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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?

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What about this matrix: $$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}$$ Doesn't this matrix have rank 1? –  user31646 May 17 '12 at 19:26
    
@Pratik: No, that matrix has rank $2$. –  Jonas Meyer May 17 '12 at 19:40
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@Pratik: No, it has rank 2. That matrix sends the basis $$e_1=\begin{pmatrix}1\\0\\0\end{pmatrix},\quad e_2=\begin{pmatrix}0\\1\\0\end{pmatrix},\quad e_3=\begin{pmatrix}0\\0\\1\end{pmatrix}$$ to the vectors $$v_1=\begin{pmatrix}0\\0\\-1\end{pmatrix},\quad v_2=\begin{pmatrix}0\\0\\0\end{pmatrix},\quad v_3=\begin{pmatrix}1\\0\\0\end{pmatrix}$$ which span the subspace $$\{(a,0,b)\in\mathbb{R}^3\mid a,b\in\mathbb{R}\}$$ which is of dimension 2. Thus, the rank of the matrix is 2. –  Zev Chonoles May 17 '12 at 19:41
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3 Answers

up vote 16 down vote accepted

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.

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That is neat. Thanks! –  Naga Oct 27 '10 at 3:43
    
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 '10 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 '13 at 18:06
    
even if there are some arrays which are equal to each other! –  jack Oct 31 '13 at 18:09
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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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Edit: 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.

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