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

<p>$$P_XP_1 = P_1$$</p>

<p>Since the columns of $X_1$ are in $X$ it follows that</p>

<p>$$P_X X_1 = X_1$$</p>

<p>because in general it holds that </p>

<p>$$P_X X = X$$</p>

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

<p>$$P_X P_1 = P_XX_1(X_1'X_1)^{-1}X_1' = (P_XX_1)(X_1'X_1)^{-1}X_1'$$</p>

<p>$$=X_1(X_1'X_1)^{-1}X_1' = P_1$$</p>

CC BY-SA 4.0
2018-08-13T02:28:17.107Z
" Thanks. I knew it was true but it's nice to see proof.
mark leeds
CC BY-SA 4.0
2018-08-13T09:33:58.923Z
Yes. This looks like something from the econometrics course. Aah nostalgia.
erik
CC BY-SA 4.0
2018-08-13T01:53:03.837Z
2018-08-13T01:53:03.837Z
Alecos Papadopoulos
<p>Let $\mathbf{y} = \mathbf{X_1}\mathbf{\beta_1}+\mathbf{X_2}\mathbf{\beta_2}+\mathbf{u}$. </p>

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

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

<p>Could you check if this is okay? </p>

CC BY-SA 4.0
2018-08-11T14:13:31.570Z
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 $M_{x}$.
mark leeds
CC BY-SA 4.0
2018-08-11T19:48:06.160Z
I think, $M_x$ is $I-X(X'X)^{-1}X'$. Do you know what $P_1$ is?
erik
CC BY-SA 4.0
2018-08-12T00:48:52.677Z
$P_1$ is $X_1 (X_1'X_1)^{-1} X_1'$, I guess.
chan1142
CC BY-SA 4.0
2018-08-13T02:26:12.300Z
Erik: I think you are correct and I also agree with chan1142.
mark leeds
<econometrics>
prove that $M_xM_1 = ?$
shk910
# prove that $M_xM_1 = ?$

(Asked by: username_0 on 2018-08-11)

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?

**Tags:** econometrics
> **Comments:**
> **username_1**: 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 $M_{x}$.  
> **username_2**: I think, $M_x$ is $I-X(X'X)^{-1}X'$. Do you know what $P_1$ is?  
> **username_3**: $P_1$ is $X_1 (X_1'X_1)^{-1} X_1'$, I guess.  
> **username_1**: Erik: I think you are correct and I also agree with chan1142.  

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## 1 answers

### Answer by username_4 on 2018-08-13. Score: 1

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

$$P\_X P\_1 = P\_XX\_1(X\_1'X\_1)^{-1}X\_1' = (P\_XX\_1)(X\_1'X\_1)^{-1}X\_1'$$

$$=X\_1(X\_1'X\_1)^{-1}X\_1' = P\_1$$

> **Comments:**
> **username_1**: " Thanks. I knew it was true but it's nice to see proof.  
> **username_2**: Yes. This looks like something from the econometrics course. Aah nostalgia.  

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