A matrix \(\mathbf{A}\in M_{n,n}(\mathbb{R})\) is orthogonal if and only if its column vectors form an orthonormal basis of \(\mathbb{R}^n\) with respect to the standard scalar product \(\langle\cdot{,}\cdot\rangle.\) To this end let \(\hat{\Omega} : (\mathbb{R}^n)^n \to M_{n,n}(\mathbb{K})\) denote the map which forms an \(n\times n\) matrix from \(n\) column vectors of length \(n.\) That is, \(\hat{\Omega}\) satisfies \[[\hat{\Omega}(\vec{a}_1,\ldots,\vec{a}_n)]_{ij}=[\vec{a}_j]_{i}\] for all \(1\leqslant i,j\leqslant n\) and where \([\vec{a}_j]_{i}\) denotes the \(i\)-th entry of the vector \(\vec{a}_j.\) Then, by the definition of matrix multiplication, we have for all \(1\leqslant i,j\leqslant n\) \[\begin{aligned} [\hat{\Omega}(\vec{a}_1,\ldots,\vec{a}_n)^T\hat{\Omega}(\vec{a}_1,\ldots,\vec{a}_n)]_{ij}&=\sum_{k=1}^n[\hat{\Omega}(\vec{a}_1\ldots,\vec{a}_n)^T]_{ik}[\hat{\Omega}(\vec{a}_1\ldots,\vec{a}_n)]_{kj}\\ &=\sum_{k=1}^n[\hat{\Omega}(\vec{a}_1\ldots,\vec{a}_n)]_{ki}[\hat{\Omega}(\vec{a}_1\ldots,\vec{a}_n)]_{kj}\\ &=\sum_{k=1}^n [\vec{a}_i]_k[\vec{a}_j]_k=\langle\vec{a}_i,\vec{a}_j\rangle_{\mathbf{1}_{n}}=\delta_{ij}, \end{aligned}\] as claimed.

The reader is invited to check that a corresponding statement also holds for the rows of an orthogonal matrix.

Let \(n \in \mathbb{N}\) and \(\sigma \in S_n\) be a permutation. Recall that for \(1\leqslant i\leqslant n,\) the \(i\)-th column of the permutation matrix \(\mathbf{P}_{\sigma}\) of \(\sigma\) is given by \(\vec{e}_{\sigma(i)},\) where \(\mathbf{e}=(\vec{e}_1,\ldots,\vec{e}_n)\) denotes the standard ordered basis of \(\mathbb{R}^n.\) Therefore, the columns of a permutation matrix form an ordered orthonormal basis of \(\mathbb{R}^n\) and hence permutation matrices are orthogonal by the previous remark.

A plane in \(\mathbb{R}^3\) is a subspace \(U\) of dimension \(2=3-1.\) More generally, a hyperplane in an \(n\)-dimensional vector space \(V\) is a subspace \(U\) of dimension \(n-1.\)

We can now verify that \(r_U\) is an orthogonal transformation. For all vectors \(u,v \in V\) we have \[\begin{aligned} \langle r_U(u),r_U(v)\rangle&=\langle u-2\langle e,u\rangle e,\langle v-2\langle e,v\rangle e\rangle\\ &=\langle u,v\rangle-2\langle u,e\rangle\langle e,v\rangle-2\langle e,u\rangle \langle e,v\rangle+4\langle e,u\rangle\langle e,v\rangle \langle e,e\rangle\\ &=\langle u,v\rangle, \end{aligned}\] where we use that \(\langle\cdot{,}\cdot\rangle\) is bilinear, symmetric and that \(\langle e,e\rangle=1.\) We conclude that \(r_U\) is an orthogonal transformation.

Finally, observe that with respect to the ordered basis \(\mathbf{b}=(u_1,\ldots,u_{n-1},u_n)\) of \(V\) we have \[\mathbf{M}(r_U,\mathbf{b},\mathbf{b})=\begin{pmatrix} \mathbf{1}_{n-1} & \\ & -1 \end{pmatrix}.\] Indeed, since \(u_{i} \in U\) for all \(1\leqslant i\leqslant n-1,\) we obtain \(r_U(u_i)=2\Pi^{\perp}_U(u_i)-u_i=2u_i-u_i=u_i.\) Furthermore, \(r_U(u_n)=u_n-2\langle u_n,u_n\rangle u_n=-u_n,\) as claimed. We conclude that \(\det r_U=-1.\)

(The group \(\mathrm{O}(2)\)). Recall from the exercises that if a matrix \(\mathbf{R}\in M_{2,2}(\mathbb{R})\) satisfies \(\mathbf{R}^T\mathbf{R}=\mathbf{1}_{2},\) then it is either of the form \[\begin{pmatrix} a & -b \\ b & a \end{pmatrix} \qquad \text{or} \qquad \begin{pmatrix} a & b \\ b & -a \end{pmatrix}\] for some real numbers \(a,b.\) The condition \(\mathbf{R}^T\mathbf{R}=\mathbf{1}_{2}\) implies that \(a^2+b^2=1,\) hence we can write \(a=\cos \alpha\) and \(b=\sin\alpha\) for some \(\alpha \in \mathbb{R}.\) In the second case we have \(\det \mathbf{R}=-a^2-b^2=-1,\) thus \[\mathrm{SO}(2)=\left\{\mathbf{R}_{\alpha}=\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} | \alpha \in \mathbb{R}\right\}.\] Recall also that the mapping \(f_{\mathbf{R}_{\alpha}} : \mathbb{R}^2 \to \mathbb{R}^2\) is the counter-clockwise rotation around \(0_{\mathbb{R}^2}\) by the angle \(\alpha.\) In the second case we obtain \[\begin{pmatrix} \cos \alpha & \sin \alpha \\ \sin\alpha & -\cos\alpha\end{pmatrix}=\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.\] The matrix \[\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\] is the matrix representation of the orthogonal reflection along the \(x\)-axis in \(\mathbb{E}^2\) with respect to the standard ordered basis \(\mathbf{e}\) of \(\mathbb{R}^2.\) We thus obtain a complete picture of all orthogonal transformations of \(\mathbb{E}^2.\) An orthogonal transformation of \(\mathbb{E}^2\) is either a special orthogonal transformation in which case it is a rotation around \(0_{\mathbb{R}^2}\) or else a composition of the orthogonal reflection along the \(x\)-axis and a rotation around \(0_{\mathbb{R}^2}.\)

1. The proof of Proposition 10.44
   Proposition 10.44 ➔

   Let \((V,\langle\cdot{,}\cdot\rangle)\) and \((W,\langle\!\langle\cdot{,}\cdot\rangle\!\rangle)\) be finite dimensional Euclidean spaces and \(f : V \to W\) a linear map. Then there exists a unique linear map \(f^* : W \to V\) such that \[\langle\!\langle f(v),w\rangle\!\rangle=\langle v,f^*(w)\rangle\] for all \(v \in V\) and \(w \in W.\)

   implies that an endomorphism \(f : V \to V\) of a finite dimensional Euclidean space \((V,\langle\cdot{,}\cdot\rangle)\) is self-adjoint if and only if its matrix representation with respect to an orthonormal basis \(\mathbf{b}\) of \(V\) is symmetric. In particular, in \(\mathbb{R}^n\) equipped with the standard scalar product, a mapping \(f_\mathbf{A}: \mathbb{R}^n \to \mathbb{R}^n\) for \(\mathbf{A}\in M_{n,n}(\mathbb{R})\) is self-adjoint if and only if \(\mathbf{A}\) is symmetric.

2. Let \((V,\langle\cdot{,}\cdot\rangle)\) be a finite dimensional Euclidean space and \(f : V \to V\) an orthogonal transformation. Then \(f\) is normal. Indeed, using that \(f\) is orthogonal, we obtain for all \(u,v \in V\) \[\langle f^{-1}(u),v\rangle=\langle f(f^{-1}(u)),f(v)\rangle=\langle u,f(v)\rangle\] so that the adjoint mapping of an orthogonal transformation is its inverse mapping, \(f^*=f^{-1}.\) It follows that \(f\circ f^{*}=f\circ f^{-1}=\mathrm{Id}_V=f^{-1}\circ f=f^*\circ f\) so that \(f\) is normal.