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Why does $\operatorname{rank}(\mathbf{X})$ equal $\operatorname{rank}(\mathbf{X^TX})$? Is this true in general, please? And what is $\dim(\mathbf{X^TX})$, please? Does it equal to $\dim(\mathbf{X})$ in general, please? Thank you!

Imagine I prove the first claim by assuming that $\mathbf{X}$ is an $n\times k$ matrix with $n>k$ and $\mathbf{X}$ has rank $l<k$. Then, we know $\operatorname{nullity}(\mathbf{X})+\operatorname{rank}(\mathbf{X})=\dim (\mathbf{X})=l<k.$

To this end, I need to show first that $\dim(\mathbf{X})=\dim (\mathbf{X^TX})=l$. And then show $\operatorname{nullity}(\mathbf{X})=\operatorname{nullity}(\mathbf{X^TX})$.

I do not know how to show the first claim. Any hint, please? Thank you!

Provided the first claim is true, it is easy to see the following. If $\mathbf{X}v=\bf{0}$, then $\mathbf{X^TX}v=\bf{0}$. If $\mathbf{X^TX}v=\bf{0}$, then $v^T\mathbf{X^TX}v=\bf{0}$, which implies that $\mathbf{X}v=\bf{0}$. Hence, the second claim is proved.

Shaun
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LaTeXFan
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  • What is $\mathbf{X}$? Is it square matrix? Is it arbitrary matrix over arbitrary field? Is it a matrix over $\mathbb R$? – Martin Sleziak Jan 01 '14 at 13:08
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    You are not using the rank-nullity theorem correctly. If $X$ is an $m\times n$ matrix, then the rank-nullity theorem states that $$n = \mathrm{rank}(X)+\mathrm{nullity}(X)$$ When you write $\dim (X)$, the expression makes no sense; a matrix does not have a dimension. – EuYu Jan 01 '14 at 13:34
  • @EuYu Thank you for your response. You are certainly right. And you are the only one read my question carefully. – LaTeXFan Jan 01 '14 at 21:11

1 Answers1

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Look at the null space - if: $$X^T X v=0$$ Then also: $$v^T X^T X v = |Xv|^2 =0$$ So that the null space of $X^TX$ is the same as that of $X$. The rank is therefore the same as well.

Nathaniel Bubis
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    OP mentions all of this in the question, why repeat it? On the other hand, I have no idea what the real question is supposed to be. – Marek Jan 01 '14 at 11:13
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    @Nathaniel You mean $v^{\color{red}T}X^TXv$ in the second equation – Git Gud Jan 01 '14 at 12:02