Skew-symmetric matrix

Remarks (See here:

• They constitute the Lie algebra of the orthogonal group (see orthogonal matrix and SO(3)).

• The rank of a skew-symmetric matrix is even.

Proof.

Suppose that $A$ is a skew-symmetric matrix of rank $r$ and dimension $n\times n$. Now $r$ could very well be zero, and since zero is an even number, then $A$ has an even rank. So assume instead that $r>0$. Consequently, we can pick out exactly $r$ rows, say those with the indices $i_1,i_2,...,i_r$, which span the entire row space. Given that for a skew-symmetric matrix each column is equal to $-1$ times the transpose of the corresponding row, therefore every column of the matrix can be expressed as a linear combination of the columns with indices $i_1,...,i_r$ in the exact same way that the corresponding row is expressed as a linear combination of the rows with these same indices. We know that if we remove a row/column of a matrix that is in the span of the remaining rows/columns, the rank does not change. Thus, we can remove all the $n-r$ rows and $n-r$ columns remaining and not change the rank. Due to symmetry, every time we remove a row, we remove its corresponding column. This way, we have preserved the structure of the matrix. The resultant submatrix $A^{\prime }$ has dimensions $r\times r$ and full rank $r$.

Suppose for the sake of contradiction that $r$ is odd. We know that $A^{\prime T}=-A^{\prime }$. Then the determinant of this submatrix is

$$det(A^{\prime })=det(A^{\prime T})=det(-A^{\prime })=(-1{)}^r\cdot det(-A^{\prime })=-det(A^{\prime })$$

and therefore $det(A^{\prime })=0$. This contradicts our assumption that $r\ne 0$. Hence, $r$ is an even natural number.

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• The non-zero roots of the characteristic polynomial of a real skew-symmetric matrix are purely imaginary numbers.

• A real skew-symmetric matrix is similar to a matrix

where

with ${\alpha }_i$ real numbers, $i=1,\dots ,t$.

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es