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The rotation matrix $e^{xA}$ commutes with $C$ and is orthogonal, soSince $C^n=\mathrm{diag}(1,(-1{)}^n,(-1{)}^n)$,

The orthogonal group isThe special orthogonal group is its determinant-one subgroup

Every eigenvalue of an orthogonal matrix has modulus one. A real three-by-three matrix has at least one real eigenvalue, and nonreal eigenvalues occur in a complex conjugate pair. If $Q\in SO(3)$ has such a pair $e^{\pm i\theta }$, their product is one, so the remaining eigenvalue is $\det Q=1$. If all three eigenvalues are real, each is $\pm 1$ and their product is one; an odd number of three signs with product one must include $+1$. Hence every element of $SO(3)$ has an eigenvector of eigenvalue one and represents a rotation in three dimensions about its span.

It is false that every element of $O(3)$ is either a rotation or a plane reflection. For example,is an improper orthogonal transformation combining a rotation with a reflection. Its eigenvalues are $e^{\pm i\theta }$ and $-1$, whereas a plane reflection has eigenvalues $1,1,-1$ and a rotation has determinant one.

The linear least-squares problem is to choose $x\in {\mathbb{R}}^n$ minimizing the Euclidean norm of the residual:For a QR decomposition $A=QR$, orthogonality of $Q$ gives $\Vert Ax-b{\Vert }_2=\Vert Rx-Q^{T}b{\Vert }_2$. If $A$ has full column rank and $R=\left(\genfrac{}{}{0ex}{}{R_1}{0}\right)$ is in standard form with invertible upper-triangular $R_1\in {\mathbb{R}}^{n\times n}$, the minimizer solves $R_{1}x=(Q^{T}b{)}_{1:n}$.

Applying the Gram-Schmidt process to the columns of the given matrix yieldsIndeed, $Q$ is an orthogonal matrix and $A=QR$. Moreover,Back substitution in the leading triangular system givesso the unique least-squares solution isThe unused final transformed residual component is $1$, so the minimum residual norm is $1$.

The paths of $U{X}_t$ are continuous, and linear transformation preserves independence of increments. Moreover,because $U$ is an orthogonal matrix. Thus all defining properties are preserved, proving the orthogonal invariance of Brownian motion:

For an orthogonal matrix $A$, invariance in distribution givesThe orthogonal group is transitive on each Euclidean sphere, so ${\phi }_X(u)$ depends only on $\Vert u{\Vert }^2$. DefineThe continuity of a characteristic function makes $f$ continuous. Taking $A=-I$ also gives $X\stackrel{d}{=}-X$, and henceso $f$ is real-valued. Thusas for every orthogonally invariant random vector.

Given a solution $u(t)$, construct $a_j(t)=\frac{1}{2}{e}^{u_j(t)/2}$ and hence the smooth matrices $L(t)$ and $B(t)$. Let $U(t)$ solve the linear ordinary differential equationBecause $B(t)$ is skew-symmetric,so $U(t)$ is an orthogonal matrix and is invertible. Moreover,It follows thatThus the Lax pair flow is an Isospectral Lax equation: every $L(t)$ is similar to $L(0)$ and has the same eigenvalues and characteristic polynomial.

Every symmetric function of these constant eigenvalues is a first integral. In particular, the trace invariants of a Lax equationare constant because the cyclic property of the trace givesFor example,is an explicit first integral of equation $(†)$.

By the real spectral theorem, a real symmetric matrix has an orthogonal diagonalizationWriting $y=Q^{T}x$ givesThis is nonnegative for every $x$ exactly when every ${\lambda }_i\ge 0$. Thus the quadratic-form and eigenvalue definitions of a positive semidefinite matrix agree.

When $A$ is positive semidefinite, defineThis matrix is symmetric and positive semidefinite, and its square is $A$, so it is the principal square root of a positive semidefinite matrix.

For nonsingular $M$, the matrix $M^{T}M$ is symmetric andfor every nonzero $x$. Thus $M^{T}M$ is positive definite, andexists and is nonsingular. Let $R=M{P}^{-1}$. Since $P$ is symmetric,Hence $R$ is an orthogonal matrix andthe polar decomposition of an invertible real matrix.

In three dimensions, $P$ stretches or contracts along three mutually perpendicular eigenvector directions by its positive eigenvalues. The orthogonal map $R$ is then applied: it is a rotation when $\det M>0$, and a rotation combined with a reflection when $\det M<0$.

For a nonzero vector $v\in {\mathbb{R}}^n$, a Householder reflection isIt is immediately symmetric. If $P=v{v}^T/(v^{T}v)$, then $P^2=P$, soHence $H$ is also an orthogonal matrix, with $H^{-1}=H$.

The similarity transformation can be expanded asComputing $v^{T}A$, $Av$, and the scalar $v^{T}Av$ costs $O(n^2)$ arithmetic operations, after which the remaining updates are outer products and scalar multiples, also costing $O(n^2)$. Thus $HA{H}^{-1}$ can be formed in $O(n^2)$ operations rather than by two general $O(n^3)$ matrix multiplications.

Starting with $A_1=A$, for $k=1,\dots ,n-2$ choose a Householder reflection $H_k$ that acts only on coordinates $k+1,\dots ,n$ and maps the tail of column $k$,to a multiple of its first coordinate vector. SetThis zeros all entries in column $k$ below its first subdiagonal. Because $H_k$ fixes the first $k$ coordinates, it preserves the zeros created in earlier columns. After $n-2$ steps,is an upper Hessenberg matrix.

Each $H_k$ is orthogonal and symmetric. Ifthen $Q$ is orthogonal andPart (a) shows that each similarity update costs $O(n^2)$ arithmetic operations, and there are $O(n)$ updates. The total cost is therefore

By the spectral theorem for real symmetric matrices, write $A=O^{t}DO$ with $O$ orthogonal and $D=\mathrm{diag}({\xi }_1,\dots ,{\xi }_n)$. Part a and the tensor-product property of the Fourier transform giveThusApplying the pullback rule from part c to the orthogonal change of variables, for which $|\det O|=1$, replaces $D$ by $A$ and $D^{-1}$ by $A^{-1}$. Since determinant and signature are invariant under orthogonal conjugation,

Each QR decomposition $A_k=Q_k{R}_k$ givesInduction therefore yields the first Accumulated QR factorization identity:

We next prove the corresponding factorization of a matrix power. The case $k=0$ is just $A=Q_0{R}_0$. Ifthen the similarity formula above impliesConsequently,The product ${\overline{Q}}_k$ is orthogonal, and the product ${\overline{R}}_k$ is upper triangular. Henceis a QR decomposition.

False under the hypotheses as printed. Nonzero coefficients do not ensure the two-column dominant-subspace condition.

For a counterexample, take $n=3$, eigenvaluesand an orthonormal eigenbasis $w_1,w_2,w_3$ for whichThe two displayed coefficient vectors are orthonormal and can be completed to an orthogonal matrix, so such a real symmetric matrix $A$ exists. Every $b_i$ and $c_i$ is nonzero, butIndeed, with $t=1/\sqrt{2}$,is a nonzero multiple of $w_1$, while the remaining direction in
$\mathrm{span}(A^k{e}_1,A^k{e}_2)$ isThusBy the simultaneous iteration interpretation of the QR algorithm, the leading two columns of the accumulated $Q$ factor span this same space. The leading block $B_k$ is therefore an orthogonal-coordinate representation of the compression of $A$ to this space. Sincewe havefor every relevant $k$, rather than $\{-2,2\}$. Hence the requested Hausdorff convergence does not hold.

For completeness, the intended statement becomes true if one addsLet $Z_k$ consist of the first two columns of ${\overline{Q}}_{k-1}$. The Accumulated QR factorization identity givesAfter division by $|{\lambda }_n{|}^k$, every component along
$w_1,\dots ,w_{n-2}$ tends to zero because of the strict spectral gap, while $\Delta \ne 0$ makes the two surviving dominant components independent. Therefore these two-dimensional subspaces converge to the dominant invariant subspaceThere are two-by-two orthogonal matrices $U_k$ such thatBecauseup to the harmless index convention at $k=0$, we obtainin the matrix 2-norm. Orthogonal similarity preserves the spectrum, and the stated spectral perturbation bound now gives

Over $\mathbb{C}$, the real skew-symmetric matrix $A$ is normal because $A^{\ast }=-A$. Part (a) gives an orthonormal complex eigenbasis, and the nonzero eigenvalues are purely imaginary pairs $\pm i\lambda$.

For each $\lambda >0$, choose a unit real vector $x$ in the kernel of $A^2+{\lambda }^{2}I$ and putThen $x\perp y$, $\Vert y\Vert =1$, andThus $A$ has matrixon the orthonormal basis $(x,y)$. Distinct such invariant planes are orthogonal; complete them by an orthonormal basis of $\ker A$. Taking these basis vectors as the columns of an orthogonal matrix $R$ gives

The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear partial differential operator $P(D)$ on ${\mathbb{R}}^n$ has a fundamental solution of a linear differential operator: there is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ such that $P(D)E={\delta }_0$.

Write $D=-i\partial$. After an orthogonal change of coordinates and multiplication by a nonzero constant, its polynomial symbol may be written as a monic polynomial in the last frequency,For each real ${\mu }^{\prime }$, this polynomial has $M$ complex roots counted with multiplicity. Among a fixed finite collection of horizontal lines at bounded heights, one can choose a line that stays a positive distance from all those roots. Continuity of the roots preserves the choice on a neighborhood $N({\mu }^{\prime })$. Take a countable locally finite cover by such neighborhoods, refine it to a measurable disjoint partition ${\mathbb{R}}^{n-1}={⨆}_j{\Delta }_j$, and let $c_j$ be the chosen height on ${\Delta }_j$. The resulting Hörmander staircasehas bounded heights and may be chosen so that $|P({\xi }^{\prime },s+i{c}_j)|\ge 1$ on each step.

For a test function $\phi$, defineThe Paley–Wiener–Schwartz theorem gives rapid decay in the real frequency directions and at most a fixed exponential factor in the bounded imaginary direction. Together with $|P|\ge 1$, this proves that the integral defines a continuous distribution. Applying $P(D)$ cancels the denominator. The remaining integrand is entire in $z$, so the Cauchy integral theorem shifts every horizontal contour to the real axis; the partition then recombines into ${\mathbb{R}}^{n-1}$. The Fourier inversion theorem giveswhich proves the theorem.

Suppose $A^T=A$. For eigenvectors $v_i,v_j$ with distinct real eigenvalues, symmetry givesThus $({\lambda }_i-{\lambda }_j)v_i^T{v}_j=0$, and the eigenvectors are orthogonal. Divide each by its nonzero Euclidean norm to obtain an orthonormal basis. The matrix $P$ having these normalized columns satisfies $P^{T}P=I$, so it is an orthogonal matrix and $P^{-1}=P^T$. As in part (a),This proves the needed orthogonal diagonalization from symmetry and distinct eigenvalues, without invoking an unproved spectral theorem.

$A$ 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.

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

Represent the quadratic form by the symmetric matrixThe spectral theorem for real symmetric matrices gives an orthogonal matrix $Q$ and real eigenvalues ${\lambda }_1,{\lambda }_2$ such that $Q^{T}MQ=\mathrm{diag}({\lambda }_1,{\lambda }_2)$. In the orthogonal change of coordinates $\mathbf{x}=Q\mathbf{u}$, the equation becomes ${\lambda }_1{u}_1^2+{\lambda }_2{u}_2^2=1$.

If both eigenvalues are positive, this is an ellipse with semiaxes ${\lambda }_1^{-1/2}$ and ${\lambda }_2^{-1/2}$. Conversely, a zero eigenvalue makes the set either empty or unbounded along that coordinate, while two negative eigenvalues give the empty set. Opposite signs give an unbounded hyperbola. A negative eigenvalue together with a zero one also gives the empty set. None of those cases is an ellipse. Therefore the exact condition is that $M$ be a symmetric positive-definite matrix.

The sum and product of its eigenvalues are its matrix trace and determinant:Two real numbers are both positive precisely when their sum and product are positive. ThusAn orthogonal transformation preserves area. Using i gives the quadratic-form ellipse area