# Proof that the rank of a skew-symmetric matrix is at least $2$

Is there a succinct proof for the fact that the rank of a non-zero skew-symmetric matrix ($A = -A^T$) is at least 2? I can think of a proof by contradiction: Assume rank is 1. Then you express all other rows as multiple of the first row. Using skew-symmetric property, this matrix has to be a zero matrix.

Why does such a matrix have at least 2 non-zero eigenvalues?

For a skew symmetric (real) matrix, the eigenvalues are all purely imaginary. This is because if $Av = \lambda v$, then we have $\lambda \langle v,v\rangle = \langle \lambda v, v\rangle = \langle Av,v\rangle = \langle v, -Av \rangle = \langle v, -\lambda v\rangle = -\overline{\lambda} \langle v,v\rangle$, so we conclude that $\lambda = -\overline{\lambda}$, i.e., that $\lambda$ is purely imaginary. Here, we're using an Hermitian inner product.

For a real matrix, complex eigenvalues come in conjugate pairs, so the rank must be even.

• If you don't assume the matrix is real, but only that the base field has characteristic $\neq 2$, the result is still true, see my answer.
– Plop
Oct 27, 2010 at 12:51
• @Naga I think this proof has a problem.because $\lambda$ is a n x n diagonal matrix so the rank is equal to n
– jack
Oct 31, 2013 at 18:06
• even if there are some arrays which are equal to each other!
– jack
Oct 31, 2013 at 18:09
• How does it follow that rank is even? Dec 5, 2019 at 14:40
• @Hrit: The rank can be computed as the number (counting multiplicity) of non-zero eigenvalues. The first paragraph establishes that all the eigenvalues are purely imaginary. The complex conjugate of a non-zero purely imaginary number is a different number, so you can pair up all the non-zero eigenvalues without having any left over (and nothing gets paired with itself). This tells you the rank is $2$ times the number of pairs, so is manifestly even. Dec 5, 2019 at 16:17

The following answers the first part of the OP's question, without using the concept of eigenvalues. It works on all fields (including $$\mathbb{R}$$) with characteristic $$\ne2$$.*

Every rank-$$1$$ matrix can be written as $$A=uv^\top$$ for some nonzero vectors $$u$$ and $$v$$ (so that every row of $$A$$ is a scalar multiple of $$v^\top$$). If $$A$$ is skew-symmetric, we have $$A=-A^\top=-vu^\top$$. Hence every row of $$A$$ is also a scalar multiple of $$u^\top$$. It follows that $$v=ku$$ for some nonzero scalar $$k$$. But then $$vu^\top=-uv^\top$$ implies that $$kuu^\top=-kuu^\top$$ or $$2kuu^\top=0$$, which is impossible because both $$k$$ and $$u$$ are nonzero and the characteristic of the field is not $$2$$. Therefore, skew-symmetric matrices cannot be rank-1 matrices, and vice versa.

When the underlying field has characteristic 2, the notions of symmetric matrices and skew-symmetric matrices coincide. Hence every nonzero matrix of the form $$uu^\top$$ with nonzero vector $$u$$ is a rank-1 skew-symmetric matrix.

Remark. In most modern textbooks, a matrix $$A$$ in a field of characteristic $$2$$ is said to be skew-symmetric if $$A$$ has a zero diagonal and $$A^T=-A$$. This modern definition is better because the discrepancy between skew-symmetric matrix and alternating bilinear form now vanishes. With this definition, symmetric matrices and skew-symmetric matrices are different notions and a matrix of the form $$uu^\top$$ cannot be skew-symmetric unless it is zero.

There exists an invertible matrix $P$ such that $^t P A P$ is diagonal with blocks equal to $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ or $0$ (it is a simple exercise in bilinear forms), so that the rank of $A$ is necessarily even.