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?


3 Answers 3


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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Plop
    Oct 27, 2010 at 12:51
  • $\begingroup$ @Naga I think this proof has a problem.because $\lambda$ is a n x n diagonal matrix so the rank is equal to n $\endgroup$
    – jack
    Oct 31, 2013 at 18:06
  • $\begingroup$ even if there are some arrays which are equal to each other! $\endgroup$
    – jack
    Oct 31, 2013 at 18:09
  • 1
    $\begingroup$ How does it follow that rank is even? $\endgroup$
    – Not Euler
    Dec 5, 2019 at 14:40
  • $\begingroup$ @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. $\endgroup$ 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.


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