3

Let $F$ be a field where $char(F)\neq2$ and let $A$ be a skew-symmetric matrix over $F$. Prove that rank of $A$ is even.

I think the best way to prove it, is using induction on size of $A$.

for $n=1$, matrix $A$ should be zero matrix so $rank(A)=0$ and it's even.

for an $(n+1)\times (n+1)$ matrix, by eliminating the last row and the last column we will have an $n \times n$ skew-symmetric matrix which by the induction hypothesis has even rank. Now how should I prove that by adding back the eliminated row and column the rank of matrix again will be even? I've stuck here!

Please do not mark the question as duplicate with Rank of skew-symmetric matrix. At least the approach that I'm trying to prove this question with is different. No one has proved this statement by induction, This question is not duplicate really!

Community
  • 1
F.K
  • 1,184
  • 1
    Over $\Bbb R$, by the Spectral Theorem all of the nonzero eigenvalues of $A$ are imaginary, so the nonzero eigenvalues come in conjugate pairs. Then, it's enough to know that skew-symmetric matrices are diagonalizable. – Travis Willse Apr 11 '16 at 12:24
  • With which involution do you endow the space of matrices? Is it simply the transpose, are we supposing F has involution and the matrices have the conjugate-transpose, are we taking the symplectic involution? – Jose Brox Apr 11 '16 at 12:25
  • @JoseBrox In the absence of more qualification, I've only ever heard "skew-symmetric" refer to the usual transpose. – Travis Willse Apr 11 '16 at 12:30
  • This answer addresses this question: http://math.stackexchange.com/a/375599/155629 (I will not mark this question as a duplicate of the question there, as it implicitly just asks about matrices over $\Bbb R$.) – Travis Willse Apr 11 '16 at 12:31
  • 1
    @JoseBrox by skew-symmetric, I mean a matrix say $A$ which $A^T=-A$ – F.K Apr 11 '16 at 12:36
  • 1
    Note that this Question asks both about a more general setting (char $\mathbb{F} \neq 2$) and a specific proof tactic (induction). Even if the specific answer to this Question is one of several answers on other posts, it deserves (in my opinion) to be given here. We close as exact duplicate based on the Question being the same, not on existence of applicable Answers. – hardmath Apr 11 '16 at 15:07
  • @akech: Although the answer you link to addresses the char$(\mathbb{F})\neq 2$ context, it does so only to proving the question asked there, Is the rank of a skew-symmetric matrix never equal to $1$? While this might constitute something of a base case for induction, it falls short of establishing that ranks of skew-symmetric matrices are are even. – hardmath Apr 11 '16 at 15:33
  • @Travis I usually work with Lie algebras, and the skew-symmetric elements of an algebra with involution (over a field with involution) are an important subset of them (here we work with the involution abstractly). – Jose Brox Apr 11 '16 at 15:37

0 Answers0