Largest rank submatrix of a skew symmetric matrix 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.
 A: Another proof: For any $I$ and $J$ two subsets of $\{1,2,\ldots,n\}$ of the same cardinality, let $D(I,J)$ be the minor in rows $I$ and columns $J$. Let $Pf(I)$ be the Pfaffian $\sqrt{D(I,I)}$. We set $D(\emptyset, \emptyset) = Pf(\emptyset) =1$.
Lemma: Every $D(I,J)$ can be written as a quadratic polynomial in the $Pf(K)$'s.
Proof of Lemma: I claim that
$$D(I,J) = \sum_{S \subset J \ |J| \ \mbox{even}} (-1)^{|S|/2} \ Pf(I \cup (J \setminus S)) \ Pf(S).$$
(There may be some sign errors.)
Every term on the right can be visualized as corresponding to a perfect matching on the indices $J \sqcup I$. Focus on one particular term. Let $(j_1, j_1')$, $(j_2, j_2')$, ..., $(j_r, j_r')$ be the matchings which lie within $J$. This term will appear only when $S$ is a union of some set of these matchings. The total contribution of this term is thus $\sum_{k=0}^r (-1)^k \binom{r}{k}$. This is $1$ for $r=0$ and $0$ for $r>0$.
Application of lemma:  Suppose that $Pf(K)=0$ for all $K$ with $|K| > r$. Let $|I|=|J|>r$. Write $D(I,J)$ as a quadratic polynomial in the $Pf(K)$'s. Then, by degree considerations, for every monomial $Pf(K) Pf(K')$ in that quadratic polynomial, one of $Pf(K)$ and $Pf(K')$ is $0$. So imposing that all the principal minors of size $>r$ be $0$ implies that the same is true for non-principal minors.
A: 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.
