Property of real skew symmetric matrices

If you don't know what a skew-symmetric matrix is, don't worry; I didn't know what it meant either until I found this problem. It's literally a matrix A for which \(A^T=-A\). In this article, we have to prove that the rank of any real skew symmetric matrix is even. From what I've been told, the rank of any skew symmetric matrix, in any field with characteristic different than 2, is even, but for now I only know how to prove this property for the \(M_n {(R)}\). Without further ado, let's get straight into the problem.

First of all, let us prove a few lemmas:

1. If A is a real skew symmetric matrix, then it is diagonalizable.

Through Schur's theorem, we see that there is a matrix \(S\) in \(M_n (R)\) that satisfies the following properties: \(SS^{T} = I_n\) and \(SAS^{-1} = J\), where J is an upper triangular matrix. By writing \(S^{-1}=S^{T}\) and the same relation with S, we have that \(-(S^{-1})^{T}A^{T}S^{T}=-J^{T}\), and \(J^{T}\) is a lower triangular matrix. J is a diagonal matrix, because the entries under the diagonal, 0s, are equal to the negatives of the values above the diagonal. Thus, A is diagonalizable.

2. The eigenvalues of a real skew symmetric matrix are either pairs of complex conjugates or 0.

If \(\lambda\) is an eigenvalue of A, then \(AX=\lambda X\), for a certain vector X. By transposing, we get that \(X^{T}A^{T}=\lambda X^{T}\). Conjugate the whole thing, then multiply to the right by \(X\) and replace \(A^{T} = -A\) and \(AX=\lambda X\), and you get that \(\lambda\) is a purely imaginary number or 0. In order to prove that the conjugates are eigenvalues too, we just look at the characteristic polynomial of A is \(\lambda\), and conjugate it. Since A is real, it's characteristic polynomial has real coefficients, so the conjugate of \(\lambda\) is also a root.

Now we are ready to solve the problem. Since A is diagonalizable, then there exists an S such that \(SAS^{-1}=J\), where J is the Jordan canonical form of A. Since the eigenvalues of A come in pairs of conjugates, the rank of A is equal to 2 times the number of non-zero eigenvalues, so it is even.

As with most matrix problems that use this theory, they are somewhere on the boundary between college Mathematics and high school Mathematics. As far as I know, the high school curriculum doesn't include Jordan forms or Schur's theorem in any country (perhaps except for some countries in Asia). Apologies for the fact that this article is a bit smaller, but I've got something really interesting on the way; one of my all time favorite problems is coming in the next article.

Cris.