<p>For every <span class="math-container">$k \times 1$</span> vector <span class="math-container">$\mathbf c\neq \mathbf 0$</span>, we have that <span class="math-container">$\mathbf X \mathbf c \neq \mathbf 0$</span>.
Choose <span class="math-container">$\mathbf e_i$</span> to be a standard basis <span class="math-container">$k \times 1$</span> vector having <span class="math-container">$1$</span> in the <span class="math-container">$i$</span>-th position, and <span class="math-container">$0$</span> everywhere else. <span class="math-container">$\mathbf X\mathbf e_i \neq \mathbf 0$</span> is a <span class="math-container">$n \times 1$</span> vector, and equal to the <span class="math-container">$i$</span>-th column of <span class="math-container">$\mathbf X$</span>.
So in block matrix notation we have that</p>
<p><span class="math-container">$$\mathbf X = \left[ \begin{matrix} \mathbf X\mathbf e_1\;\; \ldots\;\;\mathbf X\mathbf e_k\end{matrix}\right] $$</span></p>
<p>Note that the set of vectors <span class="math-container">$\mathbf X\mathbf e_i$</span> also form a basis.
Now,</p>
<p><span class="math-container">$$\mathbf C \mathbf X =\left[ \begin{matrix} \mathbf C\mathbf X\mathbf e_1\;\; \ldots\;\;\mathbf C\mathbf X\mathbf e_k\end{matrix}\right]$$</span></p>
<p>Since <span class="math-container">$\mathbf C$</span> is non-singular (and so of full column rank), and <span class="math-container">$n \times n$</span>, it follows that the <span class="math-container">$n \times 1$</span> vectors <span class="math-container">$\mathbf C(\mathbf X \mathbf e_i) \neq \mathbf 0$</span>. Moreover they are linearly independent by virtue of the full set of basis vectors used. So <span class="math-container">$\mathbf C\mathbf X$</span> is of full column rank.</p>

CC BY-SA 3.0
2015-09-05T15:00:46.577Z
@user200947 You're welcome.
Alecos Papadopoulos
CC BY-SA 4.0
2015-08-29T17:50:21.180Z
2022-09-17T12:58:38.337Z
2022-09-17T12:58:38.337Z
Alecos Papadopoulos
<p>Hi I am reading Hayashi's econometrics and have got stuck in the review question 1.5.1. The question asks us to prove that the multicollinearity assumption of the CLRM is satisfied by the transformed data matrix CX where C is (n by n) invertible matrix and X is an (n by k) matrix with rank = k. We have to show that the rank of CX = k. I know this is a theorem in matrix algebra and we can prove that rank CX = Rank X = k. But Hayashi suggests in the hint to the question another way to prove it and says that we must show that for all non-zero vectors c it must be that Xc is not equal to zero. Now, I could easily show this from the the fact that X has a full rank. But I get stuck in showing the final result that rank (CX) = rank (X) = k. Any suggestions on how to go about this, will be very helpful. It might something very straightforward that I am missing. Thanks a lot in advance. P.S. In case you are interested in looking at the question it is on page 59 of chapter 1.<a href="http://press.princeton.edu/chapters/s6946.pdf" rel="nofollow">http://press.princeton.edu/chapters/s6946.pdf</a></p>

<econometrics><statistics>
Multicollinearity in Transformed model: GLS
user200947
# Multicollinearity in Transformed model: GLS

(Asked by: username_0 on 2015-08-29)

Hi I am reading Hayashi's econometrics and have got stuck in the review question 1.5.1. The question asks us to prove that the multicollinearity assumption of the CLRM is satisfied by the transformed data matrix CX where C is (n by n) invertible matrix and X is an (n by k) matrix with rank = k. We have to show that the rank of CX = k. I know this is a theorem in matrix algebra and we can prove that rank CX = Rank X = k. But Hayashi suggests in the hint to the question another way to prove it and says that we must show that for all non-zero vectors c it must be that Xc is not equal to zero. Now, I could easily show this from the the fact that X has a full rank. But I get stuck in showing the final result that rank (CX) = rank (X) = k. Any suggestions on how to go about this, will be very helpful. It might something very straightforward that I am missing. Thanks a lot in advance. P.S. In case you are interested in looking at the question it is on page 59 of chapter 1.<http://press.princeton.edu/chapters/s6946.pdf>

**Tags:** econometrics, statistics


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## 1 answers

### Answer by username_1 on 2015-08-29. Score: 2 | Accepted answer

For every $k \times 1$ vector $\mathbf c\neq \mathbf 0$, we have that $\mathbf X \mathbf c \neq \mathbf 0$.
Choose $\mathbf e\_i$ to be a standard basis $k \times 1$ vector having $1$ in the $i$-th position, and $0$ everywhere else. $\mathbf X\mathbf e\_i \neq \mathbf 0$ is a $n \times 1$ vector, and equal to the $i$-th column of $\mathbf X$.
So in block matrix notation we have that

$$\mathbf X = \left[ \begin{matrix} \mathbf X\mathbf e\_1\;\; \ldots\;\;\mathbf X\mathbf e\_k\end{matrix}\right] $$

Note that the set of vectors $\mathbf X\mathbf e\_i$ also form a basis.
Now,

$$\mathbf C \mathbf X =\left[ \begin{matrix} \mathbf C\mathbf X\mathbf e\_1\;\; \ldots\;\;\mathbf C\mathbf X\mathbf e\_k\end{matrix}\right]$$

Since $\mathbf C$ is non-singular (and so of full column rank), and $n \times n$, it follows that the $n \times 1$ vectors $\mathbf C(\mathbf X \mathbf e\_i) \neq \mathbf 0$. Moreover they are linearly independent by virtue of the full set of basis vectors used. So $\mathbf C\mathbf X$ is of full column rank.

> **Comments:**
> **username_1**: @username_0 You're welcome.  

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