Q:

why does the squared norm projection of a standard normal follow a chi squared distribution

Let $Z \sim \mathcal{N} ({\bf 0}, I_d)$ be a $d$-dimensional
multivariate normal RV, $n$ be a unit vector where all components are non-zero, then $Z$ projected on to the
plane normal to $n$ is $Z - \langle n, Z \rangle n = (I_d - n n^T) Z$.
Our lecture notes say that $\| (I_d - n n^T) Z \|^2$ is a chi-squared distribution with $d - 1$ degrees of freedom due to $Z$ being invariant under rotations, but I cannot see why the chi-squared distribution claim follows.

A:

I've come up with this answer but perhaps there's also an intuitive understanding of the above?
Let $A := I_d - n n^T$. Then $A$ has eigenvalue $1$ with multiplicity $d - 1$ and $0$ with multiplicity $1$. To see this suppose $A x = \lambda x$ where $\lambda \in \mathbb{R}, x \in \mathbb{R}^d$, then $(1 - \lambda) x = n n^T x$. If $\lambda = 1$ then since $n \ne 0$ we have $n^T x = 0$, and since no component of $n$ is 0 there are $d - 1$ linearly independent $x$s by rank-nullity. Otherwise $n^T x \ne 0$, and $x = c n$ for some $c \ne 0 \in \mathbb{R}$, then since $\| n \| = 1$ and no component of $n$ is $0$ we have $\lambda = 1$.
Now $A$ is symmetric so it is diagonalizable using orthogonal matrices, so for some $Q \in \text{SO} (d)$ $\| Q^T A Q Z \|^2 = \| A Q Z \|^2 \overset{d}{=} \| A Z \|^2$ since $Q Z \overset{d}{=} Z$. The diagonals give us $\chi^2_{d - 1}$.

