Orthogonal matrix

A real square matrix is orthogonal when $R^{T}R=I$, equivalently when its columns form an orthonormal basis.

An invertible matrix over the complex numbers preserving the symmetric bilinear form $v^{T}w$: $A^{T}A=I$. The transpose here does not include complex conjugation, unlike the defining condition for a unitary matrix. For example, $\left(\begin{array}{cc}\cosh t& i\sinh t\\ -i\sinh t& \cosh t\end{array}\right)$ is complex orthogonal for real $t$, but is not unitary when $t\ne 0$.

Every two-by-two complex orthogonal matrix has form $A=\left(\begin{array}{cc}a& b\\ -\epsilon b& \epsilon a\end{array}\right)$, where $\epsilon =\det A\in \{+1,-1\}$ and $a^2+b^2=1$. This follows by equating $A^{-1}$ to $A^T$ in the two-by-two inverse formula, and direct multiplication proves the converse. A complete parametrization is $a=\cosh t$, $b=i\sinh t$ with complex $t$; restricting $t$ to real values selects only a subfamily.

An orientation-reversing orthogonal transformation. In three dimensions it is a rotation about an axis followed by a reflection perpendicular to that axis. Its eigenvalues are $-1,e^{i\theta },e^{-i\theta }$, giving $\mathrm{tr}R=-1+2\cos \theta$. A plane reflection is the case $\theta =0$, while inversion through the origin is the case $\theta =\pi$.

An orientation-preserving orthogonal transformation. In three dimensions it has an invariant axis and acts as a planar rotation through angle $\theta$ perpendicular to that axis. Its eigenvalues are $1,e^{i\theta },e^{-i\theta }$, so $\mathrm{tr}R=1+2\cos \theta$. In higher dimensions a proper rotation can involve several mutually orthogonal planes and need not fix an axis.

For a real three-dimensional linear map $A$, preservation of every surface area is equivalent to being an orthogonal matrix. For necessity, diagonalize the symmetric matrix $A^{T}A$ in an orthonormal basis, with nonnegative eigenvalues ${\lambda }_i$. The three unit coordinate parallelograms have image areas $\sqrt{{\lambda }_1{\lambda }_2}$, $\sqrt{{\lambda }_1{\lambda }_3}$, and $\sqrt{{\lambda }_2{\lambda }_3}$. Setting all three equal to $1$ forces every ${\lambda }_i=1$. Sufficiency follows from preservation of dot products and $\Vert \mathbf{u}\times \mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2\Vert \mathbf{v}{\Vert }^2-(\mathbf{u}\cdot \mathbf{v}{)}^2$. This specifically uses dimension three; in dimension two, preservation of planar area alone only requires $|\det A|=1$.

Suppose real symmetric matrices $E,F$ satisfy $E^2=F^2=I$ and $EF=-FE$. The involutions are orthogonal matrices, and $J=EF$ satisfies $J^T=-J$, $J^2=-I$. The matrix exponential therefore reduces to $R(t)=e^{tJ}=\cos tI+\sin tJ$. Multiplication gives $R(t{)}^{T}R(t)=I$ and the conjugation identitiesIndeed $JE=-F$, $EJ=F$, $JF=E$, and $FJ=-E$, which give both identities by expansion. The angle in this two-dimensional space of matrices is twice the parameter in the orthogonal matrix. These formulas hold in any finite dimension where the hypotheses can be satisfied.

For a real orthogonal involution, $A^{-1}=A=A^T$, so it is a symmetric matrix. Its eigenvalues are $\pm 1$. In two dimensions the possibilities are the identity, the half-turn $-I$, and reflectionsThe reflection fixes the line making angle $\theta /2$ with the first coordinate axis and negates its perpendicular direction. This follows either from the two corresponding eigenvectors or by rotating $\mathrm{diag}(1,-1)$. A nonidentity element of order two in a faithful representation must be either the half-turn or a reflection; the identity matrix is excluded by faithfulness.

An orthogonal transformation of a real inner-product space is a linear map $Q$ satisfying $Q^{T}Q=I$. It preserves inner products, lengths, angles, and therefore Gaussian curvature when applied to an embedded surface in Euclidean space.

Let $b_1,\dots ,b_n$ be an arbitrary basis of a real inner product space, with Gram matrix $G=(⟨b_i,b_j⟩)$. If a linear map $T$ has matrix $A$ in this basis, thenConsequently $T$ is an orthogonal transformation exactly when $A^{T}GA=G$. The simpler orthogonal matrix condition $A^{T}A=I$ applies directly when the chosen basis is an orthonormal basis, so that $G=I$.

For example, take a two-dimensional vector space with weighted Euclidean inner product having $G=\mathrm{diag}(1,4)$. Thensatisfies $A^{T}GA=G$, but $A^{T}A=\mathrm{diag}(1/4,4)\ne I$. Thus it represents an orthogonal transformation even though it is not an orthogonal matrix.