I have this question I met in my studies of matrix theory on which I need help:
Let A be positive semidefinite and real of order $ n \times n $. Let $ A[\alpha] $ be a principal submatrix of A (the submatrix of A indexed by rows and columns given by the vector $ \alpha $) and we are asked to prove the following equalities of rank $$\DeclareMathOperator\rk{rank} \rk A[\alpha|1,2,...,n] = \rk A[1,2,...,n|\alpha] = \rk A[\alpha] $$ and we are given the following hint: we are told to assume $ \alpha=\{1,...,k\} $ and as A is positive semidefinite we may factorize it as $ A=BB^T $ where B is the block 2x1 matrix of the form $ B=\begin{pmatrix}X\\Y\end{pmatrix} $ where X is a $ k \times k $ matrix.
To be honest, I have no real idea how to use the clue and why they ask us to make the assumption about $ \alpha=\{1,2,...,k\} $ and how to proceed to the general equality we are asked to prove. I appreciate all helps.
Here's an idea: take $B = \pmatrix{X\\Y}$ where $X$ is a $k \times n$ matrix, where $B$ is such that $BB^T = A$. In particular, we have $$ A = \pmatrix{XX^T & XY^T\\YX^T & YY^T} $$
Let $P = \pmatrix{I & 0}$, where $I$ is the $k \times k$ identity matrix. Notice that for $\alpha = \{1,\dots,k\}$, we have $$ A[\alpha] = PAP^T = (PB)(PB)^T = XX^T $$ Conclude that $A[\alpha]$ has the same rank as $PB = X$. On the other hand, $$ A[\alpha \mid 1,\dots,n] = PA = (PB)B^T = XB^T $$ thus, we conclude that the rank of $A[\alpha \mid 1,\dots,n]$ is at most the rank of $X$, which is to say that $$ \operatorname{rank}(A[\alpha \mid 1,\dots,n]) \leq \operatorname{rank}(A[\alpha]) $$ I think you'll find that the reverse inequality, $\operatorname{rank}(A[\alpha \mid 1,\dots,n]) \geq \operatorname{rank}(A[\alpha])$, is a bit more obvious.