Push-through rule for square root of specific projection matrix

Question

Is the following a valid identity?

$$ (I - XX^T)^{1/2} X = X (I - X^T X)^{1/2} $$

Without the square root, it is quite easy to see that

$$ (I - XX^T) X = X (I - X^T X) $$

holds but I have trouble verifying that the upper equation holds.

Answer 1

A very general rule exists for these things.

Let $A,B\,$ be any two matrices such that the product $AB$ is a square matrix, and let $f(z)$ be an analytic function of a scalar argument.

Then the following is true $$f(AB)\cdot A = A\cdot f(BA)$$

The proof consists of a series expansion of the function coupled with the associativity of matrix multiplication.

In this particular case, $\,A=X,\,B=X^T\,$ and $\,f(z) = \sqrt{1-z}$

Answer 2

Making use of the SVD $X = U \Sigma V^T$, one arrives at

$$ (I - X X^T)^{1/2} X = U (I - \Sigma^2)^{1/2} \Sigma V $$

and

$$ X (I - X^T X)^{1/2} = U \Sigma (I - \Sigma^2)^{1/2} V $$

where the diagonal entries may be swapped, $\Sigma (I - \Sigma^2)^{1/2} = (I - \Sigma^2)^{1/2} \Sigma$. Hence the identity holds!