How to solve $X'X=A$

Question

I am doing some statistical calculations.

Suppose I know $X'X=\begin{pmatrix} 50& 0 & 0 & 0\\ 0& 60 &0 & 0\\ 0& 0& 80& 20\\ 0& 0 &20 & 80 \end{pmatrix}$

$X'$ is the transpose of $X$ here.

My question is: Can I solve $X$ from this equation? or some function forms of $X$?

The Lyapunov Equation seems not work here. Matlab, R, c++ or any numberic soltion will be appreciated too.

Thanks.

There is not a unique solution here. For instance, let $O$ be some orthogonal matrix with the same dimension as the number of rows of $X$ and let $Z = OX.$ Then $Z'Z = X'O'OX = X'X.$ — spaceisdarkgreen, Oct 09 '17 at 04:11
Thanks, we don't know $X$ here, therefore, $Z$ cannot be decided either here. I guess there may have a lot of solutions, then can we get one of them? — Deep North, Oct 09 '17 at 04:34

score 1 · Accepted Answer · answered Oct 09 '17 at 04:34

1

As stated in the comment, there is no unique solution.

There is a name to such decomposition, cholesky decomposition.

One possible solution can be obtained by eigenvalue decomposition. where $A$ is your symmetric matrix.

$$A=UDU^T=UD^\frac12D^\frac12U^T=(D^\frac12U)^T(D^\frac12U)$$

where $U$ is unitary and $D$ is diagonal. Then you can choose $X$ to be $D^\frac12U$ to be a feasible solution.

You might also like to check out the function chol in R or Matlab.

answered Oct 09 '17 at 04:34

Siong Thye Goh

Thanks, I will look this solution. – Deep North Oct 09 '17 at 04:39
Thanks for the answer, is there more stricted solution that $U$ is a real matrix and not necessarily square? – Deep North Oct 09 '17 at 06:01
Well, you can also patch zero. $\begin{bmatrix} U & 0 \end{bmatrix}\begin{bmatrix} D & 0 \ 0 & 0\end{bmatrix}\begin{bmatrix} U^T \ 0 \end{bmatrix}=UDU^T$ – Siong Thye Goh Oct 09 '17 at 06:12
That's right, then how to restrict U to real numbers only? – Deep North Oct 09 '17 at 06:21
$U$ consists of orthonormal eigenvectors. They are real numbers. – Siong Thye Goh Oct 09 '17 at 06:22
Ok, that will change to spectral decomposition? – Deep North Oct 09 '17 at 06:26
$UDU^T$ is indeed just spectral decomposition. – Siong Thye Goh Oct 09 '17 at 06:27

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