Suppose $P=X(X'X)^{-1}X'$ and $X$ can be descomposed as $X= [X_1 X_2]$, where $X$ is a matrix. Then is true that $PX_1=X_1$.
My proof:
$X'X=[X_1^2+X_2^2]$ Then $[X_1^2+X_2^2]^{-1}=1/(X_1^2+X_2^2)$. Thus $(X_1^2+X_2^2)X_1/(X_1^2+X_2^2)=X_1$
Is this proof correct?
Best Answer
ok, So $X$ is tall skinny matrix, typically with many many more rows than columns.
Suppose, for example that $X$ is a $100\times5$ matrix. Then $X^\top X$ is a $5\times5$ matrix. If $X_1$ is a $100\times3$ matrix and $X_2$ is $100\times2,$ then what is meant by $X_1^2+X_2^2,$ let alone by its reciprocal?
If $x$ is any member of the column space of $X$, then $Px=x.$
This is proved as follows: $x = Xu$ for some suitable column vector $u$.
Then $Px = \Big(X(X^\top X)^{-1} X^\top\Big) (Xu) = X(X^\top X)^{-1}(X^\top X) u = Xu = x.$
(Similarly if $x$ is orthogonal to the column space of $X$, then $Px=0.$ The proof of that is much simpler.)
Now observe that the columns of $X_1$ are in the column space of $X.$