implies Y^T &= X(X^TX)^{-T} = X(X^TX)^{-1}\\ Y^TY &= X(X^TX)^{-1}(X^TX)^{-1}X^T\\ \implies (Y^TY)^{-1}Y^T &= [X(X^TX)^{-1}(X^TX)^{-1}X^T]^{-1}X(X^TX)^{-1} \end{align}

Any ideas on how that last expression reduces to $$X$$? Matrix algebra proofs appreciated.

Unless $$X$$ is a square and nonsingular, it is not equal to $$(Y^TY)^{-1}Y^T$$.

When you write $$Y=(X^TX)^{-1}X^T$$, you are assuming that $$X$$ has full column rank (otherwise $$X^TX$$ is not invertible). It follows that $$X$$ is a "tall" matrix, i.e. $$X$$ is $$m\times n$$ for some $$m\ge n$$. Hence $$Y$$ is a "fat" matrix. So, when $$m>n$$, $$Y$$ has deficient column rank and $$Y^TY$$ cannot possibly be invertible.

The correct expression of $$X$$ in terms of $$Y$$ should be $$X=Y^T(YY^T)^{-1}$$: \begin{aligned} Y^T(YY^T)^{-1} &=\left((X^TX)^{-1}X^T\right)^T\left[(X^TX)^{-1}X^T\left((X^TX)^{-1}X^T\right)^T\right]^{-1}\\ &=X(X^TX)^{-1}\left[(X^TX)^{-1}X^TX(X^TX)^{-1}\right]^{-1}\\ &=X(X^TX)^{-1}\left[(X^TX)^{-1}\right]^{-1}\\ &=X. \end{aligned}

\begin{align} [X(X^TX)^{-1}(X^TX)^{-1}X^T]^{-1}X(X^TX)^{-1}&=X\\ \Leftrightarrow X(X^TX)^{-1} &= [X(X^TX)^{-1}(X^TX)^{-1}X^T]X\\ \Leftrightarrow X(X^TX)^{-1} &= X(X^TX)^{-1}\end{align}