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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.

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    $\begingroup$ It's true, for both symmetric and skew-symmetric matrices, over an arbitrary field (and also hermitian...), and known as "Principal minor theorem". For a generalization, see Theorem 2.9 here (V. Kodiyalam, T. Lam, R. Swan, Determinantal ideals, Pfaffian ideals, and the principal minor theorem. Noncommutative rings, group rings, diagram algebras and their applications, 35-60, Contemp. Math., 456, Amer. Math. Soc., Providence, RI, 2008). I haven't found a more classical reference. $\endgroup$
    – YCor
    Jul 16, 2019 at 11:31

2 Answers 2

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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.

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  • $\begingroup$ Nice, I think this will works for any matrix $M$ whose $0$-eigenspace is the same as the generalized $0$-eigenspace, that is the set of all vectors killed by a power of $M$. $\endgroup$ Apr 22, 2010 at 14:22
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    $\begingroup$ Indeed, the "rank by principal minors" computes the multiplicity of 0 in the characteristic polynomial. $\endgroup$
    – damiano
    Apr 22, 2010 at 14:30
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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.

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