Rank of a psd matrix and principal minors

Question

Let $M\in\mathbf{R}^{n\times n}$ be a symmetric matrix and assume that $M$ is positive semidefinite. Let $M_i$ be the leading principal ${i\times i}$ minor and assume that $det (M_i)\neq 0$. Now if rank $det (M_{i+1})=0$ thus this sufficie to deduce that rank $M=i$?

Answer 1

No, consider the matrix $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$$ and the case $i = 1$. The leading principal $1 \times 1$ minor has $\det M_1 = 1$ and $\det M_2 = 0$ but $\text{rank } M = 2 \neq 1$.

The conditions stated do guarantee, however, that $\det M_{i + 1} = i$.