Is the following statement true?


Given a skew symmetric matrix M, among all of its largest rank sub-matrix, there must be one that is the principal submatrix of M.



Largest rank submatrix of a skew symmetric matrix
Yes. Suppose that you have a general $n \times n$ matrix *M*. The coefficient of $(-x)^k$ in the characteristic polynomial is the sum of all the principal minors of the matrix of size *n-k*. From this it follows that the "rank computed by the principal minors" can differ from the actual rank of the matrix only if the matrix is not diagonalizable. Since skew-symmetric matrices are diagonalizable, the result follows.


