<p><span class="math-container">$\boldsymbol X^*_1$</span> is the residual of the <a href="https://en.wikipedia.org/wiki/Projection_matrix" rel="nofollow noreferrer">projection</a> of <span class="math-container">$\boldsymbol X_1 $</span> on <span class="math-container">$\boldsymbol X_2 $</span>. The 'residual maker' matrix is <span class="math-container">$\boldsymbol I - \boldsymbol X_2 \left(\boldsymbol X_2^T\boldsymbol X_2\right)^{-1}\boldsymbol X_2^T$</span>. Therefore, <span class="math-container">$$ \boldsymbol X_1^* = \left(\boldsymbol I -\boldsymbol X_2 \left(\boldsymbol X_2^T\boldsymbol X_2\right)^{-1}\boldsymbol X_2^T\right)\boldsymbol X_1.$$</span></p>

CC BY-SA 4.0
2019-08-14T09:30:11.210Z
Thanks! Can you please look at my other question too? https://economics.stackexchange.com/questions/30496/specification-bias-estimated-variance-is-biased-estimator-of-true-variance-of
Elina Gilbert
CC BY-SA 4.0
2019-08-14T08:38:01.840Z
2019-08-14T09:21:31.250Z
2019-08-14T09:21:31.250Z
ahorn
<p>Given the partitioned regression equation (into <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span>), I want to transform <span class="math-container">$X_1$</span>, say <span class="math-container">$X^*_1$</span>, such that <span class="math-container">$X_2$</span> and <span class="math-container">$X^*_1$</span> becomes orthogonal ie. <span class="math-container">$X_2^T$</span>. <span class="math-container">$X_1^*$</span>= 0.
A matrix can be transformed as <span class="math-container">$X_1^*$</span> = <span class="math-container">$X_1$</span>.p where P is transformation matrix.</p>

<p>If we premultiply <span class="math-container">$X_1^*$</span> = <span class="math-container">$X_1$</span>.p with <span class="math-container">$X_2^T$</span> we will get <span class="math-container">$X_2^T$</span>.<span class="math-container">$X_1$</span>.p = <span class="math-container">$X_2^T$</span>.<span class="math-container">$X_1^*$</span>= 0. How can I find P?</p>

<p>Part (ii) is easy to solve. Parameters of <span class="math-container">$X_2^T$</span> and <span class="math-container">$X_1^*$</span> will depend on their separate regression with Y. </p>

<p><a href="https://i.sstatic.net/NBPGf.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/NBPGf.jpg" alt="Question"></a></p>

<p><a href="https://i.sstatic.net/jdrg7.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/jdrg7.jpg" alt="Greene, Econometrics Analysis, Chapter 3,pg.49"></a></p>

<regression>
Transforming a matrix of explanatory variable in regression
Elina Gilbert
# Transforming a matrix of explanatory variable in regression

(Asked by: username_0 on 2019-08-12)

Given the partitioned regression equation (into $X\_1$ and $X\_2$), I want to transform $X\_1$, say $X^\*\_1$, such that $X\_2$ and $X^\*\_1$ becomes orthogonal ie. $X\_2^T$. $X\_1^\*$= 0.
A matrix can be transformed as $X\_1^\*$ = $X\_1$.p where P is transformation matrix.

If we premultiply $X\_1^\*$ = $X\_1$.p with $X\_2^T$ we will get $X\_2^T$.$X\_1$.p = $X\_2^T$.$X\_1^\*$= 0. How can I find P?

Part (ii) is easy to solve. Parameters of $X\_2^T$ and $X\_1^\*$ will depend on their separate regression with Y.

[![Question](https://i.sstatic.net/NBPGf.jpg)](https://i.sstatic.net/NBPGf.jpg)

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**Tags:** regression


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## 1 answers

### Answer by username_1 on 2019-08-14. Score: 2 | Accepted answer

$\boldsymbol X^\*\_1$ is the residual of the [projection](https://en.wikipedia.org/wiki/Projection_matrix) of $\boldsymbol X\_1 $ on $\boldsymbol X\_2 $. The 'residual maker' matrix is $\boldsymbol I - \boldsymbol X\_2 \left(\boldsymbol X\_2^T\boldsymbol X\_2\right)^{-1}\boldsymbol X\_2^T$. Therefore, $$ \boldsymbol X\_1^\* = \left(\boldsymbol I -\boldsymbol X\_2 \left(\boldsymbol X\_2^T\boldsymbol X\_2\right)^{-1}\boldsymbol X\_2^T\right)\boldsymbol X\_1.$$

> **Comments:**
> **username_0**: Thanks! Can you please look at my other question too? https://economics.stackexchange.com/questions/30496/specification-bias-estimated-variance-is-biased-estimator-of-true-variance-of  

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