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My teacher gave me a problem, but he only give me the $(\pmb{X}'\pmb{X})^{-1}$ matrix.

If I have only $(\pmb{X}'\pmb{X})^{-1}$, how can I find $\pmb{X}$ (the design matrix)?
I think this is an algebra problem, but now I can't think a solution.

igorkf
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1 Answers1

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It is in general impossible. $(X^TX)^{-1} \in \mathbb{R}^{k\times k}$ and $X \in \mathbb{R}^{n\times k}$ with $n > k$. There are many matrices $K \in \mathbb{R}^{n\times k}$ such that $(K^TK) = (X^TX)$. Do you know something else about X?

user158565
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