In Patterson and Thompson, 1971 I find the following claim which I cannot prove myself, nor have I been able to find a proof:
"As $S$ is idempotent and symmetric [and of size $n\times n$ and with rank $=n-t$], it can be expressed in the form $AA'$, where $A$ is an $n \times (n – t)$ matrix such that $A'A = I$."
My questions are: How do I prove this, and can the matrix $A$ be computed in a straightforward way?
I note that almost the same question was asked here and the answer accepted. However, if I understand it correctly that answer claims to prove the statement false, and that does not seem reasonable to me. I believe that answer considers square matrices $A$? Anyway, any help appreciated.
Best Answer
Supposing we are working on the field $\mathbb{R}$, $S$ can be written as $PDP^T$, where $P$ is an orthogonal matrix and $D$ is diagonal. We further suppose that the $n-t$ nonzero diagonal elements of $D$ are at the top-left.
Let's remark that since $PDP^T=S=S^2=PD^2P^T$, any eigenvalue $\lambda$ of $D$ satisfies $\lambda^2=\lambda$, so that $\lambda=1$ or $\lambda=0$. Therefore, any nonzero eigenvalue of $S$ is $1$, and we get that
$$S=QQ^T,$$
where $Q$ is the $n\times (n-t)$ matrix obtained by keeping the first $n-t$ columns of $P$. Furthermore, we have $Q^TQ=I_{n-t}$ since $P$ is orthogonal.