Suppose there is a rank-$K$ matrix $\mathbf{X}$ that has dimensions $G \times K$ with $G > K$. The $K \times G$ matrix $\mathbf{L}$ is defined as:
$\mathbf{L} \equiv (\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'$
The question is: can I solve to find $\mathbf{X}$ as a function of $\mathbf{L}$? Is the solution unique? Is the following ever a solution?
$\mathbf{X} = \mathbf{L}'(\mathbf{L}\mathbf{L}')^{-1}$
I'm sure there must be a simple answer to this but I don't know what to look up.
On
I think I've worked out the answer.
We start with $\mathbf{L} \equiv (\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'$. This implies $\mathbf{L}\mathbf{X} = \mathbf{I}$ as Fede notes. As a result we can write \begin{align} \mathbf{L}\mathbf{L}' &= (\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'\mathbf{L}' \\ &=(\mathbf{X}'\mathbf{X})^{-1} \end{align} because $\mathbf{X}'\mathbf{L}' = (\mathbf{L}\mathbf{X})' = \mathbf{I}' =\mathbf{I}$. We can then substitute this into the definition for $\mathbf{L}$ \begin{align} \mathbf{L} &\equiv \mathbf{L}\mathbf{L}'\mathbf{X}' \\ (\mathbf{L}\mathbf{L}')^{-1}\mathbf{L} &= \mathbf{X}' \\ \mathbf{X} &= \mathbf{L}'(\mathbf{L}\mathbf{L}')^{-1} \end{align} where the inversion of $\mathbf{L}\mathbf{L}'$ is possible because it is dimension $K \times K$ and rank $K$ (the latter because $X$ is rank $K$).
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Beware of the different sizes of the matrices. Remember that for a matrix to be invertible it needs to be square first. In this spirit, while $(X'X)$ might be invertible, $X$ alone is not. You can try the following: \begin{align*} L&= (X'X)^{-1}X' \\ LX&= (X'X)^{-1}X'X \\ LX&= I_{k} \\ L'LX &=L'\\ X&=(L'L)^{-1}L' \end{align*} I invite you to plug this expression for $X$ back into the first line and check for yourself it is valid.
And as a recommendation for your future questions: it is customary to include what you have tried towards solving your question to help those who answer identify where your difficulties lie and structure their answers accordingly.