A square matrix can be invertible without respecting the geometry of the space it acts on. There are linear maps of determinant $1$ that preserve oriented volume in $\mathbb{R}^n$ but bend right angles into slanted angles. The orthogonal group is designed to separate genuine Euclidean symmetries from merely invertible or volume-preserving linear maps. From the point of view of [group theory](/page/Group), the orthogonal group is a child of the general notion of a group: its operation is matrix multiplication, its identity is the identity matrix, and its inverses are matrix inverses. What makes it special is the additional geometric constraint that the Euclidean [inner product](/page/Inner%20Product) must be preserved. The first failure is a shear. It has determinant $1$, so a determinant-only test would treat it like a rotation. Looking at two perpendicular vectors shows why that test is too weak. [example: Area Preservation Does Not Preserve Angles] Let $A:\mathbb{R}^2\to\mathbb{R}^2$ be the [linear map](/page/Linear%20Map) determined by $Ae_1=e_1$ and $Ae_2=e_1+e_2$, where $e_1=(1,0)$ and $e_2=(0,1)$. Since the columns of the standard matrix of a linear map are the images of the standard basis vectors, the first column of $A$ is $Ae_1=(1,0)$ and the second column is $Ae_2=e_1+e_2=(1,0)+(0,1)=(1,1)$. Hence \begin{align*} A=\begin{pmatrix}1&1\cr 0&1\end{pmatrix}. \end{align*} For a $2\times 2$ matrix $\begin{pmatrix}a&b\cr c&d\end{pmatrix}$, the determinant is $ad-bc$. Therefore \begin{align*} \det A=(1)(1)-(1)(0)=1-0=1. \end{align*} So this shear preserves signed area. Now compare what $A$ does to the standard perpendicular basis vectors. Their original Euclidean inner product is \begin{align*} \langle e_1,e_2\rangle=\langle (1,0),(0,1)\rangle=(1)(0)+(0)(1)=0+0=0. \end{align*} The images are \begin{align*} Ae_1=e_1=(1,0). \end{align*} Also, \begin{align*} Ae_2=e_1+e_2=(1,0)+(0,1)=(1+0,0+1)=(1,1). \end{align*} Their new inner product is \begin{align*} \langle Ae_1,Ae_2\rangle=\langle (1,0),(1,1)\rangle=(1)(1)+(0)(1)=1+0=1. \end{align*} Thus two input vectors with inner product $0$ are sent to two output vectors with inner product $1$, so perpendicularity is not preserved. The same map changes the length of $e_2$. Since $Ae_2=(1,1)$, its squared length is \begin{align*} |Ae_2|^2=\langle Ae_2,Ae_2\rangle=\langle (1,1),(1,1)\rangle=(1)(1)+(1)(1)=1+1=2. \end{align*} Meanwhile, \begin{align*} |e_2|^2=\langle e_2,e_2\rangle=\langle (0,1),(0,1)\rangle=(0)(0)+(1)(1)=0+1=1. \end{align*} Since $|Ae_2|^2=2$ and $|e_2|^2=1$, the vector $e_2$ does not keep its length under $A$. Determinant $1$ records signed area preservation here, but it does not force preservation of lengths or angles; the missing condition is preservation of the Euclidean inner product. [/example] This page develops the matrix equation that rules out this failure. We start with the definition, then translate it into preservation of lengths and angles, separate rotations from reflections by determinant, study the action on spheres, and finish with the infinitesimal [Lie algebra](/page/Lie%20Algebra) viewpoint. ## Definition The shear example shows that determinant alone does not recognise Euclidean geometry. The missing datum is not area but the whole dot-product structure: a genuine linear symmetry must preserve the Euclidean inner product between every pair of vectors. The primary definition therefore builds the group directly from that preservation law. [definition: Orthogonal Group] For $n\in\mathbb{N}$, let $M_n(\mathbb{R})$ denote the set of real $n\times n$ matrices, and let $GL(n,\mathbb{R})$ denote the group of invertible matrices in $M_n(\mathbb{R})$. For a matrix $A$, let $A^\top$ denote its transpose, and let $I_n$ denote the $n\times n$ identity matrix. The orthogonal group of degree $n$ over $\mathbb{R}$ is the group \begin{align*} O(n)=O_n(\mathbb{R}):=\{A\in GL(n,\mathbb{R}):A^\top A=I_n\}, \end{align*} with group operation $m:O(n)\times O(n)\to O(n)$ given by \begin{align*} m(A,B)=AB. \end{align*} [/definition] The inverse of an element of $O(n)$ is its transpose, so the definition is rigid enough to make composition and reversal stay inside the same class of symmetries. The rest of the page explains why the equation $A^\top A=I_n$ is exactly the algebraic form of preserving lengths, angles, orthonormal bases, and orientation data. ## Inner Products and Lengths ### The Euclidean Inner Product The defining equation of $O(n)$ becomes geometric only after naming the pairing it preserves. In Euclidean space, that pairing is the dot product: it measures length when both inputs agree and angle when they do not. [definition: Euclidean Inner Product] For $n \in \mathbb{N}$, the Euclidean inner product on $\mathbb{R}^n$ is the map $\langle \cdot,\cdot\rangle:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$ defined by \begin{align*} \langle x,y\rangle=\sum_{i=1}^n x_i y_i. \end{align*} [/definition] The associated Euclidean norm is $|x|=\sqrt{\langle x,x\rangle}$. This norm statement is not an extra axiom; it is the standard way the inner product produces lengths. Requiring a linear map to preserve every dot product appears to be an infinite family of conditions, one for each pair of vectors. Matrix multiplication gives a compact way to express exactly that obstruction: the columns must interact with dot products as the standard basis does. [definition: Orthogonal Matrix] Let $n \in \mathbb{N}$. Write $M_n(\mathbb{R})$ for the set of all real $n\times n$ matrices. A matrix $A\in M_n(\mathbb{R})$ is orthogonal if \begin{align*} A^\top A=I_n. \end{align*} [/definition] ### Preservation Laws The defining equation of $O(n)$ is useful because it packages several geometric properties into one algebraic condition. The next theorem is needed to justify the definition: it proves that $A^\top A=I_n$ is not an arbitrary equation but exactly the preservation of the Euclidean inner product. [quotetheorem:8281] This theorem turns a matrix condition into a geometric preservation law. The next natural question is whether it is enough to test only lengths rather than all pairwise inner products, since lengths are often easier to see. [quotetheorem:8282] The norm formulation says that $O(n)$ is the group of linear isometries of Euclidean space fixing the origin. The next example prevents a common overreach: not every distance-preserving map of Euclidean space is orthogonal. [example: A Translation Is Not Orthogonal] For $x=(x_1,x_2)$, define the translation \begin{align*} T(x)=x+(1,0)=(x_1+1,x_2). \end{align*} We first show that $T$ preserves distances between pairs of points, and then show that it is not an element of $O(2)$ because it is not linear. Let $y=(y_1,y_2)$. Then \begin{align*} T(y)=y+(1,0)=(y_1+1,y_2). \end{align*} Subtracting the two images gives \begin{align*} T(x)-T(y)=(x_1+1,x_2)-(y_1+1,y_2)=((x_1+1)-(y_1+1),x_2-y_2). \end{align*} In the first coordinate, \begin{align*} (x_1+1)-(y_1+1)=x_1+1-y_1-1=x_1-y_1. \end{align*} Therefore \begin{align*} T(x)-T(y)=(x_1-y_1,x_2-y_2)=x-y. \end{align*} The squared distance between the images is \begin{align*} |T(x)-T(y)|^2=|(x_1-y_1,x_2-y_2)|^2=(x_1-y_1)^2+(x_2-y_2)^2. \end{align*} The original squared distance is \begin{align*} |x-y|^2=|(x_1-y_1,x_2-y_2)|^2=(x_1-y_1)^2+(x_2-y_2)^2. \end{align*} Hence \begin{align*} |T(x)-T(y)|^2=|x-y|^2. \end{align*} Both $|T(x)-T(y)|$ and $|x-y|$ are nonnegative, so equality of their squares gives \begin{align*} |T(x)-T(y)|=|x-y|. \end{align*} Thus $T$ preserves the Euclidean distance between every pair of points in $\mathbb{R}^2$. However, $T$ is not linear. A linear map must send the zero vector to the zero vector, but \begin{align*} T(0,0)=(0,0)+(1,0)=(1,0)\ne(0,0). \end{align*} Additivity fails as well. With $e_1=(1,0)$ and $e_2=(0,1)$, \begin{align*} e_1+e_2=(1,0)+(0,1)=(1,1). \end{align*} So \begin{align*} T(e_1+e_2)=T(1,1)=(1,1)+(1,0)=(2,1). \end{align*} On the other hand, \begin{align*} T(e_1)=T(1,0)=(1,0)+(1,0)=(2,0). \end{align*} Also, \begin{align*} T(e_2)=T(0,1)=(0,1)+(1,0)=(1,1). \end{align*} Adding these two images gives \begin{align*} T(e_1)+T(e_2)=(2,0)+(1,1)=(3,1). \end{align*} Since $(2,1)\ne(3,1)$, we have \begin{align*} T(e_1+e_2)\ne T(e_1)+T(e_2). \end{align*} So $T$ is a distance-preserving map of the Euclidean [metric space](/page/Metric%20Space) $\mathbb{R}^2$, but it is not an element of $O(2)$: orthogonal transformations are linear maps represented by $2\times 2$ matrices, while this translation does not even preserve the origin. [/example] Translations belong to the larger Euclidean group, while $O(n)$ records the part of a rigid motion that fixes the origin. To compute inside $O(n)$, we need a test that uses the entries of a matrix directly. ### Orthonormal Columns The columns of a matrix are the images of the standard basis vectors. Since every vector is built from the standard basis, a natural question is whether preserving all inner products is already forced by what happens to those finitely many basis vectors. [quotetheorem:8283] The theorem says that an orthogonal matrix is an ordered [orthonormal basis](/page/Orthonormal%20Basis) written as columns. The next example shows how this criterion turns membership in $O(n)$ into a small collection of dot-product checks. [example: Checking Orthogonality by Columns] Let $A$ be the $3\times 3$ matrix with columns $v_1=(2/3,2/3,-1/3)$, $v_2=(-1/3,2/3,2/3)$, and $v_3=(2/3,-1/3,2/3)$. We test orthogonality by checking that these columns form an orthonormal set. The squared length of the first column is \begin{align*}|v_1|^2=(2/3)^2+(2/3)^2+(-1/3)^2=4/9+4/9+1/9=9/9=1.\end{align*} The squared length of the second column is \begin{align*}|v_2|^2=(-1/3)^2+(2/3)^2+(2/3)^2=1/9+4/9+4/9=9/9=1.\end{align*} The squared length of the third column is \begin{align*}|v_3|^2=(2/3)^2+(-1/3)^2+(2/3)^2=4/9+1/9+4/9=9/9=1.\end{align*} Now compute the pairwise inner products. For the first two columns, \begin{align*}\langle v_1,v_2\rangle=(2/3)(-1/3)+(2/3)(2/3)+(-1/3)(2/3)=-2/9+4/9-2/9=0/9=0.\end{align*} For the first and third columns, \begin{align*}\langle v_1,v_3\rangle=(2/3)(2/3)+(2/3)(-1/3)+(-1/3)(2/3)=4/9-2/9-2/9=0/9=0.\end{align*} For the second and third columns, \begin{align*}\langle v_2,v_3\rangle=(-1/3)(2/3)+(2/3)(-1/3)+(2/3)(2/3)=-2/9-2/9+4/9=0/9=0.\end{align*} Thus each column has length $1$, and distinct columns have inner product $0$. By the *[Column Criterion for Orthogonal Matrices](/theorems/8283)*, this implies $A\in O(3)$. It remains to determine whether $A$ preserves orientation. Reading the entries from the three columns gives \begin{align*}A=\begin{pmatrix}2/3&-1/3&2/3\cr 2/3&2/3&-1/3\cr -1/3&2/3&2/3\end{pmatrix}.\end{align*} For a $3\times 3$ matrix, expansion along the first row gives \begin{align*}\det A=(2/3)\left((2/3)(2/3)-(-1/3)(2/3)\right)-(-1/3)\left((2/3)(2/3)-(-1/3)(-1/3)\right)+(2/3)\left((2/3)(2/3)-(2/3)(-1/3)\right).\end{align*} The first minor is \begin{align*}(2/3)(2/3)-(-1/3)(2/3)=4/9-(-2/9)=4/9+2/9=6/9=2/3.\end{align*} The second minor is \begin{align*}(2/3)(2/3)-(-1/3)(-1/3)=4/9-1/9=3/9=1/3.\end{align*} The third minor is \begin{align*}(2/3)(2/3)-(2/3)(-1/3)=4/9-(-2/9)=4/9+2/9=6/9=2/3.\end{align*} Substituting these three values into the [cofactor expansion](/theorems/398) gives \begin{align*}\det A=(2/3)(2/3)-(-1/3)(1/3)+(2/3)(2/3)=4/9-(-1/9)+4/9=4/9+1/9+4/9=9/9=1.\end{align*} We have shown $A\in O(3)$ and $\det A=1$. Since $SO(3)=\{A\in O(3):\det A=1\}$, it follows that $A\in SO(3)$. This matrix is therefore an orientation-preserving orthogonal transformation of $\mathbb{R}^3$. [/example] The column viewpoint also suggests how orthogonal matrices act on directions: the first column can be any point of the unit sphere, while the later columns must be chosen in successive perpendicular spaces. ## Determinant, Orientation, and Reflections ### Orientation Sign For a general linear map, the determinant measures both volume scaling and orientation. Orthogonal matrices preserve lengths and hence preserve Euclidean volume in absolute value, so the determinant can no longer vary freely; only the orientation sign remains. [definition: Determinant Sign of an Orthogonal Matrix] For $n\in\mathbb{N}$, the determinant sign map on the orthogonal group is the restriction \begin{align*} \det:O(n)\to\{-1,1\} \end{align*} of the determinant map on $GL(n,\mathbb{R})$. For $A\in O(n)$, the determinant sign of $A$ is $\det A$. [/definition] This terminology is useful because in $O(n)$ the determinant carries no size information. Once the determinant has been reduced to a sign, it raises a structural question: which orthogonal transformations lie in the orientation-preserving part, and does that part form a group in its own right? [definition: Special Orthogonal Group] For $n\in\mathbb{N}$, the special orthogonal group of degree $n$ over $\mathbb{R}$ is the determinant-one part of the orthogonal group: \begin{align*} SO(n)=\{A\in O(n):\det A=1\}. \end{align*} The group operation $m:SO(n)\times SO(n)\to SO(n)$ is matrix multiplication, given by \begin{align*} m(A,B)=AB. \end{align*} [/definition] Equivalently, if $SL(n,\mathbb{R})$ denotes the special linear group of real $n\times n$ matrices with determinant $1$, then \begin{align*} SO(n)=O(n)\cap SL(n,\mathbb{R}). \end{align*} Thus $SO(n)$ consists of the orthogonal transformations with positive determinant. More generally, a linear map in $\mathbb{R}^n$ scales unoriented Euclidean volume by $|\det A|$, while determinant $1$ preserves oriented volume. The natural structural question is whether an orthogonal matrix can carry any determinant data beyond orientation. Since orthogonality forces $|\det A|=1$, the determinant can only be $1$ or $-1$; the point is to identify this sign as a [group homomorphism](/page/Group%20Homomorphism) and to locate $SO(n)$ as the subgroup it kills. This turns the determinant sign from a numerical invariant into an organizing map for the whole orthogonal group. The formal result below packages three facts needed later: $SO(n)$ is stable under multiplication and inversion, it sits normally inside $O(n)$, and the quotient remembers exactly whether a transformation preserves or reverses orientation. The quoted theorem uses standard group notation. Here $O_n(\mathbb{R})$ and $\mathrm{SO}_n(\mathbb{R})$ are the same groups denoted above by $O(n)$ and $SO(n)$; $A^\top$ means $A^\top$, and $I$ means the identity matrix $I_n$. The symbol $\unlhd$ means "is a [normal subgroup](/page/Normal%20Subgroup) of", $\cong$ means "is isomorphic to", and $C_2$ denotes the [cyclic group](/page/Cyclic%20Group) with two elements. The quotient $O_n(\mathbb{R})/\mathrm{SO}_n(\mathbb{R})$ is the group whose two classes record the two possible determinant signs. This group-level viewpoint is needed before introducing concrete orientation-reversing maps. Once $SO(n)$ has been identified as the normal determinant-one subgroup, every orthogonal transformation outside it can be understood as carrying the single remaining obstruction: the sign of the determinant. [quotetheorem:805] The quotient statement means that, after forgetting all continuous geometric data, only the orientation sign remains. The next definition introduces the basic orientation-reversing building blocks. ### Reflections To produce an orientation-reversing orthogonal map, one should change exactly one perpendicular direction while leaving the whole hyperplane beside it fixed. A unit normal vector supplies the distinguished direction, and the inner product extracts the component of a vector along that normal. [definition: Hyperplane Reflection] Let $u\in\mathbb{R}^n$ satisfy $|u|=1$. The hyperplane reflection with normal vector $u$ is the linear map \begin{align*} R_u:\mathbb{R}^n&\to\mathbb{R}^n,\qquad x\mapsto x-2\langle x,u\rangle u. \end{align*} [/definition] This map fixes the hyperplane $u^\perp$ pointwise and sends $u$ to $-u$. The next example makes the formula concrete in the plane. [example: A Reflection Matrix] In $\mathbb{R}^2$, take the unit normal vector $u=e_1=(1,0)$. For $x=(x_1,x_2)$, the hyperplane reflection formula gives \begin{align*} R_{e_1}(x)=x-2\langle x,e_1\rangle e_1. \end{align*} The Euclidean inner product with $e_1$ is \begin{align*} \langle (x_1,x_2),(1,0)\rangle=x_1\cdot 1+x_2\cdot 0=x_1+0=x_1. \end{align*} Substituting this value gives \begin{align*} R_{e_1}(x_1,x_2)=(x_1,x_2)-2x_1(1,0). \end{align*} Since \begin{align*} 2x_1(1,0)=(2x_1,0), \end{align*} we get \begin{align*} R_{e_1}(x_1,x_2)=(x_1,x_2)-(2x_1,0)=(x_1-2x_1,x_2-0)=(-x_1,x_2). \end{align*} Now evaluate the reflection on the standard basis. Since $e_1=(1,0)$, \begin{align*} R_{e_1}(e_1)=R_{e_1}(1,0)=(-1,0)=-e_1. \end{align*} Since $e_2=(0,1)$, \begin{align*} R_{e_1}(e_2)=R_{e_1}(0,1)=(0,1)=e_2. \end{align*} The matrix of a linear map has as its columns the images of the standard basis vectors, so the matrix of $R_{e_1}$ is \begin{align*} A=\begin{pmatrix}-1&0\cr 0&1\end{pmatrix}. \end{align*} Because $A$ is diagonal, its transpose is the same matrix: \begin{align*} A^\top=\begin{pmatrix}-1&0\cr 0&1\end{pmatrix}. \end{align*} Therefore \begin{align*} A^\top A=\begin{pmatrix}-1&0\cr 0&1\end{pmatrix}\begin{pmatrix}-1&0\cr 0&1\end{pmatrix}. \end{align*} Multiplying the entries gives \begin{align*} A^\top A=\begin{pmatrix}(-1)(-1)+0\cdot 0&(-1)\cdot 0+0\cdot 1\cr 0\cdot(-1)+1\cdot 0&0\cdot 0+1\cdot 1\end{pmatrix}. \end{align*} The four entries are \begin{align*} (-1)(-1)+0\cdot 0=1+0=1, \end{align*} \begin{align*} (-1)\cdot 0+0\cdot 1=0+0=0, \end{align*} \begin{align*} 0\cdot(-1)+1\cdot 0=0+0=0, \end{align*} and \begin{align*} 0\cdot 0+1\cdot 1=0+1=1. \end{align*} Hence \begin{align*} A^\top A=\begin{pmatrix}1&0\cr 0&1\end{pmatrix}=I_2. \end{align*} Thus $A\in O(2)$. The same reflection preserves the length of every vector. For $x=(x_1,x_2)$, we have $R_{e_1}(x)=(-x_1,x_2)$, so \begin{align*} |R_{e_1}(x)|^2=\langle (-x_1,x_2),(-x_1,x_2)\rangle=(-x_1)(-x_1)+x_2x_2=x_1^2+x_2^2. \end{align*} The original squared length is \begin{align*} |x|^2=\langle (x_1,x_2),(x_1,x_2)\rangle=x_1x_1+x_2x_2=x_1^2+x_2^2. \end{align*} Therefore \begin{align*} |R_{e_1}(x)|^2=|x|^2. \end{align*} Both $|R_{e_1}(x)|$ and $|x|$ are nonnegative, so taking the nonnegative square root of both sides gives \begin{align*} |R_{e_1}(x)|=|x|. \end{align*} Finally, for a $2\times 2$ matrix $\begin{pmatrix}a&b\cr c&d\end{pmatrix}$, the determinant is $ad-bc$. Here \begin{align*} A=\begin{pmatrix}-1&0\cr 0&1\end{pmatrix}, \end{align*} so $a=-1$, $b=0$, $c=0$, and $d=1$. Hence \begin{align*} \det A=(-1)(1)-(0)(0)=-1-0=-1. \end{align*} This reflection preserves Euclidean lengths and satisfies $A^\top A=I_2$, but it reverses orientation: it lies in $O(2)$ and outside $SO(2)$. [/example] A single reflection changes only one normal direction, so it is not obvious that such simple maps can account for all of $O(n)$. The structural question is whether an arbitrary orthogonal transformation can be reduced one direction at a time by composing with suitable reflections. [quotetheorem:8284] This theorem explains why reflections are fundamental. It also connects the orthogonal group with Householder transformations in numerical linear algebra, where reflections are used to simplify matrices without changing Euclidean lengths. ## Low-Dimensional Geometry ### The Plane The abstract definition becomes vivid in dimensions $2$ and $3$. In the plane, choosing the image of $e_1$ on the unit circle almost determines an orientation-preserving orthogonal map; the image of $e_2$ must be the perpendicular unit vector with the same orientation. [definition: Planar Rotation Matrix] For $\theta\in\mathbb{R}$, the planar rotation matrix of angle $\theta$ is the linear map \begin{align*} R_\theta:\mathbb{R}^2&\to\mathbb{R}^2,\qquad (x_1,x_2)\mapsto (x_1\cos\theta-x_2\sin\theta,x_1\sin\theta+x_2\cos\theta). \end{align*} [/definition] The two image vectors form an oriented orthonormal basis, so $R_\theta\in SO(2)$. The remaining question is whether every matrix in $SO(2)$ arises this way, or whether the plane contains orientation-preserving orthogonal maps that are not captured by a single angle. [quotetheorem:8285] This theorem identifies $SO(2)$ with the circle group. The next example records how the group law becomes addition of angles. [example: Composition of Plane Rotations] For $\theta,\phi\in\mathbb{R}$ and $x=(x_1,x_2)\in\mathbb{R}^2$, we compute $R_\theta R_\phi$ and show that composition adds the angles. By the definition of the planar rotation matrix, \begin{align*} R_\phi(x)=(x_1\cos\phi-x_2\sin\phi,\;x_1\sin\phi+x_2\cos\phi). \end{align*} Write $R_\phi(x)=(y_1,y_2)$. Then \begin{align*} y_1=x_1\cos\phi-x_2\sin\phi. \end{align*} Also, \begin{align*} y_2=x_1\sin\phi+x_2\cos\phi. \end{align*} Applying $R_\theta$ to $(y_1,y_2)$ gives \begin{align*} R_\theta(R_\phi(x))=(y_1\cos\theta-y_2\sin\theta,\;y_1\sin\theta+y_2\cos\theta). \end{align*} Substituting the two expressions for $y_1$ and $y_2$ gives \begin{align*} R_\theta(R_\phi(x))=((x_1\cos\phi-x_2\sin\phi)\cos\theta-(x_1\sin\phi+x_2\cos\phi)\sin\theta,\;(x_1\cos\phi-x_2\sin\phi)\sin\theta+(x_1\sin\phi+x_2\cos\phi)\cos\theta). \end{align*} For the first coordinate, distributing the scalar factors gives \begin{align*} (x_1\cos\phi-x_2\sin\phi)\cos\theta-(x_1\sin\phi+x_2\cos\phi)\sin\theta=x_1\cos\phi\cos\theta-x_2\sin\phi\cos\theta-x_1\sin\phi\sin\theta-x_2\cos\phi\sin\theta. \end{align*} Collecting the terms containing $x_1$ and $x_2$ gives \begin{align*} x_1\cos\phi\cos\theta-x_2\sin\phi\cos\theta-x_1\sin\phi\sin\theta-x_2\cos\phi\sin\theta=x_1(\cos\phi\cos\theta-\sin\phi\sin\theta)-x_2(\sin\phi\cos\theta+\cos\phi\sin\theta). \end{align*} Since multiplication in $\mathbb{R}$ is commutative, \begin{align*} \cos\phi\cos\theta-\sin\phi\sin\theta=\cos\theta\cos\phi-\sin\theta\sin\phi. \end{align*} By the angle-sum identity for cosine, \begin{align*} \cos\theta\cos\phi-\sin\theta\sin\phi=\cos(\theta+\phi). \end{align*} Similarly, \begin{align*} \sin\phi\cos\theta+\cos\phi\sin\theta=\cos\theta\sin\phi+\sin\theta\cos\phi. \end{align*} By the angle-sum identity for sine, \begin{align*} \cos\theta\sin\phi+\sin\theta\cos\phi=\sin(\theta+\phi). \end{align*} Therefore the first coordinate is \begin{align*} x_1\cos(\theta+\phi)-x_2\sin(\theta+\phi). \end{align*} For the second coordinate, distributing the scalar factors gives \begin{align*} (x_1\cos\phi-x_2\sin\phi)\sin\theta+(x_1\sin\phi+x_2\cos\phi)\cos\theta=x_1\cos\phi\sin\theta-x_2\sin\phi\sin\theta+x_1\sin\phi\cos\theta+x_2\cos\phi\cos\theta. \end{align*} Collecting the terms containing $x_1$ and $x_2$ gives \begin{align*} x_1\cos\phi\sin\theta-x_2\sin\phi\sin\theta+x_1\sin\phi\cos\theta+x_2\cos\phi\cos\theta=x_1(\cos\phi\sin\theta+\sin\phi\cos\theta)+x_2(-\sin\phi\sin\theta+\cos\phi\cos\theta). \end{align*} Commuting real factors gives \begin{align*} \cos\phi\sin\theta+\sin\phi\cos\theta=\sin\theta\cos\phi+\cos\theta\sin\phi. \end{align*} By the angle-sum identity for sine, \begin{align*} \sin\theta\cos\phi+\cos\theta\sin\phi=\sin(\theta+\phi). \end{align*} Also, \begin{align*} -\sin\phi\sin\theta+\cos\phi\cos\theta=\cos\theta\cos\phi-\sin\theta\sin\phi. \end{align*} By the angle-sum identity for cosine, \begin{align*} \cos\theta\cos\phi-\sin\theta\sin\phi=\cos(\theta+\phi). \end{align*} Therefore the second coordinate is \begin{align*} x_1\sin(\theta+\phi)+x_2\cos(\theta+\phi). \end{align*} Combining the two coordinates, \begin{align*} R_\theta(R_\phi(x))=(x_1\cos(\theta+\phi)-x_2\sin(\theta+\phi),\;x_1\sin(\theta+\phi)+x_2\cos(\theta+\phi)). \end{align*} By the definition of $R_{\theta+\phi}$, \begin{align*} R_{\theta+\phi}(x)=(x_1\cos(\theta+\phi)-x_2\sin(\theta+\phi),\;x_1\sin(\theta+\phi)+x_2\cos(\theta+\phi)). \end{align*} Thus \begin{align*} R_\theta(R_\phi(x))=R_{\theta+\phi}(x). \end{align*} Since this equality holds for every $x\in\mathbb{R}^2$, \begin{align*} R_\theta R_\phi=R_{\theta+\phi}. \end{align*} The same formula with $\theta$ and $\phi$ interchanged gives \begin{align*} R_\phi R_\theta=R_{\phi+\theta}. \end{align*} Since addition in $\mathbb{R}$ is commutative, \begin{align*} \phi+\theta=\theta+\phi. \end{align*} Therefore \begin{align*} R_{\phi+\theta}=R_{\theta+\phi}. \end{align*} Combining the preceding equalities gives \begin{align*} R_\phi R_\theta=R_{\phi+\theta}=R_{\theta+\phi}=R_\theta R_\phi. \end{align*} Thus plane rotations commute under composition, which is the concrete reason $SO(2)$ is abelian; this special commutativity does not persist for rotations about different axes in $\mathbb{R}^3$. [/example] The plane also has orientation-reversing orthogonal transformations. After rotations account for determinant $1$, the remaining determinant $-1$ component must be understood: the question is whether every such map is just a reflection across some line through the origin. [quotetheorem:8286] The classification of $O(2)$ is complete because every orthonormal basis in the plane is determined by its first vector and its orientation. In dimension $3$, a new geometric feature appears: an orientation-preserving orthogonal map has an axis. ### Three Dimensions Rotations in space are harder to classify by a single angle because an axis is also needed. Algebraically, that axis should appear as a nonzero vector fixed by the matrix, but it is not immediate that every orientation-preserving orthogonal map in three dimensions must have one. [quotetheorem:8287] The fixed vector is the rotation axis. This theorem hints at the higher-dimensional picture: orthogonal transformations decompose into rotations on mutually perpendicular planes, together with possible sign changes on remaining lines. ## Actions on Spheres and Stabilisers [Orthogonal matrices preserve norms](/theorems/8282), so vectors of length $1$ stay vectors of length $1$. To study directions without also tracking scale, we isolate the locus of all unit vectors as the natural space on which $O(n)$ acts. [definition: Unit Sphere] For $n\ge 2$, the unit sphere in $\mathbb{R}^n$ is \begin{align*} S^{n-1}=\{x\in\mathbb{R}^n:|x|=1\}. \end{align*} [/definition] The sphere records all directions in Euclidean space. A basic symmetry question is whether the orthogonal group can move any chosen unit direction to any other, or whether some directions carry invariant geometric information. [quotetheorem:8288] Transitivity says that any direction can be moved to any other direction. The next object to understand is the subgroup that remains after one direction has been fixed. [definition: Stabiliser of a Unit Vector] Let $n\ge 2$ and let $u\in S^{n-1}$. The stabiliser of $u$ in $O(n)$ is \begin{align*} O(n)_u=\{A\in O(n):Au=u\}. \end{align*} [/definition] Fixing $u$ leaves the perpendicular hyperplane $u^\perp$ invariant, so every element of the stabiliser still has freedom to rotate or reflect the directions orthogonal to $u$. The structural question is whether this is the entire stabiliser. To use stabilisers recursively, we need a precise identification of $O(n)_u$ with the orthogonal group of the $(n-1)$-dimensional space $u^\perp$: the action on $u^\perp$ should determine the whole stabiliser element, and every orthogonal motion of $u^\perp$ should extend back to one of $\mathbb{R}^n$. [quotetheorem:8289] This is the recursive structure of orthogonal geometry. The next example displays the stabiliser explicitly for the last coordinate vector. [example: The Stabiliser of $e_n$] Let $n\ge 2$ and write each vector of $\mathbb{R}^n$ as $(x',x_n)$, where $x'=(x_1,\dots,x_{n-1})\in\mathbb{R}^{n-1}$. We compute the stabiliser \begin{align*}O(n)_{e_n}=\{A\in O(n):Ae_n=e_n\}.\end{align*} The result is that $O(n)_{e_n}$ consists exactly of the maps \begin{align*}(x',x_n)\mapsto (Bx',x_n)\end{align*} with $B\in O(n-1)$. Suppose first that $A\in O(n)_{e_n}$, and let $v_1,\dots,v_n$ be the columns of $A$. Multiplying a matrix by $e_n$ selects its last column, so $Ae_n=e_n$ gives \begin{align*}v_n=e_n.\end{align*} Since $A\in O(n)$, we have \begin{align*}A^\top A=I_n.\end{align*} The $(i,j)$-entry of $A^\top A$ is the dot product of the $i$th and $j$th columns: \begin{align*}(A^\top A)_{ij}=v_i^\top v_j=\langle v_i,v_j\rangle.\end{align*} Thus \begin{align*}\langle v_i,v_j\rangle=(I_n)_{ij}.\end{align*} In particular, $\langle v_i,v_i\rangle=1$ for every $i$, and $\langle v_i,v_j\rangle=0$ whenever $i\ne j$. For $1\le j\le n-1$, we have $j\ne n$, so \begin{align*}0=\langle v_j,v_n\rangle=\langle v_j,e_n\rangle.\end{align*} Write $v_j=((v_j)',(v_j)_n)$ with $(v_j)'\in\mathbb{R}^{n-1}$. Since $e_n=(0,\dots,0,1)$, \begin{align*}\langle v_j,e_n\rangle=\langle ((v_j)',(v_j)_n),(0,\dots,0,1)\rangle=0+\cdots+0+(v_j)_n\cdot 1=(v_j)_n.\end{align*} Therefore $(v_j)_n=0$, so for each $1\le j\le n-1$ there is a unique vector $w_j\in\mathbb{R}^{n-1}$ such that \begin{align*}v_j=(w_j,0).\end{align*} Let $B$ be the $(n-1)\times(n-1)$ matrix whose columns are $w_1,\dots,w_{n-1}$. For $1\le i,j\le n-1$, \begin{align*}\langle w_i,w_j\rangle_{\mathbb{R}^{n-1}}=\langle (w_i,0),(w_j,0)\rangle_{\mathbb{R}^n}=\langle v_i,v_j\rangle_{\mathbb{R}^n}.\end{align*} The last quantity is $(I_n)_{ij}$, and because $i,j\le n-1$, this is the same number as $(I_{n-1})_{ij}$. Hence \begin{align*}(B^\top B)_{ij}=w_i^\top w_j=\langle w_i,w_j\rangle=(I_{n-1})_{ij}.\end{align*} Since this holds for every $1\le i,j\le n-1$, \begin{align*}B^\top B=I_{n-1}.\end{align*} Therefore $B\in O(n-1)$. Now take $(x',x_n)\in\mathbb{R}^n$, with $x'=(x_1,\dots,x_{n-1})$. Using the columns of $A$, \begin{align*}A(x',x_n)=x_1v_1+\cdots+x_{n-1}v_{n-1}+x_nv_n.\end{align*} Substituting $v_j=(w_j,0)$ for $j<n$ and $v_n=e_n=(0,\dots,0,1)$ gives \begin{align*}A(x',x_n)=x_1(w_1,0)+\cdots+x_{n-1}(w_{n-1},0)+x_n(0,\dots,0,1).\end{align*} The first $n-1$ coordinates are $x_1w_1+\cdots+x_{n-1}w_{n-1}=Bx'$, and the last coordinate is $x_n$. Hence \begin{align*}A(x',x_n)=(Bx',x_n).\end{align*} Thus every element of $O(n)_{e_n}$ has the form $(x',x_n)\mapsto (Bx',x_n)$ for some $B\in O(n-1)$. Conversely, let $B\in O(n-1)$ and define $A_B:\mathbb{R}^n\to\mathbb{R}^n$ by \begin{align*}A_B(x',x_n)=(Bx',x_n).\end{align*} In block matrix form, \begin{align*}A_B=\begin{pmatrix}B&0\cr 0&1\end{pmatrix}.\end{align*} Since $B\in O(n-1)$, \begin{align*}B^\top B=I_{n-1}.\end{align*} Also, \begin{align*}A_B^\top=\begin{pmatrix}B^\top&0\cr 0&1\end{pmatrix}.\end{align*} Multiplying the two block matrices gives \begin{align*}A_B^\top A_B=\begin{pmatrix}B^\top B&0\cr 0&1\end{pmatrix}.\end{align*} Substituting $B^\top B=I_{n-1}$ yields \begin{align*}A_B^\top A_B=\begin{pmatrix}I_{n-1}&0\cr 0&1\end{pmatrix}=I_n.\end{align*} Thus $A_B\in O(n)$. Since $e_n=(0,\dots,0,1)=(0,1)$ in the decomposition $\mathbb{R}^{n-1}\times\mathbb{R}$, \begin{align*}A_Be_n=A_B(0,1)=(B0,1)=(0,1)=e_n.\end{align*} Therefore $A_B\in O(n)_{e_n}$. Define \begin{align*}\Phi:O(n-1)\to O(n)_{e_n}\end{align*} by $\Phi(B)=A_B$. For $B,C\in O(n-1)$ and $(x',x_n)\in\mathbb{R}^n$, \begin{align*}A_BA_C(x',x_n)=A_B(Cx',x_n)=(B(Cx'),x_n)=((BC)x',x_n)=A_{BC}(x',x_n).\end{align*} Since the two maps agree on every vector $(x',x_n)$, \begin{align*}\Phi(B)\Phi(C)=A_BA_C=A_{BC}=\Phi(BC).\end{align*} So $\Phi$ is a group homomorphism. The map $\Phi$ is injective. If $\Phi(B)=\Phi(C)$, then $A_B=A_C$. For every $x'\in\mathbb{R}^{n-1}$, \begin{align*}(Bx',0)=A_B(x',0)=A_C(x',0)=(Cx',0).\end{align*} Equality of ordered pairs gives $Bx'=Cx'$ for every $x'\in\mathbb{R}^{n-1}$. Therefore $B$ and $C$ have the same action on every vector of $\mathbb{R}^{n-1}$, so $B=C$. The map $\Phi$ is surjective because the first part proved that every $A\in O(n)_{e_n}$ has the form \begin{align*}A(x',x_n)=(Bx',x_n)=A_B(x',x_n)\end{align*} for some $B\in O(n-1)$. Hence every $A\in O(n)_{e_n}$ equals $\Phi(B)$ for some $B\in O(n-1)$. Thus $\Phi$ is an injective and surjective group homomorphism, so it is a group isomorphism: \begin{align*}O(n-1)\cong O(n)_{e_n}.\end{align*} Fixing the last coordinate direction leaves exactly the freedom to apply an arbitrary orthogonal transformation on the perpendicular coordinate hyperplane $\mathbb{R}^{n-1}\times\{0\}$. [/example] The orbit-stabiliser principle now gives a concrete way to encode the sphere using matrices. Identify two matrices $A,B\in O(n)$ when they send the last basis vector to the same point of the sphere: \begin{align*} A\sim B\quad\text{if and only if}\quad Ae_n=Be_n. \end{align*} Equivalently, $A\sim B$ precisely when $B^{-1}A$ fixes $e_n$, so the difference between $A$ and $B$ lies in the stabiliser identified above with $O(n-1)$. The map from equivalence classes to the sphere is \begin{align*} [A]\longmapsto A e_n. \end{align*} This map is well-defined by the definition of $\sim$. It is surjective by transitivity of the $O(n)$-action on $S^{n-1}$, and it is injective because equal images are exactly the [equivalence relation](/page/Equivalence%20Relation) just defined. Thus, at the level needed here, the sphere is the set of possible images of $e_n$ under orthogonal matrices, with the stabiliser $O(n)_{e_n}\cong O(n-1)$ accounting for the non-uniqueness: \begin{align*} S^{n-1}\cong O(n)/O(n)_{e_n}\cong O(n)/O(n-1). \end{align*} This orbit-and-stabiliser viewpoint is one reason orthogonal groups appear throughout geometry. Spheres, Stiefel manifolds, Grassmannians, and frame bundles can all be described by orthogonal symmetry and stabilisers. ## Lie Structure and Infinitesimal Orthogonality The equation $A^\top A=I_n$ is polynomial in the entries of $A$, and a standard regular-level-set argument shows that it cuts out a smooth matrix group. Near the identity, the relevant question is which tangent directions preserve orthogonality to first order. Differentiating a path identity suggests the condition $X^\top+X=0$. The next definition names the matrices satisfying this infinitesimal orthogonality relation. [definition: Skew-Symmetric Matrix] For $n\ge 1$, a matrix $X\in M_n(\mathbb{R})$ is skew-symmetric if \begin{align*} X^\top=-X. \end{align*} [/definition] Skew-symmetric matrices are infinitesimal rotations, but to use them systematically we need a space of allowable velocity matrices closed under the commutator operation. The obstruction is that an arbitrary matrix velocity would immediately deform lengths; the Lie algebra consists precisely of the velocities that satisfy the first-order orthogonality constraint. [definition: Lie Algebra $\mathfrak{so}(n)$] For $n\ge 1$, the Lie algebra of $O(n)$ and $SO(n)$ is \begin{align*} \mathfrak{so}(n)=\{X\in M_n(\mathbb{R}):X^\top=-X\}, \end{align*} with Lie bracket $[\cdot,\cdot]:\mathfrak{so}(n)\times\mathfrak{so}(n)\to\mathfrak{so}(n)$ given by \begin{align*} [X,Y]=XY-YX. \end{align*} [/definition] Although $O(n)$ has two connected components when $n\ge 2$, the identity component determines the Lie algebra. The remaining issue is to justify that the first-order calculation has not merely produced a plausible candidate: every tangent vector to an orthogonal path through $I_n$ must be skew-symmetric, and every skew-symmetric matrix must actually arise from such a path. [quotetheorem:8290] The matrix exponential supplies paths from infinitesimal data. The next example shows this mechanism in the smallest nonzero Lie algebra of rotations. [example: The Lie Algebra $\mathfrak{so}(2)$] Let $X\in\mathfrak{so}(2)$, and write \begin{align*} X=\begin{pmatrix}x_{11}&x_{12}\cr x_{21}&x_{22}\end{pmatrix}. \end{align*} The condition $X\in\mathfrak{so}(2)$ means $X^\top=-X$. Since \begin{align*} X^\top=\begin{pmatrix}x_{11}&x_{21}\cr x_{12}&x_{22}\end{pmatrix} \end{align*} and \begin{align*} -X=\begin{pmatrix}-x_{11}&-x_{12}\cr -x_{21}&-x_{22}\end{pmatrix}, \end{align*} we have \begin{align*} \begin{pmatrix}x_{11}&x_{21}\cr x_{12}&x_{22}\end{pmatrix}=\begin{pmatrix}-x_{11}&-x_{12}\cr -x_{21}&-x_{22}\end{pmatrix}. \end{align*} Equating entries gives $x_{11}=-x_{11}$, $x_{22}=-x_{22}$, $x_{21}=-x_{12}$, and $x_{12}=-x_{21}$. From $x_{11}=-x_{11}$, adding $x_{11}$ to both sides gives $2x_{11}=0$, hence $x_{11}=0$. Similarly, $x_{22}=-x_{22}$ gives $2x_{22}=0$, hence $x_{22}=0$. If $a=x_{21}$, then $x_{12}=-x_{21}=-a$, so every element of $\mathfrak{so}(2)$ has the form \begin{align*} X=\begin{pmatrix}0&-a\cr a&0\end{pmatrix}. \end{align*} This matrix sends each standard basis vector into the perpendicular coordinate direction. Indeed, \begin{align*} Xe_1=\begin{pmatrix}0&-a\cr a&0\end{pmatrix}\begin{pmatrix}1\cr 0\end{pmatrix}=\begin{pmatrix}0\cdot 1+(-a)\cdot 0\cr a\cdot 1+0\cdot 0\end{pmatrix}=\begin{pmatrix}0\cr a\end{pmatrix}=ae_2. \end{align*} Also, \begin{align*} Xe_2=\begin{pmatrix}0&-a\cr a&0\end{pmatrix}\begin{pmatrix}0\cr 1\end{pmatrix}=\begin{pmatrix}0\cdot 0+(-a)\cdot 1\cr a\cdot 0+0\cdot 1\end{pmatrix}=\begin{pmatrix}-a\cr 0\end{pmatrix}=-ae_1. \end{align*} Set \begin{align*} J=\begin{pmatrix}0&-1\cr 1&0\end{pmatrix}. \end{align*} Then \begin{align*} aJ=a\begin{pmatrix}0&-1\cr 1&0\end{pmatrix}=\begin{pmatrix}0&-a\cr a&0\end{pmatrix}=X. \end{align*} Multiplying $J$ by itself gives \begin{align*} J^2=\begin{pmatrix}0&-1\cr 1&0\end{pmatrix}\begin{pmatrix}0&-1\cr 1&0\end{pmatrix}=\begin{pmatrix}0\cdot 0+(-1)\cdot 1&0\cdot(-1)+(-1)\cdot 0\cr 1\cdot 0+0\cdot 1&1\cdot(-1)+0\cdot 0\end{pmatrix}. \end{align*} The four entries are $0\cdot 0+(-1)\cdot 1=-1$, $0\cdot(-1)+(-1)\cdot 0=0$, $1\cdot 0+0\cdot 1=0$, and $1\cdot(-1)+0\cdot 0=-1$. Hence \begin{align*} J^2=\begin{pmatrix}-1&0\cr 0&-1\end{pmatrix}=-I_2. \end{align*} Thus $J^0=I_2$, $J^1=J$, $J^2=-I_2$, and \begin{align*} J^3=J^2J=(-I_2)J=-J. \end{align*} For every $k\ge 0$, the even powers satisfy \begin{align*} J^{2k}=(J^2)^k=(-I_2)^k=(-1)^kI_2, \end{align*} and the odd powers satisfy \begin{align*} J^{2k+1}=J^{2k}J=(-1)^kI_2J=(-1)^kJ. \end{align*} Using the matrix exponential series, \begin{align*} e^{tX}=e^{atJ}=\sum_{m=0}^{\infty}\frac{(atJ)^m}{m!}. \end{align*} Because $at$ is a scalar and $J$ commutes with its own powers, $(atJ)^m=(at)^mJ^m$ for every $m\ge 0$. Therefore \begin{align*} e^{atJ}=\sum_{m=0}^{\infty}\frac{(at)^mJ^m}{m!}. \end{align*} Splitting this series into its even and odd terms gives \begin{align*} e^{atJ}=\sum_{k=0}^{\infty}\frac{(at)^{2k}J^{2k}}{(2k)!}+\sum_{k=0}^{\infty}\frac{(at)^{2k+1}J^{2k+1}}{(2k+1)!}. \end{align*} Substituting $J^{2k}=(-1)^kI_2$ into the even part gives \begin{align*} \sum_{k=0}^{\infty}\frac{(at)^{2k}J^{2k}}{(2k)!}=\sum_{k=0}^{\infty}\frac{(at)^{2k}(-1)^kI_2}{(2k)!}=\left(\sum_{k=0}^{\infty}\frac{(-1)^k(at)^{2k}}{(2k)!}\right)I_2. \end{align*} Substituting $J^{2k+1}=(-1)^kJ$ into the odd part gives \begin{align*} \sum_{k=0}^{\infty}\frac{(at)^{2k+1}J^{2k+1}}{(2k+1)!}=\sum_{k=0}^{\infty}\frac{(at)^{2k+1}(-1)^kJ}{(2k+1)!}=\left(\sum_{k=0}^{\infty}\frac{(-1)^k(at)^{2k+1}}{(2k+1)!}\right)J. \end{align*} Combining the even and odd parts, \begin{align*} e^{atJ}=\left(\sum_{k=0}^{\infty}\frac{(-1)^k(at)^{2k}}{(2k)!}\right)I_2+\left(\sum_{k=0}^{\infty}\frac{(-1)^k(at)^{2k+1}}{(2k+1)!}\right)J. \end{align*} The Taylor series identities \begin{align*} \cos s=\sum_{k=0}^{\infty}\frac{(-1)^ks^{2k}}{(2k)!} \end{align*} and \begin{align*} \sin s=\sum_{k=0}^{\infty}\frac{(-1)^ks^{2k+1}}{(2k+1)!} \end{align*} with $s=at$ give \begin{align*} e^{tX}=e^{atJ}=\cos(at)I_2+\sin(at)J. \end{align*} Now \begin{align*} \cos(at)I_2=\begin{pmatrix}\cos(at)&0\cr 0&\cos(at)\end{pmatrix} \end{align*} and \begin{align*} \sin(at)J=\sin(at)\begin{pmatrix}0&-1\cr 1&0\end{pmatrix}=\begin{pmatrix}0&-\sin(at)\cr \sin(at)&0\end{pmatrix}. \end{align*} Adding the matrices entry by entry gives \begin{align*} e^{tX}=\begin{pmatrix}\cos(at)&0\cr 0&\cos(at)\end{pmatrix}+\begin{pmatrix}0&-\sin(at)\cr \sin(at)&0\end{pmatrix}=\begin{pmatrix}\cos(at)&-\sin(at)\cr \sin(at)&\cos(at)\end{pmatrix}. \end{align*} This is the planar rotation matrix $R_{at}$. Thus the scalar $a$ is the angular velocity encoded by the infinitesimal generator $X$, and exponentiating for time $t$ gives the finite rotation through angle $at$. [/example] The finite rotation matrix can therefore be recovered from an infinitesimal generator. In higher dimensions, skew-symmetric matrices generate simultaneous rotations in orthogonal coordinate planes after a suitable orthogonal change of basis. ## Compactness and the Standard Action After the geometric classification examples, it is useful to keep only the algebraic structure that belongs directly to the definition. The central action is the one on the [vector space](/page/Vector%20Space) whose inner product defines the group, so the next definition names this action as a representation. [definition: Standard Representation of $O(n)$] The standard representation of $O(n)$ is the group homomorphism $\rho:O(n)\to GL(\mathbb{R}^n)$ defined by $\rho(A)(x)=Ax$ for all $A\in O(n)$ and $x\in\mathbb{R}^n$. [/definition] The representation preserves the Euclidean inner product by construction, and many later constructions use it as the bridge from abstract group multiplication to concrete linear algebra. There is a small but important point to check: when we replace a matrix by its induced linear map on $\mathbb{R}^n$, we should not accidentally identify two different group elements. The next theorem records that this does not happen for the standard action: the action on vectors remembers the whole orthogonal matrix. [quotetheorem:8291] Faithfulness is what lets us treat elements of $O(n)$ interchangeably as matrices and as transformations of Euclidean space. For example, if an orthogonal transformation fixes every vector in a basis, then all of its columns are forced, so the transformation is the identity; more generally, knowing the image of every vector determines the matrix entries. This is special to the chosen action: a different representation of the same group could have a kernel and forget part of the group. Here no information is lost, so later invariant constructions can be phrased either in terms of matrices or in terms of their action on vectors. The next structural property explains why this faithful matrix group is more rigid than $GL(n,\mathbb{R})$: it is compact. [quotetheorem:8292] Compactness is one of the main reasons $O(n)$ behaves better than $GL(n,\mathbb{R})$. The next example illustrates the standard compact-group technique: average an object over all orthogonal changes of coordinates to keep only invariant data. [example: Averaging a Quadratic Form over $O(n)$] Let $B\in M_n(\mathbb{R})$ be symmetric, let $q(x)=\langle Bx,x\rangle$, and let $\mu$ be normalised Haar measure on the compact group $O(n)$. Define \begin{align*}F(x)=\int_{O(n)}q(Ax)\,d\mu(A)=\int_{O(n)}\langle BAx,Ax\rangle\,d\mu(A).\end{align*} For $U\in O(n)$, set $C=AU$. Right multiplication by $U$ is a measure-preserving bijection of $O(n)$ by right invariance of Haar measure, so \begin{align*}F(Ux)=\int_{O(n)}q(AUx)\,d\mu(A)=\int_{O(n)}q(Cx)\,d\mu(C)=F(x).\end{align*} Thus $F$ is unchanged by every orthogonal change of coordinates. For each $A\in O(n)$, using $\langle y,z\rangle=y^\top z$ and $B^\top=B$ gives \begin{align*}\langle BAx,Ax\rangle=(BAx)^\top(Ax)=x^\top A^\top B^\top Ax=x^\top A^\top BAx.\end{align*} Therefore, by linearity of integration in the matrix entries, \begin{align*}F(x)=x^\top Mx,\end{align*} where \begin{align*}M=\int_{O(n)}A^\top BA\,d\mu(A).\end{align*} For every $A\in O(n)$, \begin{align*}(A^\top BA)^\top=A^\top B^\top(A^\top)^\top=A^\top BA,\end{align*} so each integrand is symmetric. Taking transpose entry by entry under the integral gives $M^\top=M$. We show that $M$ is a scalar matrix. Let $D_i$ be the diagonal sign-change matrix sending $e_i$ to $-e_i$ and fixing every $e_k$ with $k\ne i$. Its columns are the standard basis vectors with one sign changed, so every column has length $1$ and distinct columns have inner product $0$; hence $D_i\in O(n)$. Applying the invariance of $F$ with $U=D_i$ gives \begin{align*}x^\top Mx=F(x)=F(D_ix)=(D_ix)^\top M(D_ix).\end{align*} Take $x=e_i+e_j$ with $i\ne j$. Since $M$ is symmetric, $M_{ji}=M_{ij}$. Hence \begin{align*}(e_i+e_j)^\top M(e_i+e_j)=e_i^\top Me_i+e_i^\top Me_j+e_j^\top Me_i+e_j^\top Me_j=M_{ii}+M_{ij}+M_{ji}+M_{jj}=M_{ii}+2M_{ij}+M_{jj}.\end{align*} Also $D_i(e_i+e_j)=-e_i+e_j$, so \begin{align*}(-e_i+e_j)^\top M(-e_i+e_j)=e_i^\top Me_i-e_i^\top Me_j-e_j^\top Me_i+e_j^\top Me_j=M_{ii}-M_{ij}-M_{ji}+M_{jj}=M_{ii}-2M_{ij}+M_{jj}.\end{align*} Equating these two expressions gives \begin{align*}M_{ii}+2M_{ij}+M_{jj}=M_{ii}-2M_{ij}+M_{jj}.\end{align*} Subtracting $M_{ii}+M_{jj}$ from both sides gives \begin{align*}2M_{ij}=-2M_{ij}.\end{align*} Adding $2M_{ij}$ to both sides gives \begin{align*}4M_{ij}=0,\end{align*} and therefore $M_{ij}=0$. Thus every off-diagonal entry of $M$ is zero. Now let $P_{ij}$ be the permutation matrix exchanging $e_i$ and $e_j$ and fixing the other standard basis vectors. Its columns are a reordering of the standard basis, so $P_{ij}\in O(n)$. Applying the invariance of $F$ with $U=P_{ij}$ gives \begin{align*}M_{ii}=e_i^\top Me_i=F(e_i)=F(P_{ij}e_i)=(P_{ij}e_i)^\top M(P_{ij}e_i)=e_j^\top Me_j=M_{jj}.\end{align*} Thus all diagonal entries of $M$ are equal. Since all off-diagonal entries are zero, there is a scalar $\lambda$ such that \begin{align*}M=\lambda I_n.\end{align*} It remains to determine $\lambda$. Taking traces in the definition of $M$ and using linearity of trace and integration gives \begin{align*}\operatorname{tr}(M)=\operatorname{tr}\left(\int_{O(n)}A^\top BA\,d\mu(A)\right)=\int_{O(n)}\operatorname{tr}(A^\top BA)\,d\mu(A).\end{align*} For $A\in O(n)$, the equation $A^\top A=I_n$ gives $A^{-1}=A^\top$, hence $AA^\top=I_n$. By cyclicity of trace, \begin{align*}\operatorname{tr}(A^\top BA)=\operatorname{tr}(BAA^\top)=\operatorname{tr}(BI_n)=\operatorname{tr}(B).\end{align*} Since $\mu$ is normalised, $\mu(O(n))=1$, so \begin{align*}\operatorname{tr}(M)=\int_{O(n)}\operatorname{tr}(B)\,d\mu(A)=\operatorname{tr}(B)\mu(O(n))=\operatorname{tr}(B).\end{align*} On the other hand, \begin{align*}\operatorname{tr}(M)=\operatorname{tr}(\lambda I_n)=\lambda\operatorname{tr}(I_n)=\lambda n.\end{align*} Therefore \begin{align*}n\lambda=\operatorname{tr}(B),\end{align*} so \begin{align*}\lambda=\frac{\operatorname{tr}(B)}{n}.\end{align*} Substituting $M=(\operatorname{tr}(B)/n)I_n$ into $F(x)=x^\top Mx$ gives \begin{align*}F(x)=x^\top\left(\frac{\operatorname{tr}(B)}{n}I_n\right)x=\frac{\operatorname{tr}(B)}{n}x^\top I_nx=\frac{\operatorname{tr}(B)}{n}x^\top x=\frac{\operatorname{tr}(B)}{n}|x|^2.\end{align*} Averaging over all orthogonal coordinate changes removes the directional part of the quadratic form and keeps exactly its trace contribution. [/example] The example signals how orthogonal symmetry removes directional information and keeps rotationally invariant quantities such as norms, dot products, and traces. ## Beyond and Connected Topics The orthogonal group is the first classical matrix group where group theory, linear algebra, geometry, and topology meet in a single definition. As a child of the general [group](/page/Group) concept, it gives a concrete family whose subgroup structure, actions, quotient spaces, and representations can all be studied by direct calculation. The nearest algebraic continuation is the study of matrix groups such as $GL(n)$, $SL(n)$, and symplectic groups. The special orthogonal group sits inside $SL(n,\mathbb{R})$, but the inclusion forgets the inner product. This contrast explains why determinant preservation and inner-product preservation are different kinds of rigidity. Representation theory uses $O(n)$ and $SO(n)$ as central compact examples. Their standard representations lead to tensor powers, symmetric powers, exterior powers, and invariant theory. Orthogonal invariance is also the algebraic source of many familiar scalar quantities, such as $|x|^2$, dot products, and traces. Differential geometry treats $O(n)$ as the structure group of orthonormal frames on a Riemannian manifold. Replacing arbitrary bases by orthonormal bases reduces $GL(n)$-symmetry to $O(n)$-symmetry, and orientation reduces the structure group further to $SO(n)$. Lie theory studies $SO(n)$ through the Lie algebra $\mathfrak{so}(n)$. In dimension $3$, this connects rotations of physical space with the cross product and angular velocity. In higher dimensions, skew-symmetric matrices decompose into rotation blocks and reveal the local structure of the group. The internal notes [Cambridge IA Groups](/page/Cambridge%20IA%20Groups), [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra), and [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules) provide the group-theoretic and linear-algebraic background for this page. For more advanced algebraic structure and invariant viewpoints, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra) is a later algebraic reference, especially when matrix groups are studied through polynomial equations. ## References Androma, [Cambridge IA Groups](/page/Cambridge%20IA%20Groups). Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Artin, *Algebra* (1991). Hall, *Lie Groups, Lie Algebras, and Representations* (2003). Roman, *Advanced Linear Algebra* (2008).