In mathematics, particularly in linear algebra, a skew-symmetric (or antisymmetric or antimetric) matrix is a square matrix whose transpose equals its negative. That is, it satisfies the condition

${\displaystyle A{\text{ skew-symmetric}}\quad \iff \quad A^{\textsf {T}}=-A.}$

In terms of the entries of the matrix, if ${\textstyle a_{ij}}$ denotes the entry in the ${\textstyle i}$-th row and ${\textstyle j}$-th column, then the skew-symmetric condition is equivalent to

${\displaystyle A{\text{ skew-symmetric}}\quad \iff \quad a_{ij}=-a_{ji}.}$

The matrix $${\displaystyle A={\begin{bmatrix}0&2&-45\\-2&0&-4\\45&4&0\end{bmatrix}}}$$ is skew-symmetric because $${\displaystyle A^{\textsf {T}}={\begin{bmatrix}0&-2&45\\2&0&4\\-45&-4&0\end{bmatrix}}=-A.}$$

Throughout, we assume that all matrix entries belong to a field ${\textstyle \mathbb {F} }$ whose characteristic is not equal to 2. That is, we assume that 1 + 1 ≠ 0, where 1 denotes the multiplicative identity and 0 the additive identity of the given field. If the characteristic of the field is 2, then a skew-symmetric matrix is the same thing as a symmetric matrix.

As a result of the first two properties above, the set of all skew-symmetric matrices of a fixed size forms a vector space. The space of ${\textstyle n\times n}$ skew-symmetric matrices has dimension ${\textstyle {\frac {1}{2}}n(n-1).}$

Let ${\displaystyle {\mbox{Mat}}_{n}}$ denote the space of ${\textstyle n\times n}$ matrices. A skew-symmetric matrix is determined by ${\textstyle {\frac {1}{2}}n(n-1)}$ scalars (the number of entries above the main diagonal); a symmetric matrix is determined by ${\textstyle {\frac {1}{2}}n(n+1)}$ scalars (the number of entries on or above the main diagonal). Let ${\textstyle {\mbox{Skew}}_{n}}$ denote the space of ${\textstyle n\times n}$ skew-symmetric matrices and ${\textstyle {\mbox{Sym}}_{n}}$ denote the space of ${\textstyle n\times n}$ symmetric matrices. If ${\textstyle A\in {\mbox{Mat}}_{n}}$ then $${\displaystyle A={\tfrac {1}{2}}\left(A-A^{\mathsf {T}}\right)+{\tfrac {1}{2}}\left(A+A^{\mathsf {T}}\right).}$$