Sub-matrix positive definite $\implies$ Matrix positive semidefinite Let $M$ be a symmetric matrix $n\times n$ such that $\operatorname{rank}(M)=p$. The sub-matrix $M_0$ spanned by the rows $\{1,\ldots,p\}$ and columns $\{1,\ldots,p \}$ is such that $\operatorname{rank}(M_0)=p$. Prove that

$M_0$ positive definite $\implies$ $M$ positive semi-definite ?

I appreciate any help. Thank you
 A: Write the matrix out as
$$
M = \pmatrix{M_0&B\\B^T & C}
$$
$M_0$ is (symmetric) positive semidefinite, so we can select an invertible $P$ so that $P^TM_0P = I$.  We then note that
$$
\pmatrix{P\\&I} M \pmatrix{P\\&I}^T = \pmatrix{I_{p\times p}&PB\\(PB)^T & C} =: \pmatrix{I&D\\D^T & C} =: \bar M
$$ 
and it suffices to show that this new matrix $\bar M$ is positive semidefinite. 
Note that $rk(\bar M) = p$.  It follows that the $n-p$ bottom rows are a linear combination of the first $p$ rows.  That is, there exists some matrix $R$ such that 
$$
R\pmatrix{I & D} = \pmatrix{D^T&C}
$$
However, $R \pmatrix{I&D} = \pmatrix{R & RD}$, so we must have $R = D^T$ since the left blocks are equal.  Thus, our matrix $\bar M$ is now
$$
\bar M = \pmatrix{I & D\\D^T & D^TD} = \pmatrix{I & D}^T \pmatrix{I & D}
$$
So, $M$ is positive semidefinite.
A: Somewhat similar to the previous answer, but a little bit shorter: do Gauss elimination of $B^T$ in the block matrix $M = \pmatrix{M_0&B\\B^T & C}$ as
$$
\pmatrix{I&0\\-B^TM_0^{-1} & I}\pmatrix{M_0&B\\B^T & C}=\pmatrix{M_0&B\\0 & -B^TM_0^{-1}B+C}.
$$
The rank condition implies that $-B^TM_0^{-1}B+C=0$ giving $C=B^TM_0^{-1}B$. Thus
$$
\pmatrix{M_0&B\\B^T & C}=\pmatrix{M_0&B\\B^T & B^TM_0^{-1}B}=\pmatrix{I\\B^TM_0^{-1}}M_0\pmatrix{I&M_0^{-1}B}
$$
is indeed positive semidefinite.
