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The relevant inner products areHence $1$ and $x$ are orthogonal vectors. Dividing by their norms gives the orthonormal basisof their linear span $U$.

$A$ is not an orthogonal matrix. Its first column has squared Euclidean normso $A^{T}A\ne I$. The reason this is compatible with ii is that $1,x,x^2$ is not an orthonormal basis for the given inner product. Its Gram matrix isThe correct condition on this matrix representation of a linear map iswhich follows either by direct multiplication or by the inner product preservation just proved. This is the rule for an orthogonal transformation in a nonorthonormal basis.

Put $p=(3x^2-2)/\sqrt{2}$ and use the orthonormal basis $\mathcal{B}=(1,\sqrt{3/2}x,p)$ from iii. The columns of the given matrix giveConsequentlyThe images of $\mathcal{B}$ are therefore $(p,\sqrt{3/2}x,-1)$, again an orthonormal basis. By i, $T$ is an orthogonal transformation. Its matrix in this orthonormal basis is

If $T$ is an orthogonal transformation and $e_1,e_2,e_3$ is an orthonormal basis, thenThe images form an orthonormal set, hence are linearly independent and form an orthonormal basis in dimension three.

Conversely, suppose the images $T{e}_1,T{e}_2,T{e}_3$ of one chosen orthonormal basis form an orthonormal basis. Write $u={\sum }_i{a}_i{e}_i$ and $v={\sum }_j{b}_j{e}_j$. By linearity of $T$ and bilinearity,Thus mapping one orthonormal basis to an orthonormal basis is equivalent to preserving every inner product. In the converse, a single chosen orthonormal basis suffices.

If $T^2=I$, every $v\in V$ has the decompositionwhose first term belongs to $E_1$ and second to $E_{-1}$. Their intersection is zero by ii, so $V=E_1\oplus E_{-1}$ is a direct sum. Choosing bases of the two eigenspaces gives a diagonal matrix with entries $1$ and $-1$.

Since $T^2\ne T$ while $T^2=I$, we have $T\ne I$. Thus $\dim E_{-1}$ is $1$, $2$, or $3$. Reordering the basis gives exactlyFor $-I_3$, $T$ is point inversion through the origin. For $\mathrm{diag}(1,1,-1)$, it fixes the plane $E_1$ pointwise and reverses the complementary line $E_{-1}$: if $v=p+\ell$ in this direct sum, then $Tv=p-\ell$. For $\mathrm{diag}(1,-1,-1)$, it fixes the line $E_1$ pointwise and reverses the complementary plane $E_{-1}$.

These are instances of oblique symmetry of complementary subspaces, allowing oblique eigenspaces. They are a Euclidean reflection in a hyperplane and a rotation in three dimensions through $\pi$ about a line, respectively, when the two eigenspaces are perpendicular. The hypotheses do not require an orthogonal transformation or an orthonormal basis, so that additional metric property cannot be inferred just from the displayed diagonal forms.

Let $v\in S$. Choose a unit vector $u$ perpendicular to $v$, and put $w=v\times u$. The cross product identities show that $(v,u,w)$ is an orthonormal basis with positive orientation: $u\times (v\times u)=v$, soThus the matrix $A_v$ with these columns belongs to the special orthogonal group $G$ and satisfies $A_{v}N=v$. For any $v,z\in S$, the element $A_z{A}_v^{-1}$ sends $v$ to $z$. Therefore the group action on the unit sphere is transitive.

Use the point ${\mathbf{p}}_0=(-2,-2,4)$ and the orthonormal basis of directions in the cutting planeBoth vectors are orthogonal to the normal vector $(1,1,2)$, so every point of the plane has unique coordinatesSubstituting into the cone equation givesConsequently the intersection has equationBecause these are orthonormal coordinates, their lengths are the actual lengths in the plane. Thus this is an ellipse with semi-major axis $\sqrt{24}$ and semi-minor axis $\sqrt{8}$, not merely a projected ellipse. By a, its eccentricity is

A real square matrix $A$ is an orthogonal matrix whenHere $A^T$ is the transpose and $I$ the identity matrix; equivalently, its columns form an orthonormal basis. If this holds, the squared Euclidean norm obeysso the linear transformation preserves length.

Conversely, suppose $|A\mathbf{v}|=|\mathbf{v}|$ for every vector $\mathbf{v}$. Put $B=A^{T}A-I$, a symmetric matrix. Then ${\mathbf{v}}^{T}B\mathbf{v}=0$ for every vector. Taking $\mathbf{v}={\mathbf{e}}_i$, the standard basis vectors, gives $B_{ii}=0$. Taking $\mathbf{v}={\mathbf{e}}_i+{\mathbf{e}}_j$ for $i\ne j$ gives $2B_{ij}=0$. Thus $B=0$, proving that the linear transformation preserves length if and only if $A$ is an orthogonal matrix.

Choose the first orthonormal basis vector to be ${\mathbf{e}}_1=\mathbf{a}$. If $\mathbf{b}$ is not parallel to $\mathbf{a}$, choose ${\mathbf{e}}_2$ along $\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{a}$; otherwise choose any unit vector orthogonal to $\mathbf{a}$. Complete to an orthonormal basis with ${\mathbf{e}}_3$. The matrix with rows ${\mathbf{e}}_i^T$ is an orthogonal matrix, and expressing the three unit vectors in this basis givesThe last two components of ${\mathbf{c}}^{\prime }$ have polar coordinates of the displayed form; we may choose $w\ge 0$. If they vanish, $\theta$ is arbitrary.

An orthogonal transformation preserves dot products. Taking dot products with ${\mathbf{a}}^{\prime }$, and then using the unit Euclidean norms, givesThe remaining dot product givesAn orthogonal matrix has determinant $\pm 1$, so it preserves the absolute scalar triple product. In the new coordinates this scalar triple product is $yw\sin \theta$. ConsequentlyTaking the nonnegative square root provesNo division by $y$ or $w$ was used, so the proof includes parallel vectors. Equivalently, the expression under the square root is the Gram determinant of the three unit vectors.

By the spectral theorem for real symmetric matrices, there is an orthogonal matrix $Q$ such thatFor clarity, the diagonalization can also be justified recursively. The characteristic polynomial has a complex root of a polynomial by the fundamental theorem of algebra. By ii its eigenvalue is real; taking a nonzero real or imaginary part of an associated eigenvector in ${\mathbb{C}}^n$ gives a real eigenvector, which can be normalized. Its orthogonal complement is invariant, because $⟨Ax,z⟩=⟨x,Az⟩=\lambda ⟨x,z⟩=0$. Restriction to this linear subspace is again symmetric, and induction on dimension gives an orthonormal basis of eigenvectors.

All ${\lambda }_i\ge 0$ by ii. The permitted invariance of determinant and matrix trace under change of basis, applied to the diagonal matrix, yields

Let $B=A^{T}A$. It is a symmetric matrix, and ${\mathbf{v}}^{T}B\mathbf{v}=\Vert A\mathbf{v}{\Vert }^2\ge 0$. By the spectral theorem for real symmetric matrices, choose an orthonormal basis ${\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3$ of eigenvectors of $B$, with nonnegative eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$.

For each $i\ne j$, take the unit parallelogram parametrized by $s{\mathbf{v}}_i+t{\mathbf{v}}_j$, $0\le s,t\le 1$. Its surface area is $1$. Moreover,The image parallelogram consequently has area $\sqrt{{\lambda }_i{\lambda }_j}$. By the assumption this equals $1$ for all three pairs, givingNo eigenvalue can be zero. Dividing the equations and using positivity yields ${\lambda }_1={\lambda }_2={\lambda }_3=1$. HenceThus testing all surfaces is stronger than necessary: the three unit parallelograms in these principal directions already determine the result. This is the three-dimensional characterization of linear surface-area isometries.

Decompose $\mathbf{v}$ by orthogonal projection onto the straight line spanned by $\mathbf{n}$ and its orthogonal complement:The first term is fixed by the rotation. If $\mathbf{w}\ne 0$, the vectors $\mathbf{w}/|\mathbf{w}|$ and $(\mathbf{n}\times \mathbf{w})/|\mathbf{w}|$ form an orthonormal basis of the perpendicular plane, oriented in the positive sense about $\mathbf{n}$. A planar rotation through $\theta$ therefore sendsHere $\mathbf{n}\times \mathbf{w}=\mathbf{n}\times \mathbf{v}$. Adding the fixed parallel component proves the Rodrigues rotation formula in the required form:When $\mathbf{w}=0$, both perpendicular terms vanish, so the same formula fixes $\mathbf{v}$ as required. Positive angles use the right-hand convention about $\mathbf{n}$.

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

The quadratic form isThe symmetric matrix has eigenvalues $2,8$, with respective orthogonal eigenvectors $(1,-1{)}^T,(1,1{)}^T$. Normalize them to obtain the orthonormal basisThe given change of basis meansSubstitution, or orthogonal diagonalization of a real symmetric matrix, givesThus the semi-major and semi-minor axes have lengths $2\sqrt{2}$ and $\sqrt{2}$. The eccentricity and enclosed area areThe orthogonal change of basis preserves area, so this is also the area in the original Cartesian coordinates.

Express the position vector in the suggested orthonormal basis as $\mathbf{r}=X{\mathbf{e}}_1+Y{\mathbf{e}}_2+Z{\mathbf{e}}_3$. ThenThis orthogonal transformation preserves $x^2+y^2+z^2=X^2+Y^2+Z^2$ and has Jacobian determinant of absolute value one, so $dV=dXdYdZ$. The basis actually reverses orientation; the absolute value of the Jacobian determinant is why no minus sign enters the volume integral. Also $x-y=\sqrt{2}Z$, so the half-space becomes $Z\ge 1/2$.

At each fixed $Z\in [1/2,1]$, the section is the disk $X^2+Y^2\le 1-Z^2$, of area $\pi (1-Z^2)$. Its reflection symmetry in $X$ makes the integral of $X$ vanish. Hence the required triple integral isThe negative sign is also consistent with the cap lying in the direction $(1,-1,0)$, where its average $y$ coordinate is negative.

The real spectral theorem states that a symmetric matrix has an orthonormal basis of real eigenvectors, and hence is a diagonalisable matrix. Here is a proof in this setting.

For $n\ge 1$, the quadratic form $q(x)=x^{T}Ax$ attains a maximum on the unit sphere, by compactness and the extreme value theorem. Choose a unit maximizer $u$ and put $\lambda =u^{T}Au$. For any $w\perp u$, the Rayleigh quotient along $u+tw$ has a maximum at $t=0$, soThe last equality uses $A=A^T$ and $u^{T}w=0$. Thus $Au$ is orthogonal to every $w\perp u$, so $Au=\lambda u$.

The orthogonal complement $u^{\perp }$ is an invariant subspace, because for $x\perp u$,The restriction of $A$ to $u^{\perp }$ is again symmetric with respect to the induced inner product. By mathematical induction on dimension, it has an orthonormal basis of eigenvectors; adjoining $u$ gives one for ${\mathbb{R}}^n$. If $Q$ has those basis vectors as columns, then $Q^{T}AQ$ is a real diagonal matrix and $Q^{-1}=Q^T$.

Write $N=(n_1,n_2,n_3)$. Restriction to the plane $z=0$ givesThis is an equation of degree at most two in two Cartesian coordinates, the general algebraic form of a conic, with possible degenerate conics and an empty set of real points. No determinant hypothesis is needed for this substitution.

For an arbitrary plane, choose a point ${\mathbf{x}}_0$ and an orthonormal basis $\mathbf{e},\mathbf{f}$ of its direction linear subspace. Put $B=(\mathbf{e}\mathbf{f})$ and $\mathbf{x}={\mathbf{x}}_0+B\mathbf{\xi }$, where $\mathbf{\xi }=(s,t{)}^T$. With $S=(M+M^T)/2$, the symmetric part of a matrix, the restricted equation isThis proves the plane-section result whenever “conic” allows the appropriate degeneracies. A rigid motion taking the chosen plane to $z=0$ gives the same argument.

The source's assumption $\det M\ne 0$ does not guarantee a nondegenerate conic. Even with $M=I$ and $f=x^2+y^2+z^2-1$, the section $z=1$ is a single point, and $z=2$ has no real points. With the nonsingular symmetric matrix giving $f=2xz+y^2$, the section $z=0$ is the repeated straight line $y=0$.

There is a further literal omission: the source does not assume $M$ is symmetric. Forone has $\det M=2$ but $f=2xz$. Thus the section $z=0$ is the entire plane, rather than a conic. Only $S$ controls the quadratic form; invertibility of an arbitrary $M$ cannot remove this exception. If the intended hypothesis is that $M$ is symmetric and nonsingular, the quadratic restriction to a plane cannot vanish identically: otherwise $B^{T}MB=0$, so $M$ would map a two-dimensional linear subspace injectively into its one-dimensional orthogonal complement, which is impossible. Degenerate and empty sections still remain possible.

Adding and subtracting the two coordinates shows that $A(1,1{)}^T=(c+s)(1,1{)}^T$ and $A(1,-1{)}^T=(c-s)(1,-1{)}^T$. The normalized eigenvectors can therefore be chosen asTheir inner product is $(1-1)/2=0$, and their squared norms are one, proving orthonormality. If $s=0$, every nonzero vector is an eigenvector; this chosen orthonormal basis still works.

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.

For an arbitrary vector $\mathbf{x}$, expansion in the orthonormal basis gives $x_i=(a_i{a}_j+b_i{b}_j+c_i{c}_j)x_j$, with summation over $j$. Since this holds for every $\mathbf{x}$, the resolution of the identity in an orthonormal basis isLet $P_a=\mathbf{a}{\mathbf{a}}^T$, and similarly for $b,c$. The three orthogonal projection matrices sum to $I$ and have pairwise products zero. ThusAny ray meeting all three ideal perpendicular mirrors therefore returns in the opposite direction, independently of reflection order. This is the geometry of a corner reflector, a kind of retroreflector used in lunar laser ranging.