Is the rank of an $n\times n$ Hermitian matrix $k-1$ if all of its principal minors of degree $k$ are zero? Is the rank of an $n\times n$ Hermitian matrix $\boldsymbol{H}$ equal to $k-1$ ($k<n$) if


*

*all its $k\times k$ principal minors are zero, and

*it has a nonzero $(k-1)\times(k-1)$ principal minor?


I am looking for a proof or counter example. Does anyone know of such a proof? 
I am really struggling with the above statement. I know that if a Hermitian matrix has rank $k$ then there must at least be one principal minor of $\boldsymbol{H}$ of dimension $k$ that is not zero. See :
K.D. Ikramov, "On the ranks of principal submatrices of diagonalizable matrices", Journal of Mathematical Sciences, Vol. 157, No. 5, 2009, 595--596.
But how do you prove the above statement from this knowledge (or maybe it is not possible)?
 A: This is a well known result and can be found, for example, here.
You argument is not quite correct, but can be saved if you use some elementary row transformations (which doesn't affect the rank) in order to make the other $n-k$ rows equal to zero. Then note that the corresponding column transformations also make the corresponding columns equal to zero. Now use your argument. 
A: That there exists a principal minor of order $k-1$ which is non-zero suggests that there exists at least $k-1$ independent rows/columns
in $\boldsymbol{H}$. The rank of $\boldsymbol{H}$ is thus at least $k-1$. Assume that the rank of $\boldsymbol{H}$ is larger than $k$. If
the rank of $\boldsymbol{H}$ is larger than $k$, $\boldsymbol{H}$ must at least have $k$ independent rows/columns. Now assume that these $k$ independent
rows are selected and that their indexes are $i_1 \cdots i_k$. Now select the columns of $\boldsymbol{H}$ that correspond to the same indexes. Since,
$\boldsymbol{H}$ is Hermitian the columns will also be linearly independent [Ikramov2009]. Let $\boldsymbol{R}$ denote the $k\times k$ principal minor formed by the cross-section of 
the selected rows and columns. The principal minor $\boldsymbol{R}$ should be non-zero due to its construction. This contradicts the fact that all principal
minors of order $k$ are zero implying that the rank of $\boldsymbol{H}$ is $k-1$.
