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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.

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    $\begingroup$ $\frac 1 {x^2} = 1$, $x$ can be 1 or -1. $\endgroup$ – user158565 Jul 13 '19 at 12:09
  • $\begingroup$ I didn't understand why you simply sayed this. Can you be more specific, please? $\endgroup$ – igorkf Jul 13 '19 at 12:13
  • $\begingroup$ Think $x$ is 1 $\times$ 1 matrix. $\endgroup$ – user158565 Jul 13 '19 at 12:19
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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?

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  • $\begingroup$ My teacher gave us the SSE (say 200) and the number of observartions (say 15). He want the unbiased $\sigma^2$. I thought use the fact of $\frac{SSE}{n-k} = \hat{\sigma}^2$, but he wants this in matrix form! $\endgroup$ – igorkf Jul 13 '19 at 12:01
  • $\begingroup$ Is SSE sum of squared errors? $\endgroup$ – Grada Gukovic Jul 13 '19 at 12:04
  • $\begingroup$ Yes, SSE is the sum of squared errors $\endgroup$ – igorkf Jul 13 '19 at 12:04
  • $\begingroup$ Both SSE and the variance are numbers, not matrices. You need $\frac{SSE}{n-k}$. That part with the matrix form is probably just a confusion. $\endgroup$ – Grada Gukovic Jul 13 '19 at 12:07
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    $\begingroup$ @igor given your comments your question seems to omit relevant information. Please edit to describe the circumstances. $\endgroup$ – Glen_b -Reinstate Monica Jul 14 '19 at 0:58

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