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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

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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

$$\mathbf X = \left[ \begin{matrix} \mathbf X\mathbf e_1\;\; ...\;\;\mathbf X\mathbf e_k\end{matrix}\right] $$

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

$$\mathbf C \mathbf X =\left[ \begin{matrix} \mathbf C\mathbf X\mathbf e_1\;\; ...\;\;\mathbf C\mathbf X\mathbf e_k\end{matrix}\right]$$

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 $\mathbf CX$ is of full column rank.

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  • $\begingroup$ @user200947 You're welcome. $\endgroup$ – Alecos Papadopoulos Sep 5 '15 at 15:00

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