Orthogonal Maps & Orthogonal Matrices

In geometry we care about transformations that keep lengths and angles intact. A linear map $f:\mathbb{R}^n\to\mathbb{R}^n$ is orthogonal when it preserves the inner product: $\langle f(\mathbf u),f(\mathbf v)\rangle=\langle \mathbf u,\mathbf v\rangle\quad\text{for all }\mathbf u,\mathbf v\in\mathbb{R}^n$ (Definition 8.10). On $\mathbb{R}^n$ with the dot product this reads $f(\mathbf u)\cdot f(\mathbf v)=\mathbf u\cdot\mathbf v$. The matrix of an orthogonal map with respect to an orthonormal basis is called an orthogonal matrix; because a change of orthonormal basis is itself orthogonal (Remark 8.7), the condition is basis-independent and can be stated directly on matrices as $(A\mathbf u)\cdot(A\mathbf v)=\mathbf u\cdot\mathbf v$. A warning on names: an orthogonal matrix has orthonormal columns, not merely orthogonal ones — the terminology is standard but slightly misaligned, and an orthogonal projection onto a proper subspace is not an orthogonal map.

The defining condition has several faces. Length: Theorem 8.17 says $f$ is orthogonal iff $\lVert f(\mathbf x)\rVert=\lVert \mathbf x\rVert$ for all $\mathbf x$ — orthogonal maps are precisely the length-preserving linear maps. One direction is immediate from $\lVert f(\mathbf x)\rVert=\sqrt{f(\mathbf x)\cdot f(\mathbf x)}=\sqrt{\mathbf x\cdot\mathbf x}=\lVert \mathbf x\rVert$; the converse recovers the dot product from norms by expanding $\lVert f(\mathbf x)+f(\mathbf y)\rVert^2$. Angle: Corollary 8.2 then gives $\angle(f(\mathbf x),f(\mathbf y))=\angle(\mathbf x,\mathbf y)$, because $\cos$ of the angle is assembled from the dot product and the two norms, all of which are preserved. Orthonormal images: Theorem 8.18 says $f$ is orthogonal iff $f(\mathbf e_1),\dots,f(\mathbf e_n)$ form an orthonormal basis of $\mathbb{R}^n$ — orthogonal maps carry one orthonormal basis to another, and that property characterises them.

At the matrix level the clean test is Theorem 8.19 / 8.21: an $n\times n$ matrix $A$ is orthogonal iff its columns form an orthonormal basis, iff $A^{T}A=I$, iff $A^{-1}=A^{T}$. The reason: the $(i,j)$ entry of $A^{T}A$ is exactly $\mathbf v_i\cdot\mathbf v_j$ for the columns $\mathbf v_1,\dots,\mathbf v_n$, so $A^{T}A=I$ says precisely $\mathbf v_i\cdot\mathbf v_j=\delta_{ij}$ (unit length, mutually perpendicular). This makes orthogonal matrices wonderfully easy to invert: no elimination, just transpose. Orthogonal matrices are moreover closed under products and inverses (Theorem 8.20) — composing length-preserving maps keeps length preserved — so they form the orthogonal group $O(n)$ (Remark 8.8).

Taking determinants in $A^{T}A=I$ gives $1=\det(A^{T}A)=\det(A^{T})\det(A)=\det(A)^2$, so every orthogonal matrix satisfies $\det A=\pm 1$ (Corollary 8.3). The sign splits the group in two. When $\det A=+1$ the map is orientation-preserving; these form the special orthogonal group $SO(n)$. In the plane they are exactly the rotations $R_\theta=\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix},$ orthogonal for every angle $\theta$ (Example 8.9) with $\det R_\theta=\cos^2\theta+\sin^2\theta=1$. When $\det A=-1$ the map reverses orientation; the $2\times 2$ examples are reflections, such as $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ (reflection across the $x$-axis) with $\det=-1$. Both kinds leave every length and angle unchanged.