I'm stuck on the question when the residual vectors of different regressions are numerically equal according to the above mentioned theorem.
Let \begin{equation} y=Xb+e=X_{1}b_1+X_{2}b_2+e \label{eq:1} \end{equation} where $b_1, b_2$ are the OLS estimates of the "true" $\beta_1, \beta_2$, respectively. By pre-multiplying \ref{eq:1} by $M_1=X_1(X'_1X_1)^{-1}X'_1$ we obtain
\begin{equation}M_{1}y=M_{1}X_{2}b_2+M_{1}e \leftrightarrow \tilde{y}=\tilde{X_2}b_2+e\end{equation} Thus, the residuals of the regression of $\tilde{y}$ on $\tilde{X_2}$ are the same.
Considering the model $\tilde{y}=X_{1}b_1+X_{2}b_2+residual$, is it possible to say that $residual$ must equal $e$, because premultiplying this model by $M_1$ one obtains $\tilde{y}=\tilde{X_2}b_2+M_1*residual$, where $M_1*residual$ must be $e$ because we are regressing $\tilde{y}$ on $\tilde{X_2}$?
Pretty sure I'm making a mistake here. If this was true, then by the same reasoning, one would obtain that the residuals from $\tilde{y}$ on $X_2$ were also $e$, which is wrong I guess.
Thanks in advance!