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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.

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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.

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