If I know that a matrix $G = (X^{T}X)^{-1}$, how can I recover what $X$ is?

Question

If I have a matrix $G$ where I know that $G=(X^{T}X)^{-1}$, is there a way to find $X$? Specifically, I would like to find $G$ where $G$ is:

$$G = \begin{bmatrix} 0.125 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0.125 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0.375 & -0.125 & 0 & -0.25 \\ 0 & 0 & -0.125 & 0.375 & 0 & 0.25 \\ 0 & 0 & 0 & 0 & 0.125 & 0 \\ 0 & 0 & -0.25 & 0.25 & 0 & 0.5 \\ \end{bmatrix} $$

thanks!

Answer 1

The problem of recovering a valid $X$ given the value of $X^TX$ (which is basically the same as being given $G = (X^TX)^{-1}$) is numerically important.

First of all, note that there are multiple solutions, assuming any single solution exists. That is, we can't recover $X$ without additional assumptions. In particular: if $X^TX = A$, then we also have $(UX)^T(UX) = A$ for any orthogonal matrix $U$.

If we make the assumption that $X$ must be upper-triangular, then we do have a unique solution, assuming that $A = X^TX$ is invertible. The process of finding the upper-triangular $X$ that satisfies $X^TX = A$ (given a matrix $A$) is known as Cholesky decomposition.

On the other hand, if we make the assumption that $X$ must be symmetric-positive definite, then we once again get a unique solution. This $X$ is known as the "positive (semi)definite square root" of $X^TX$.

Answer 2

The only way this could be possible is if the map $$X \to X^{T}X$$ were an injection - that is, if $X^{T}X$ were uniquely determined by $X$. In general this is very untrue. Indeed, let's look a bit closer - let $X$,$Y$ be invertible and $X^{T}X=Y^{T}Y$. Then $$(Y^{-1})^{T}X^{T}=YX^{-1} \implies ((YX^{-1})^{-1})^{T}=YX^{-1}$$ Writing $A=YX^{-1}$, we see $$(A^{-1})^{T}=A$$ Thus $X^{T}X=Y^{T}Y \iff Y=AX$ for some orthogonal matrix $A$.