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?
Best Answer
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.