# The linear transformation of the normal gaussian vectors

I am facing difficulty in proving the following statement. It is given in a research paper found on Google. I need help in proving this statement!

Let $X= AS$, where $A$ is orthogonal matrix and $S$ is gaussian. The isotopic behavior of the Gaussian $S$ which has the same distribution in any orthonormal basis.

How is $X$ Gaussian after applying $A$ on $S$?

• Since you mention a paper you found on Google, please link to the paper. – Ben Jun 14 '18 at 4:44
• Sorry I search in Private mode and now I am not able to track it. Actually it is related to Independent Component analysis in unsupervised learning. – ironman Jun 14 '18 at 4:52
• No problem - hopefully my answer helps anyway. – Ben Jun 14 '18 at 5:05
• Suggest to change the title to something a little more precise like "linear transformation of normal gaussian vectors". – JayCe Jun 14 '18 at 13:07

## 2 Answers

Since you have not linked to the paper, I don't know the context of this quote. However, it is a well-known property of the normal distribution that linear transformations of normal random vectors are normal random vectors. If $\boldsymbol{S} \sim \text{N}(\boldsymbol{\mu}, \boldsymbol{\Sigma})$ then it can be shown that $\boldsymbol{A} \boldsymbol{S} \sim \text{N}(\boldsymbol{A} \boldsymbol{\mu}, \boldsymbol{A} \boldsymbol{\Sigma} \boldsymbol{A}^\text{T})$. Formal proof of this result can be undertaken quite easily using characteristic functions.

For a little bit of visualisation, consider that the Gaussian distribution is scaled by r^2, so multiple independent axes form a Pythagorean relation when scaled by their standard deviations, from which follows that the re-scaled distribution fuzz ball becomes spherical (in n dimensions) and can be rotated about its centre at your convenience.

One of the radial measures is the Mahalanobis distance and is useful in many practical cases where the central limit is applied...