Find the distribution of $|v_0^*w_0|^2$ in this specific case

Question

Suppose I have a set of complex Gaussian (with zero mean and unit variance) i.i.d. vectors $w_0,w_1,\ldots,w_k$, each of which have dimension $n \times 1$.
We define matrix $W=[w_1,\ldots,w_k]$. For this matrix, we perform the QR decomposition as $W=Q R=[Q_1 \, Q_2] R$. I think that the columns of matrix $Q_2$ (of dimension $n \times (n-k)$) form an orthonormal basis for the null-space (which is $n-k$ dimensional) of $w_1,\ldots,w_k$.
Let $v_0=\frac{P^*w_0}{||P^*w_0||}$, where the $n \times n$ matrix $P$ represents the orthogonal projection onto the subspace defined by the columns of $Q$, and where $(\cdot)^*$ denotes the conjugate transpose.

I am looking for the distribution of $|v_0^*w_0|^2$, or equivalently of $\frac{|w_0^*Pw_0|^2}{||P^*w_0||^2}$.

My attempt:
As mentioned before, this nullspace is $n-k$ dimensional, and is independent of $w_0$. Maybe we can claim that $|v_0^*w_0|^2$ can be seen as the squared norm of the projection of $w_0$ on the nullspace of $w_1,\ldots,w_k$, thus $|v_0^*w_0|^2$ follows a $\chi^2_{2(n-k)}$.
Is it correct ? the dependency between $v_0$ and $w_0$ doesn't make my claim incorrect ?

Answer 1

I think that I found the solution.

We can write $|w_0^*P w_0|=|<w_0,Pw_0>|$, where $<\cdot,\cdot>$ is the scalar product.
We know that $<w_0,Pw_0>=||Pw_0||^2$, since $P$ is an orthogonal projection.
Thus, we obtain $\frac{|w_0^*P w_0|^2}{||Pw_0||^2}=||Pw_0||^2$, which is nothing but the squared norm of the projection of $w_0$ onto a nullspace of dimension $n-k$. Hence, $\frac{|w_0^*P w_0|^2}{||Pw_0||^2}$ follows a $\chi^2_{2(n-k)}$.