Skew-symmetric matrix

A matrix is skew-symmetric when $A^T=-A$.

Proof. If $(I+A)x=0$, the quadratic-form criterion for a skew-symmetric matrix gives $0=x^{\top }(I+A)x=\Vert x{\Vert }^2$, so $x=0$. The rank-nullity theorem makes the square matrix $I+A$ invertible. More generally $tI+A$ is invertible for every nonzero real $t$, since its null space would imply $t\Vert x{\Vert }^2=0$. Equivalently, the only possible real eigenvalue of a real skew-symmetric matrix is zero.

An odd integer number of rows and columns forces a real skew-symmetric matrix to be a singular matrix. The determinant satisfies $\det A=\det A^{\top }=\det (-A)=(-1{)}^n\det A=-\det A$, so it is zero. The argument also works over any field of characteristic different from two.

Proof. For a skew-symmetric matrix, the scalar $x^{\top }Ax$ equals its transpose $x^{\top }{A}^{\top }x=-x^{\top }Ax$, so it vanishes. Conversely, substituting standard basis vectors $e_i$ into the vanishing quadratic form gives $a_{ii}=0$. Substituting $e_i+e_j$ gives $a_{ij}+a_{ji}=0$. Thus $A^{\top }=-A$. In particular, a real quadratic form depends only on a matrix's symmetric part.

For real $n\times n$ skew-symmetric matrices, each upper-triangular entry can be chosen freely, the corresponding lower-triangular entry is its negative, and diagonal entries vanish. Thus the displayed matrix unit differences span the space and are linearly independent: inspecting entry $(i,j)$ with $i<j$ recovers its coefficient. There are $n(n-1)/2$ such pairs, giving the dimension.

Every real square matrix decomposes as $A=S+K$, where $S=(A+A^T)/2$ is its symmetric part and $K=(A-A^T)/2$ is a skew-symmetric matrix. For a real column vector $x$, $x^{T}Kx$ equals its transpose $x^T{K}^{T}x=-x^{T}Kx$, hence vanishes. Thus a real quadratic form $x^{T}Ax$ depends only on $S$.

Every real $2\times 2$ antisymmetric matrix has the form $A=sJ$, where $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Thus $A^2=-s^{2}I$, $\mathrm{tr}A=0$, and $\det A=s^2$. In particular,A convergent scalar power series applied to $A$ can therefore be reduced into an even scalar multiple of $I$ plus an odd scalar multiple of $A$. For real coefficients and $s\ne 0$, this is $f(A)=\mathrm{Re}f(is)I+[\mathrm{Im}f(is)/s]A$, provided the series converges at $\pm is$. Absolute convergence justifies separating even and odd terms. This converts many small matrix series problems into ordinary scalar series, without computing eigenvectors.

A real skew-symmetric matrix is an orthogonal matrix exactly when its square is minus the identity matrix. Indeed, $B^{\mathsf{T}}B=-B^2$. Such a matrix can exist only in even dimension: taking determinants of $B^2=-I$ would otherwise give $(\det B{)}^2=-1$. In two dimensions every example isThese are rotation matrices through opposite right angles. Their orthogonal similarity transforms preserve eigenvalues, but do not in general produce a diagonal matrix.

For a real skew-symmetric matrix $M$, $M^2$ is symmetric and $x^T{M}^{2}x=-\Vert Mx{\Vert }^2$. On an eigenspace with eigenvalue $-{\mu }^2<0$, every unit vector $u$ has unit partner $v=Mu/\mu$, orthogonal to $u$, with $Mu=\mu v$ and $Mv=-\mu u$. Its plane and its orthogonal complement are invariant under $M$. Repeating gives real orthogonal two-dimensional blocks $\left(\begin{array}{cc}0& -\mu \\ \mu & 0\end{array}\right)$ and a zero block. Thus every nonzero eigenvalue of $M^2$ has even multiplicity, the zero multiplicity has the parity of $n$, and the number of distinct eigenvalues is at most $n/2$ for even $n$ and $(n+1)/2$ for odd $n$. Distinct positive block parameters attain the bound, with one zero direction added when $n$ is odd.

A complex matrix $Z=-Z^T$ can be put by unitary congruence into two-by-two skew blocks and zero blocks; the magnitudes $|z_r|$ are its singular values, each repeated twice. One proof uses the antilinear map $A(v)=Z\bar{v}$. Because $Z^{†}=-\bar{Z}$, $A^2=-Z{Z}^{†}$. On a nonzero eigenspace of $Z{Z}^{†}$ with eigenvalue $s^2$, any unit vector $v$ has orthogonal partner $A(v)/s$, and $A(A(v)/s)=-sv$. These orthonormal pairs give the blocks; the orthogonal complement is invariant, so the procedure repeats. Phases of the paired vectors set the block phases.

For $a=(a_1,a_2,a_3{)}^T$, the map $b\mapsto a\times b$ has matrixIt has trace zero and annihilates $a$, so its determinant is zero.

The cross-product matrix $A\mathbf{x}=\mathbf{n}\times \mathbf{x}$ obeys $A\mathbf{n}=0$ and $A^2=\mathbf{n}{\mathbf{n}}^T-|\mathbf{n}{|}^{2}I$, by the vector triple product. On the orthogonal complement of a nonzero $\mathbf{n}$, this is $-|\mathbf{n}{|}^{2}I$; hence the two remaining complex eigenvalues are $\pm i|\mathbf{n}|$. In an orthonormal basis adapted to $\mathbf{n}$, $A$ is a planar quarter-turn multiplied by $|\mathbf{n}|$, with zero action along its axis.