# prove that $M_xM_1 = ?$

Let $\mathbf{y} = \mathbf{X_1}\mathbf{\beta_1}+\mathbf{X_2}\mathbf{\beta_2}+\mathbf{u}$.

Let $\mathbf{X} = [\mathbf{X_1} |\mathbf{X_2}]$. Define $\mathbf{M_i}$ to be the complementary projections for $i = 1,2$.

Then, $\mathbf{M_xM_1} = \mathbf{M_x}(\mathbf{I} - \mathbf{P_1}) = \mathbf{M_x - M_xP_1} = \mathbf{M_x-[I-P_x]P_1} = \mathbf{M_x-[P_1-P_xP_1]} = \mathbf{M_x - [P_1-P_1]} = \mathbf{M_x}$.

Could you check if this is okay?

• Hi: I don't quite follow the notation. I think $M_1=I−X_1(X_1^{T}X_1)^{-1}X_1^{T}$ and $M_2=I−X_2(X_2^{T}X_2)^{-1}X_2^{T}$ but what is $M_{x}$. Aug 11, 2018 at 14:13
• I think, $M_x$ is $I-X(X'X)^{-1}X'$. Do you know what $P_1$ is?
– erik
Aug 11, 2018 at 19:48
• $P_1$ is $X_1 (X_1'X_1)^{-1} X_1'$, I guess. Aug 12, 2018 at 0:48
• Erik: I think you are correct and I also agree with chan1142. Aug 13, 2018 at 2:26

The only assertion here that goes beyond self-evident matrix algebra and substitutions, is that

$$P_XP_1 = P_1$$

Since the columns of $X_1$ are in $X$ it follows that

$$P_X X_1 = X_1$$

because in general it holds that

$$P_X X = X$$

Namely the projection matrix operates as the identity matrix for the columns of the matrix from which it has been constructed. But then

$$P_X P_1 = P_XX_1(X_1'X_1)^{-1}X_1' = (P_XX_1)(X_1'X_1)^{-1}X_1'$$

$$=X_1(X_1'X_1)^{-1}X_1' = P_1$$

• " Thanks. I knew it was true but it's nice to see proof. Aug 13, 2018 at 2:28
• Yes. This looks like something from the econometrics course. Aah nostalgia.
– erik
Aug 13, 2018 at 9:33