<p>This kind of projections in econometrics are usually employed for partialling out some covariates from a linear regression.</p>

<p>Observe that in general $P_X\neq I$. Consider $X=\begin{bmatrix}1&amp;5\\1&amp;0\\1&amp;1\end{bmatrix}$. Then $X'X=\begin{bmatrix}3&amp;6\\6&amp;26\end{bmatrix}$ and $(X'X)^{-1}=\begin{bmatrix}26/42&amp;-6/42\\-6/42&amp;3/42\end{bmatrix}$ then $X(X'X)^{-1}X^T=\frac{1}{42}\begin{bmatrix}41&amp;-4&amp;5\\-4&amp;26&amp;20\\5&amp;20&amp;17\end{bmatrix}\neq I$.</p>

<p>First problem: $X^TP_X=X^TX(X'X)^{-1}X^T=(X^TX)(X'X)^{-1}X^T=I_{(\tilde k\times\tilde k)}X^T=X^T$</p>

<p>Similarly: $X^TM_X=X^T[I_{(\tilde k\times\tilde k)}-P_X]=X^TI_{(\tilde k\times\tilde k)}-X^TP_X=X^T-X^T=0$</p>

<p>The second problem is a direct consequence of $X^TM_X=0$, simply now you can post-multiply for $X_{(i)}$ since you are considering just one single column. The approach of $X_{(i)}=XA$ where $A$ is a vector $\tilde k\times 1$ of 0 apart from the $i$-th element which is 1. Then it should be straightforward to see: $0=(XA)^TM_X=M_XX_{(i)}$</p>

CC BY-SA 3.0
2017-10-14T14:11:18.493Z
I have never thought that kind of solving. It was very confusing for me to utilize such a matrix. Thanks a lot!
Jeffrey
CC BY-SA 4.0
2019-03-14T23:10:51.860Z
@GabMac Are there any econometrics books that present this linear regression thinking?
Linkman
CC BY-SA 3.0
2017-10-14T12:14:05.200Z
2017-10-14T12:14:05.200Z
GabMac
<p>When consider the following DGP : $y=X\beta^{*}+\epsilon$ where $\beta^{*}$ is a $\tilde k\times1\ $ vector.</p>

<p>Define the projection matrices: $P_{X}=X(X^{T}X)^{-1}X^{T}$ and $M_{X}=I-X(X^{T}X)^{-1}$.</p>

<p>first problem is prove that $X^{T}P_{X}=X^{T}$ and $ X^{T}M_{X}=0$</p>

<p>I thought this problem is too simple to think because P is just I. So, I think that there is an intention for given problem considering second problem.</p>

<p>Second problem is Prove thath if $X_{(i)}$ is the $i^{th}$ column of $X$, then, $M_{X}X_{(i)}=0$</p>

<p>My question is</p>

<ol>
<li><p>What is the point of first problem?</p></li>
<li><p>Can I solve these problem by considering $X_{(i)}=XA\ $ where A has a vector consisting of just 1 and the other columns are all zero? or Should I utilize first problem's solution?</p></li>
</ol>

CC BY-SA 3.0
2017-10-14T10:43:56.433Z
Please post separate questions separately.
Giskard
CC BY-SA 3.0
2017-10-14T12:04:36.980Z
The "first problem" is indeed immediately proven by simply inserting the expression for the projection matrices. For the second problem, use the fact that $M_X$ is symmetric.
Alecos Papadopoulos
<econometrics><linear-algebra>
How to utilize the projection matrix in econometrics?
Jeffrey
# How to utilize the projection matrix in econometrics?

(Asked by: username_0 on 2017-10-14)

When consider the following DGP : $y=X\beta^{\*}+\epsilon$ where $\beta^{\*}$ is a $\tilde k\times1\ $ vector.

Define the projection matrices: $P\_{X}=X(X^{T}X)^{-1}X^{T}$ and $M\_{X}=I-X(X^{T}X)^{-1}$.

first problem is prove that $X^{T}P\_{X}=X^{T}$ and $ X^{T}M\_{X}=0$

I thought this problem is too simple to think because P is just I. So, I think that there is an intention for given problem considering second problem.

Second problem is Prove thath if $X\_{(i)}$ is the $i^{th}$ column of $X$, then, $M\_{X}X\_{(i)}=0$

My question is

1. What is the point of first problem?
2. Can I solve these problem by considering $X\_{(i)}=XA\ $ where A has a vector consisting of just 1 and the other columns are all zero? or Should I utilize first problem's solution?

**Tags:** econometrics, linear-algebra

> **Comments:**
> **username_1**: Please post separate questions separately.  
> **username_2**: The "first problem" is indeed immediately proven by simply inserting the expression for the projection matrices. For the second problem, use the fact that $M_X$ is symmetric.  

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## 1 answers

### Answer by username_3 on 2017-10-14. Score: 2 | Accepted answer

This kind of projections in econometrics are usually employed for partialling out some covariates from a linear regression.

Observe that in general $P\_X\neq I$. Consider $X=\begin{bmatrix}1&5\\1&0\\1&1\end{bmatrix}$. Then $X'X=\begin{bmatrix}3&6\\6&26\end{bmatrix}$ and $(X'X)^{-1}=\begin{bmatrix}26/42&-6/42\\-6/42&3/42\end{bmatrix}$ then $X(X'X)^{-1}X^T=\frac{1}{42}\begin{bmatrix}41&-4&5\\-4&26&20\\5&20&17\end{bmatrix}\neq I$.

First problem: $X^TP\_X=X^TX(X'X)^{-1}X^T=(X^TX)(X'X)^{-1}X^T=I\_{(\tilde k\times\tilde k)}X^T=X^T$

Similarly: $X^TM\_X=X^T[I\_{(\tilde k\times\tilde k)}-P\_X]=X^TI\_{(\tilde k\times\tilde k)}-X^TP\_X=X^T-X^T=0$

The second problem is a direct consequence of $X^TM\_X=0$, simply now you can post-multiply for $X\_{(i)}$ since you are considering just one single column. The approach of $X\_{(i)}=XA$ where $A$ is a vector $\tilde k\times 1$ of 0 apart from the $i$-th element which is 1. Then it should be straightforward to see: $0=(XA)^TM\_X=M\_XX\_{(i)}$

> **Comments:**
> **username_0**: I have never thought that kind of solving. It was very confusing for me to utilize such a matrix. Thanks a lot!  
> **username_4**: @username_3 Are there any econometrics books that present this linear regression thinking?  

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