# Random vector whose non-invertible linear transformation is Gaussian

It is well known that affine transformations of Gaussian vectors are also Gaussian.

Let $$Y \sim \mathcal{N}( \mathbf{0}, I)$$ be an $$n$$ dimensional iid Gaussian random vector. Let $$A$$ be an $$n \times m$$ matrix that does not have a left inverse. Let $$A X \stackrel{d}{=} Y$$. What can we say about the distribution of $$X$$?

We need $$m>n$$ and $$A$$ full rank. Then, one option is $$X=A^T (AA^T)^{-1}W$$, where $$W\sim \mathcal N (0, I_n)$$. In particular, this means that $$X$$ has a degenerate Gaussian distribution as it's covariance matrix $$A^T (AA^T)^{-2}A^T$$ is singular.
• While I think this answer has some value (+1), I think there are two problems (unless I am mistaken, please let me know): (1) The question specifies that $A$ does not have a left-inverse, and you have used the Moore-Penrose pseudo-inverse. (2) I think you have $m > n$ where you should have $n < m$. That is, the existence of a left-inverse requires more rows than columns for linear independence. EDIT: Actually it seems that you have used a right-inverse, so I don't understand the answer at all? – Student May 31 at 5:31
• Yes that is a right inverse. The condition $m>n$ and full rank is needed because we cannot "create randomness" just by linear manipulations: if $X$ has less entries than $Y$, $Y$ cannot be standard normal (I think). The idea is then to put a standard normal inside of $X$ such that a later multiplication by $A$ reveals the standard normal. – Riccardo Sven Risuleo May 31 at 7:12