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In linear algebra, a skew-symmetric (or antisymmetric) matrix is a square matrix A whose transpose is also its negative; that is, it satisfies the equation:
or in component form, if A = (aij):
all i and j.
For example, the following matrix is skew-symmetric:
0 & 2 & -1 \\ -2 & 0 & -4 \\ 1 & 4 & 0\end{bmatrix}<math>
A matrix A is skew-symmetric if AT = −A
If matrices A and B are both skew-symmetric: (AB)T = BA
The "skew-symmetric component of a matrix A is the matrix B = (A − AT)/2; the "symmetric component" of A is C = (A + AT)/2; the matrix A is the sum of its symmetric and skew-symmetric components.
If A is skew-symmetric and x is vector then xTAx = 0.
All main diagonal entries of a skew-symmetric matrix have to be zero, and so the trace is zero.
Let A be a n×n skew-symmetric matrix. The determinant of A satisfies
In particular, if n is odd the determinant vanishes. The even-dimensional case is more interesting. It turns out that the determinant of A for n even can be written as the square of a polynomial in the entries of A:
This polynomial is called the Pfaffian of A and is denoted Pf(A). Thus the determinant of a real skew-symmetric matrix is always non-negative.
The eigenvalues of a skew-symmetric matrix always come in pairs ±λ (except in the odd-dimensional case where there is an additional unpaired 0 eigenvalue). For a real skew-symmetric matrix the eigenvalues are all pure imaginary and thus are of the form iλ1, −iλ1, iλ2, −iλ2, … where each of the λk are real.
Skew-symmetric matrices fall into the category of normal matrices and are thus subject to the spectral theorem, which states that any real or complex skew-symmetric matrix can be diagonalized by a unitary matrix. Since the eigenvalues a real skew-symmetric matrix are complex it is not possible to diagonalize one by a real matrix. However, it is possible to bring every skew-symmetric matrix to a block diagonal form by an orthogonal transformation. Specifically, every 2r × 2r skew-symmetric matrix can be written in the form A = R Σ RT where R is orthogonal and
\begin{matrix}0 & \lambda_1\\ -\lambda_1 & 0\end{matrix} & 0 & \cdots & 0 \\ 0 & \begin{matrix}0 & \lambda_2\\ -\lambda_2 & 0\end{matrix} & & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & \begin{matrix}0 & \lambda_r\\ -\lambda_r & 0\end{matrix} \end{bmatrix}<math> for real λk. The eigenvalues of this matrix are ±iλk. In the odd-dimensional case Σ has an additional row and column of zeros.
An alternating form φ on a vector space V over a field K is defined (if K doesn't have characteristic 2) to be a bilinear form
such that
Such a φ will be represented by a skew-symmetric matrix, once a basis of V is chosen; and conversely an n×n skew-symmetric matrix A on Kn gives rise to an alternating form xTAx.
The skew-symmetric n×n matrices form a vector space of dimension
This is the tangent space to the orthogonal group O(n) at the identity matrix. In a sense, then, skew-symmetric matrices can be thought of as infinitesimal rotations.
Another way of saying this is that the space of skew-symmetric matrices forms the Lie algebra o(n) of the Lie group O(n). The Lie bracket on this space is given by the commutator:
It is easy to check that the commutator of two skew-symmetric matrices is again skew-symmetric.
The matrix exponential of a skew-symmetric matrix A is then an orthogonal matrix R:
Since the image of the exponential map always lies in the connected component of O(n) (which is denoted SO(n)), R will have determinant +1. It turns out that every orthogonal matrix with unit determinant can be written as the exponential of some skew-symmetric matrix.