# What does it mean that the standard normal distribution is invariant under orthogonal transformation?

What does it mean that the standard normal distribution is invariant under orthogonal transformation?

This is the context where I found that statement: consider $H\subseteq \mathbb{\mathbb{R}^l}$ a $k$-dimensional linear subspace of $\mathbb{R}^l$. Let $(v_1,...,v_l)$ be an orthonormal basis of $\mathbb{R}^l$ whose first $k$ elements span $H$. Consider a random variable $Z$ taking values in $\mathbb{R}^l$ such that $Z\sim N(0,I_l)$ where $I_l$ is the identity matrix. Let $\tilde{Z}:=(\tilde{Z_1} \text{ }... \tilde{Z_l})^T$ be the coordinate vector of $Z$ with respect to the basis $(v_1,...,v_l)$. Then, $\tilde{Z}_i\sim N(0,1)$ for $i=1,...,l$ because the standard normal distribution is invariant under orthogonal transformation.

This means that if $Z\sim \mathcal{N}(0,I_l)$ and if $P$ is an orthogonal matrix, then $PZ\sim\mathcal{N}(0,I_l)$.
Here $P$ is the change of basis matrix from the original basis to the basis $(v_1,\dots,v_l)$. We have $Z=P\tilde{Z}$ and $P$ is an orthogonal matrix because $(v_1,\dots,v_n)$ is orthonormal. Thus $\tilde{Z}=P^{-1}Z=P^tZ\sim\mathcal{N}(0,I_l)$ because of course $P^t$ is also orthogonal.