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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.

A polar vector transforms as ${\mathbf{v}}^{\prime }=Q\mathbf{v}$ under an orthogonal matrix $Q$, including an orientation-reversing transformation. Displacement, velocity and force are polar vectors. An axial vector instead transforms as ${\mathbf{a}}^{\prime }=(\det Q)Q\mathbf{a}$.

An axial vector transforms as ${\mathbf{a}}^{\prime }=(\det Q)Q\mathbf{a}$ under an orthogonal matrix $Q$. It agrees with a polar vector under a proper rotation matrix and has the additional determinant sign under an orientation-reversing transformation. Angular momentum and angular velocity are axial vectors.

The SO(3) group consists of real three-by-three orthogonal matrices with determinant one. It acts by rotation in three dimensions. Its group manifold is ${\mathbb{RP}}^3$, since the Adjoint double cover from SU(2) to SO(3) identifies antipodal points of the three-sphere.

Every real symmetric matrix has an orthonormal basis of real eigenvectors. Equivalently, it admits an orthogonal diagonalization $A=O^{T}DO$ with $D$ real diagonal and $O$ an orthogonal matrix.

If real matrices satisfy $A^T=A\le 0$ and $B^T=-B$, then $e^{tA}$ is a contraction and $e^{tB}$ is an orthogonal matrix for $t\ge 0$. Hence $\Vert e^{kA/2}{e}^{kB}{e}^{kA/2}{\Vert }_2\le 1$ for every $k\ge 0$. The bound is independent of commutativity and gives unconditional stability for a centered convection-diffusion semidiscretization with homogeneous Dirichlet boundary conditions. It does not itself bound the mesh-dependent commutators that enter a time-error estimate.

A real orthogonal matrix $T$ of size three with $\det T=-1$ has an eigenvector $n$ of eigenvalue $-1$. Its orthogonal complement is invariant, and the restriction there is an orientation-preserving plane rotation. Thus $T$ is a rotation about the axis $\mathbb{R}n$ composed with a reflection across the plane perpendicular to $n$. Writing the plane angle as $\varphi$, $\mathrm{tr}T=-1+2\cos \varphi$. A pure plane reflection is the case $\varphi =0$; if $\det (T-I)\ne 0$, it cannot be a pure plane reflection.