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 ?


1 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 $\chi^2_{2(n-k)}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.