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Let $\mathbf{r}(u,v)$ parametrize $S$. Then $A\mathbf{r}$ parametrizes its image, with tangent vectors $A{\mathbf{r}}_u$ and $A{\mathbf{r}}_v$. An orthogonal matrix preserves dot products, so the Lagrange identity for the cross product gives, for any $\mathbf{p},\mathbf{q}$,The area element of a surface is therefore the same pointwise before and after applying $A$. Integrating provesThis includes orthogonal transformations with determinant $-1$: reversal of orientation does not affect the Euclidean norm in the surface area formula.

First find the fixed axis of rotation by solving $(M-I)\mathbf{n}=0$. After removing a nonzero common factor, the linear equations areThey give $n_2=n_3=-n_1$. Choose the following unit vectors:Their dot products show that they form an orthonormal basis; hence $P=({\mathbf{p}}_1{\mathbf{p}}_2{\mathbf{p}}_3)$ is an orthogonal matrix. Since $M$ is an orthogonal matrix fixing ${\mathbf{p}}_1$, it preserves the orthogonal complement of ${\mathbf{p}}_1$. Its restriction there has determinant $1$, so it is a planar rotation.

The trace gives $1+2\cos \theta =\mathrm{tr}M=9/7$, so $\cos \theta =1/7$. To determine the sign of the sine, rather than just its magnitude, calculateThus a suitable angle and change of basis areIn fact direct matrix multiplication gives

For the specified coordinate vectors,Choose the orthogonal matrix representing a quarter-turn,Then $S^{T}S=I$ and direct matrix multiplication gives $S^{T}LS=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)=M$. Thus these two shear mappings are orthogonally similar.

An orthogonal matrix satisfies $A^{-1}=A^T$, while a symmetric matrix satisfies $A^T=A$. Combining them gives $\boxed{A^{-1}=A}$. For the product with the antisymmetric matrix $B$, the rule for the transpose of a matrix product givesThus $AB$ is an antisymmetric matrix exactly when $AB=BA$. The hypotheses do not force this commutativity. For a counterexample, takeHere $A$ is a symmetric matrix and an orthogonal matrix, and $B$ is an antisymmetric matrix and an orthogonal matrix, but $AB$ is a nonzero symmetric matrix, so it is not an antisymmetric matrix. On the other hand, the product of two orthogonal matrices is always an orthogonal matrix:Therefore the answers are $\boxed{\text{(i) no; (ii) yes.}}$

The unique symmetric and antisymmetric parts of a matrix are $S=(M+M^T)/2$ and $A=(M-M^T)/2$, givingTheir matrix products with their transposes are $S^{T}S=(a+b{)}^{2}I/4$ and $A^{T}A=(a-b{)}^{2}I/4$. Each is an orthogonal matrix precisely when the corresponding factor is one. Thus $a+b=\pm 2$ and $a-b=\pm 2$, with independent sign choices, and the simultaneous possibilities are

The given symmetric matrices satisfy $E_j^T{E}_j=E_j^2=I$, so both are orthogonal matrices. Put $J=E_1{E}_2$, $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. Anticommutation givesThus $A=cI+sJ$ and $A^T=cI-sJ$, giving $A{A}^T=(c^2+s^2)I=I$. These are orthogonal rotations generated by anticommuting involutions. Also $A-A^T=2sJ$. Since $J^2=-I$ makes $J$ invertible, $A$ is symmetric exactly when $\sin (\theta /2)=0$.

For orthogonal matrices $Q_1,\dots ,Q_m$ of the same size, transpose reverses the order of multiplication. Thereforeby canceling the adjacent pairs $Q_j^T{Q}_j=I$ successively. Hence every finite product is an orthogonal matrix.

A real normal matrix satisfies $M{M}^T=M^{T}M$. For a symmetric matrix, $M^T=M$ and both products equal $M^2$. For an antisymmetric matrix, $M^T=-M$ and both equal $-M^2$. For an orthogonal matrix, both products are the identity. Thus each of the three classes consists of normal matrices.

A symmetric matrix that is also an orthogonal matrix satisfies $C=C^{-1}$. Multiplying by the commuting invertible factors gives $(I-B{)}^2=(I+B{)}^2$, hence $4B=0$. Thus there is no nonzero choice of $B$. The only symmetric case is $B=0$, giving $C=I$.

Write $P=I-B$, $Q=I+B$, so $P^T=Q$ and $Q^T=P$. The factors commute because both are polynomials in $B$. Also $P$ has a matrix inverse since it is the transpose of the invertible $Q$. Thus the Cayley transform of a real skew-symmetric matrix, with its parameter replaced by $-B$, satisfiesHence $C$ is an orthogonal matrix. Its inverse is $Q{P}^{-1}$; this lets us distinguish the symmetric and skew-symmetric Cayley transforms below.

The Euclidean norms are $\Vert v_{\pm }\Vert =2$ and $\Vert v_0\Vert =\sqrt{2}$. With normalized eigenvectors as rows,The rows form an orthonormal basis, so $R{R}^T=I$ and $R$ is an orthogonal matrix. The zero eigenvalue means $M{v}_0=0$ for a nonzero vector; thus $M$ is singular and has no matrix inverse.

The three columns of $N=\sqrt{8}A$ are $(\sqrt{3},1,2)$, $(-\sqrt{2},\sqrt{6},0)$, and $(-\sqrt{3},-1,2)$. Each has squared Euclidean norm eight. Their pairwise dot products areHence $N^{\mathsf{T}}N=8I$, so $\boxed{A^{\mathsf{T}}A=I}$. Thus $A$ is an orthogonal matrix, with $A^{-1}=A^{\mathsf{T}}$. The determinant from (i) places it in the determinant-one component.

The transpose reverses product order, so $(CB{)}^{\mathsf{T}}=B^{\mathsf{T}}{C}^{\mathsf{T}}$. If $B$ is an orthogonal matrix, then $B^{-1}=B^{\mathsf{T}}$ and the given identity becomesMultiply on the left by $B$ to obtain $C^{\mathsf{T}}+C=0$. Therefore $C$ is antisymmetric, so it is an antisymmetric matrix, namely $\boxed{C^{\mathsf{T}}=-C}$.

The $(i,j)$ entry of $B^{\mathsf{T}}B$ is ${\mathbf{e}}_i\cdot {\mathbf{e}}_j={\delta }_{ij}$. Hence $B$ is an orthogonal matrix, whose matrix inverse equals its transpose:This obtains the inverse without unnecessary cofactor calculations.

A real square orthogonal matrix $Q$ satisfiesequivalently $Q^{-1}=Q^T$. Its columns, and also its rows, form an orthonormal basis. Consequently it preserves dot products and Euclidean norms.

The preceding cycles show the action on a full basis, not only on one direction: $A$ sends ${\mathbf{e}}_1\to {\mathbf{e}}_2\to {\mathbf{e}}_3\to {\mathbf{e}}_1$, and $A^{\mathsf{T}}$ reverses each arrow. Therefore either composition fixes every basis vector, and by linearity fixes every vector. ThusIn particular $A$ is an orthogonal matrix, $A^{\mathsf{T}}=A^{-1}=A^2$, and $A^3=I$.

For either $A$ from part (i), chooseIndeed, using $A^2=-I$ and $A^{\mathsf{T}}=-A$,Its matrix trace is $-2/\sqrt{2}=-\sqrt{2}<0$. Explicitly, for the two $A$ matrices in the order displayed above, the corresponding orthogonal matrices areTheir negatives have positive matrix trace and would not meet the additional sign condition.

Since $A^2=-I$, the matrix inverse is $A^{-1}=-A$. The orthogonal matrix condition also gives $A^{-1}=A^{\mathsf{T}}$, so $A^{\mathsf{T}}=-A$. Thus $A$ is skew-symmetric, forcing $a=0$ and $c=-b$. Its columns must be unit vectors, giving $b^2=1$. Therefore the two possibilities areBoth satisfy the original conditions by direct multiplication. They are the two orthogonal skew-symmetric matrices in two dimensions, representing opposite quarter-turn rotations.

A three-dimensional rotation matrix is an orthogonal matrix with determinant $+1$. Here $\det A=b(a^2-d^2)$. For the branch $d=0$, the determinant condition gives $b=1$; for the branch $a=0$, it gives $b=-1$. HenceThe first is the identity rotation; the remaining three are half-turns. Their fixed axes are respectively the $y$ axis, the direction $(1,0,-1)$ and the direction $(1,0,1)$. The identity is included under the usual rotation matrix convention.

The original PDF requires $B$ itself to be an orthogonal skew-symmetric matrix; it does not require $C$ to be diagonal. A real two-dimensional skew-symmetric matrix has the form $B=\left(\begin{array}{cc}0& u\\ -u& 0\end{array}\right)$. The orthogonal matrix condition gives $u^2=1$, so choose $u=1$. Then $C=B^{\mathsf{T}}AB$ is forced:Direct multiplication verifies $B^{\mathsf{T}}=-B$, $B^{\mathsf{T}}B=I$ and $BC{B}^{\mathsf{T}}=A$. Choosing $-B$ gives the same $C$. Replacing $B$ by a matrix of eigenvectors to make $C$ diagonal would generally fail the printed antisymmetry requirement.