# If I have only $(\pmb{X}'\pmb{X})^{-1}$, how can I find $\pmb{X}$ (the design matrix)?

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.

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

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?
• 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! – igorkf Jul 13 '19 at 12:01
• 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. – Grada Gukovic Jul 13 '19 at 12:07