Take the 2-minute tour ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

Linear model in matrix form is

$ \mathbf{y}=\mathbf{X}\beta+\epsilon\textrm{ where }\epsilon\sim\mathbb{N}\left(0,\sigma^{2}\mathbf{I}\right). $

If $ \mathbf{K}^{\prime}\left(\mathbf{X}^{\prime}\mathbf{X}\right)^{-} \mathbf{K} $ is nonsingular, then

$ \textrm{rank}\left[\mathbf{K}^{\prime}\left(\mathbf{X}^{\prime}\mathbf{X}\right)^{-} \mathbf{K}\right] = \textrm{rank}\left(\mathbf{K}^{\prime}\right). $ (From Linear Models by Searle)

I'm struggling to understand the last expression invloving ranks. Is this a result of any theorem? I'd highly appreciate if you explaing this to me. Thanks

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Suppose $\mathbf{K}$ is a $p\times k$ matrix. $\mathbf{K}'(\mathbf{X}'\mathbf{X})^-\mathbf{K}$ is $k\times k$ and by hypothesis is nonsingular, so it has rank $k$. Now consider the rank of $\mathbf{K}'$. If it were less than $k$, then the rank of $\mathbf{K}'(\mathbf{X}'\mathbf{X})^-\mathbf{K}$ would also be less than $k$, since the rank of a product of matrices is no greater than the rank of any one of the matrices. Thus the rank of $\mathbf{K}'$ cannot be less than $k$. But it also can't be greater than $k$, since the rank of a matrix can't be greater than its least dimension. So the rank of $\mathbf{K}'$ is also $k$, and you have your result.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.