Let $P$ = $A(A^TA)^{-1}A^T$, where A is $m \times n $ 0f rank $n$. This is the projection matrix, right? Every site I've been on says that this is the projection matrix such that $P^2$ = $P$, but none explain why. Is this just a property of a projection matrix that doesn't require proof?
2026-03-27 01:43:11.1774575791
Show that $P^2$ = $P$
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Just by direct computation
$$P^2 = \left(A(A^T A)^{-1}A^T\right)\left(A(A^T A)^{-1}A^T\right) = A(A^T A)^{-1}(A^TA)(A^T A)^{-1}A^T=A(A^T A)^{-1}A^T = P.$$
In fact, this is the projection matrix onto the column space of a full column rank matrix $A$.