1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.