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Suppose we have a matrix of $n$ by $n$ dimension $M$ is that is full rank, symmetric, and positive semi-definite, in that $z^TMz \geq 0$ for all $z \in \mathbb{R}^p$. This can be thought of as a covariance matrix in statistics. If I were to take a square partition, would that square partition still be full rank? In example, suppose that:

$$ M = \begin{pmatrix} a_{11} \ldots a_{1n}\\ \ldots\\ a_{n1} \ldots a_{nn}\\ \end{pmatrix} $$

Then a square partition might be:

$$ M_{33} = \begin{pmatrix} a_{33} \ldots a_{3n}\\ \ldots\\ a_{n3} \ldots a_{nn}\\ \end{pmatrix} $$

Where I took the bottom right side of the original matrix $M$. Would this be full-rank as well?

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  • $\begingroup$ Isn't a positive semi-definite full rank matrix actually definite? Do square partitions necessarily sit symmetrically within $M$, as your example suggests? $\endgroup$ – hardmath Oct 5 '16 at 1:54
  • $\begingroup$ @hardmath Yes they do, sorry for not pointing that out, changed it thanks! $\endgroup$ – user123276 Oct 5 '16 at 1:58
  • $\begingroup$ It wont always be. Just thinking in other way. Lets take a $2\times 2$ matrix which is not full-rank. It will be possible to extend this matrix to a larger matrix say $n\times n$, $n>2$, such that this matrix is full rank and all its singular values are positive. $\endgroup$ – Seyhmus Güngören Oct 5 '16 at 2:04
  • $\begingroup$ @SeyhmusGungoren: Is there a not or im- missing in your Comment? Suppose the $1,1$ entry is zero. How could the rest of a symmetric matrix be filled out to overcome the fact that $e_1'Me_1$ is zero? $\endgroup$ – hardmath Oct 5 '16 at 2:24
  • $\begingroup$ @hardmath the eigenvectors are related to the whole matrix. if you modify just one row then it will affect all elements of the result of the multiplication and it still seems to me that it should be possible to get one sub square matrix which is not full rank while the whole big matrix is full rank. just add rows and rows, millions of randomly generated rows and eventually it will be possible to get one full rank matrix. I mean do you say that $[1,2; 2,4]$ can never be a submatrix a full rank matrix? $\endgroup$ – Seyhmus Güngören Oct 5 '16 at 2:37
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Short answer: yes.

One easily proves this by constructing vectors $v$ with zeros in the right places so that $v'Mv \gt 0$ proves the desired "square partition" is itself positive definite.

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    $\begingroup$ I see, so in my example above I would just be using the vector $v = (v_1, \ldots, v_n) = (0,0,v_3, v_4, \ldots, v_n)$ where $v_3, v_4, \ldots, v_n \in \mathbb{R}$? $\endgroup$ – user123276 Oct 5 '16 at 3:54
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    $\begingroup$ Yes, that is the idea. $\endgroup$ – hardmath Oct 5 '16 at 11:04
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Because your matrix is full-rank, it is positive-definite. And a matrix is positive-definite if and only if all its principal submatrices are positive-definite.

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