Complementary projection base

Question

I have a projection matrix of type

$P_1 = A (A^TA)^{-1}A^T$

The basis of the projected subspace is A.

If I have a projection of type $P_2 = I - [A(A^T A)^{-1}A^T]$

Its easy to proove that $A$ is a basis of $P_1$.

$P_1 A s = (A (A^T A)^{-1} A^T)As = As$ meaning this that any vector transformed by $A$, already lives in the subspace spanned by $P_1$.

I tried to do something similar with $P_2$.

$P_2 X s = X s$ and find the solution for X, but this approach only gives me something like $P_2Xs = 0$ which I think should verify for any s. Therefore I have $P_2X = 0$, which I cant solve since P_2 is not invertible.

So, how can I compute a basis for the projected subspace spanned by $P_2$?

Edited based on Bercis answer:

I know I can find a basis for $P_2$ using $QR$ decomposition. I think the basis its just the first $m$ columns with $m = rank(P_2)$.

I was expecting to have a direct approach, that require less computations, given that for $P_1$ I can get the basis directly.

Answer 1

So, it's the columns of $A$ which generates the subspace projected onto by $P_1$, but they are not necessarily linearly independent. Nevertheless, we can select a basis from them for $\mathrm{im}(P_1)=\mathrm{im}(A)$.

Since $P_1$ is idempotent ($P_1^2=P_1$), so is $P_2=I-P_1$, and we have $$v\in\mathrm{im}(P_2) \iff P_2v=v\iff P_1v=0$$ which means $v\perp\mathrm{im}(P_1)=\mathrm{im}(A)$ because $P_1^T=P_1$.

So, $P_2$ projects on the orthogonal complement of the column space of $A$, and to find a basis explicitly, you can use e.g. Gram-Schmidt orthogonalisation on an arbitrary basis extension of the selected $A$-columns.