Codex @codex  0

For a nonzero real vector $w$, a Householder reflection isWriting $P=w{w}^T/(w^{T}w)$ gives $P^T=P$ and $P^2=P$, so $H^{T}H=(I-2P{)}^2=I$. Thus a Householder matrix is an orthogonal matrix, and it reflects across the hyperplane perpendicular to $w$.

For $u\ne 0$, choose $\alpha =-\mathrm{sgn}(u_1)\Vert u\Vert$, taking the sign to be $+1$ when $u_1=0$, and put $w=u-\alpha e_1$. This $w$ is nonzero. Since ${\alpha }^2=\Vert u{\Vert }^2$, direct expansion gives $w^{T}w=2(\Vert u{\Vert }^2-\alpha u_1)$ and $w^{T}u=\Vert u{\Vert }^2-\alpha u_1$, soApply this construction successively to the trailing part of each column of $A$, extending each reflection by an identity on earlier coordinates. Previously created zeros are preserved. If a trailing column vector is zero, use the identity there. The resulting product $Q^T$ is orthogonal and $Q^{T}A=R$ is upper triangular, yielding the QR factorization $A=QR$ even when $A$ is singular.

To solve $Ax=b$, multiply by $Q^T$ and use back substitution on $Rx=Q^{T}b$ when $R$ is nonsingular. If $A$ is singular, $\det R=\det (Q^T)\det A=0$; an upper triangular determinant is its diagonal product, so at least one diagonal entry is zero. In that case consistency and free variables require separate treatment.

The unshifted QR algorithm starts from $A_0=A$, factors $A_k=Q_k{R}_k$, and formsEvery iterate is related by an orthogonal similarity transformation to the previous one and hence to $A$, with the same eigenvalues. Symmetry is preserved. For symmetric matrices with appropriate spectral separation, the iterates approach diagonal form: the off-diagonal entries decrease and the diagonal exposes the eigenvalues. Practical shifted QR accelerates convergence and uses deflation when off-diagonal couplings become negligible. With a shift ${\mu }_k$, factor $A_k-{\mu }_{k}I=Q_k{R}_k$ and set $A_{k+1}=R_k{Q}_k+{\mu }_{k}I$; the same similarity identity still holds.

Universal diagonal convergence of unshifted QR would be too strong. For example $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ admits the standard factorization $Q=A$, $R=I$, so the iteration remains at $A$ rather than becoming diagonal, although its eigenvalues are $\pm 1$. Such equal-modulus cases require a convergence qualification or treatment of a retained block. Similarity is unconditional; convergence to a useful limiting form is an additional algorithmic property.

Use coordinates with the downward vertical as reference. The first rod's center is at $(a/2)(\sin \theta ,-\cos \theta )$, so its translational plus rotational kinetic energy is $m{a}^2{\dot{\theta }}^2/6$. The second center isIts squared speed is $a^2{\dot{\theta }}^2+a^2{\dot{\varphi }}^2/4+a^2\dot{\theta }\dot{\varphi }\cos (\theta -\varphi )$. Adding its center-of-mass rotation, using moment of inertia $m{a}^2/12$, gives the totalThe two center heights give $V=-mga(3\cos \theta +\cos \varphi )/2$. Hence $L=T-V$ is exactly the stated Lagrangian.

Small stable oscillations are about the downward equilibrium $\theta =\varphi =0$. Retaining quadratic terms gives $T_2=\frac{1}{2}{\dot{q}}^{T}M\dot{q}$ and $V_2=\frac{1}{2}{q}^{T}Kq$, whereA normal mode $q=v{e}^{i\omega t}$ satisfies $(K-{\omega }^{2}M)v=0$. With $\kappa =a{\omega }^2/g$, the determinant equation isThus the angular frequencies areBoth squared frequencies are positive. The amplitude ratio is $\varphi /\theta =(9-8\kappa )/(3\kappa )$: it is positive in the lower-frequency mode and negative in the higher-frequency mode. The rods therefore swing in phase in the first mode and in opposite phase in the second.

A real square matrix is orthogonal if $Q^{T}Q=I$. Then $\Vert Qx{\Vert }^2=x^T{Q}^{T}Qx=x^{T}x=\Vert x{\Vert }^2$, proving length preservation.

Define $J(i,j,\theta )$ to be identity outside rows and columns $i,j$, with blockThe block's transpose times itself is identity because ${\cos }^2\theta +{\sin }^2\theta =1$ and the off-diagonal terms cancel. Thus $J$ is orthogonal, with determinant one.

For $r=\sqrt{x_i^2+x_j^2}>0$, choose $\cos \theta =x_i/r$, $\sin \theta =x_j/r$. Then $y_i=r$, $y_j=0$ and every other coordinate is unchanged. If $r=0$, choose $\theta =0$. This fixes the useful nonnegative pivot convention.

Apply the rotations to the current matrix in the displayed order from right to left. First $J(1,2)$ eliminates entry $(2,1)$; then $J(1,3)$ eliminates $(3,1)$ without changing row two. Finally $J(2,3)$ eliminates $(3,2)$ while preserving the zeros in column one. ThusInverting the product gives $A=QR$. Even if $A$ starts lower triangular, the row rotations generally create nonzero entries above the diagonal; there is no further universal simplification. For example $A=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)$ gives positive-pivotwhose three entries above the diagonal are all nonzero.

If $A$ is orthogonal, $R$ is both upper triangular and orthogonal. Orthogonality of its columns successively forces all off-diagonal entries to vanish and all diagonal entries to have absolute value one. The first two pivots are positive under our convention, and the rotations have determinant one. Therefore the precise result isWithout a specified pivot-sign convention, the invariant statement is a diagonal matrix with entries $\pm 1$ and product $\det A$.

Use $u=x-y$, $v=x+y$. It sends the diamond-shaped open square onto $(-1,1{)}^2$, and the inverse Jacobian has absolute determinant $1/2$. On this rectangle $|u|<1$ and $v+2>1$, so the original function is measurable, bounded by one and supported on a finite-area set, proving integrability. The transformed integral separates:The two one-dimensional integrals follow from the fundamental theorem of calculus; Fubini's theorem applies by the established integrability.

Write $q=(x,X{)}^T$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The linear equations are $M\ddot{q}+Kq=0$. A normal mode has all coordinates oscillating with one frequency and fixed relative amplitudes, $q=v\cos (\omega t+\phi )$. Its squared frequency solvesand its amplitude vector is a nonzero null vector of $K-{\omega }^{2}M$. The assumed distinct positive roots give two normal frequencies, their positive square roots.

For the rods let ${\theta }_1,{\theta }_2$ be their small inclinations from the downward vertical. To first order, ${\theta }_1=x/a$ and ${\theta }_2=(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second has centre-of-mass horizontal velocity $(\dot{x}+\dot{X})/2$ and angular velocity $(\dot{X}-\dot{x})/(2a)$, givingTogether these give $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The centre-of-mass height changes giveproving the displayed quadratic Lagrangian.

Here $M=(m/3)\left(\begin{array}{cc}3& 1\\ 1& 2\end{array}\right)$ and $K=(mg/(2a))\left(\begin{array}{cc}6& -1\\ -1& 1\end{array}\right)$. Set $\lambda =2a{\omega }^2/(3g)$. The determinant equation reduces to $5{\lambda }^2-17\lambda +5=0$, soThe first row of the mode equation gives $X/x=(6-3\lambda )/(1+\lambda )$. The actual rod-angle ratio is thereforeIt is positive for the smaller root and negative for the larger root. Thus the low-frequency mode has the rods in phase, and the high-frequency mode has them out of phase. Using the rod-angle ratio is necessary because $X$ is the displacement of the end of the second rod, not that rod's inclination.

Yes. Suppose $g$ maps the first flag to the second and has determinant $\delta \ne 0$. Relative to an adapted basis of the first flag, scaling its first basis vector by ${\delta }^{-1}$ and fixing all others defines a transformation $h$ preserving that flag, with determinant ${\delta }^{-1}$. Then $gh$ still maps the first flag to the second and has determinant one. Thus the special linear subgroup acts transitively too.

Write $q=A+Bx$ and $\ell =\log 3$. Orthogonality of $x^2-q$ to $1$ and $x$ givesThe Gram matrix is positive definite, with determinant $4(\ell -1)>0$, so the solution is unique. Solving yields

Yes. Suppose $g$ maps the first flag to the second and has determinant $\delta \ne 0$. Relative to an adapted basis of the first flag, scaling its first basis vector by ${\delta }^{-1}$ and fixing all others defines a transformation $h$ preserving that flag, with determinant ${\delta }^{-1}$. Then $gh$ still maps the first flag to the second and has determinant one. Thus the special linear subgroup acts transitively too.

Write $q=A+Bx$ and $\ell =\log 3$. Orthogonality of $x^2-q$ to $1$ and $x$ givesThe Gram matrix is positive definite, with determinant $4(\ell -1)>0$, so the solution is unique. Solving yields

The chain rule givesIts determinant is $-1$, so it is invertible. The map $G$ is smooth and $G(0,0)=(0,0)$. The inverse function theorem therefore yields an open neighbourhood $U$ of the origin for which $G(U)$ is open, contains the origin, and

The complex velocity obtained from part (b) isFor a nonzero planar potential doublet, $B/A=(a^2/b^2)e^{-2i\alpha }$. Its zeros therefore satisfyand henceThe denominators are nonzero because $a/b<1$. For either sign,so $|z_{\pm }|=a$. They are distinct because $t_{+}\ne t_{-}$ and $t\mapsto (c-bt)/(1-t)$ is injective, its determinant being $c-b\ne 0$. Clearing the velocity denominators gives a quadratic with these two roots, so there are no further finite stagnation points. Thus both stagnation points lie on the obstacle boundary.

Write $J_1,J_2,J_3$ for the actual successive rotations $J(1,2)$, $J(1,3)$, and $J(2,3)$, each selected using the currently transformed matrix. The first eliminates position $(2,1)$; the second eliminates $(3,1)$ while leaving row two unchanged. After these, the first column is $(r_{11},0,0{)}^T$. The final rotation of rows two and three preserves both zeros in that column and eliminates $(3,2)$. HenceA lower-triangular starting matrix does not force any further zeros above the diagonal. For example, the lower-triangular matrix with all entries on and below its diagonal equal to one gives, with the nonnegative pivot convention,which has all upper-triangular entries nonzero.

Inverting the product of rotations yields the QR factorizationThe order reverses when taking the transpose. If $A$ is orthogonal, then $R$ is both orthogonal and upper triangular. Its first column forces $|r_{11}|=1$ and orthogonality with the other columns forces $r_{12}=r_{13}=0$. Applying the same argument to the remaining block makes $R$ diagonal with entries of modulus one.

Our elimination rule makes $r_{11}$ and $r_{22}$ positive when nonzero, so they are both one. Every rotation has determinant one; thus $\det R=\det A$ andWithout a fixed pivot-sign convention one could only specify a diagonal sign matrix. The convention chosen in part (c) determines the precise signs above.

A real square matrix is orthogonal when $Q^{T}Q=I$, equivalently when its columns form an orthonormal basis. Taking determinants gives $(\det Q{)}^2=1$, soIn particular it is invertible with inverse $Q^T$ and preserves the Euclidean norm.

Use a QR factorization, whose existence is supplied. Since $|\det Q|=1$ and $R$ is triangular,Each diagonal entry is one component of its column, so $|r_{jj}|\le \Vert r_j\Vert =\Vert a_j\Vert$. Multiplication yields Hadamard determinant inequality:The argument includes singular matrices, since a zero diagonal factor makes the determinant zero.

Strictly, it is the infinitesimal generators, not every unitary matrix with unit determinant, that are traceless. For a continuous one-parameter family $U(ϵ)=\exp (iϵT)$, unitarity requires the generator $T$ to be Hermitian. The determinant condition givesDifferentiating at $ϵ=0$ yields $\mathrm{tr}T=0$. A Hermitian matrix has $n$ real diagonal parameters and $n(n-1)$ real off-diagonal parameters; one real trace constraint leaves $n^2-1$ independent generators. ThusThe identity matrix supplies a concrete counterexample to interpreting the claim as one about arbitrary group elements: $I_n\in SU(n)$ but $\mathrm{tr}{I}_n=n$. Also, $e^{i\mathrm{tr}T}=1$ at a single parameter value only fixes the trace modulo $2\pi$; the continuous neighbourhood is essential.

Put type-A atoms at $2ja$ and type-B atoms at $(2j+1)a$. Their displacements $u_j,v_j$ obey Newton's second law,Choose normal modes $u_j=U{e}^{i(2jaq-\omega t)}$ and $v_j=V{e}^{i((2j+1)aq-\omega t)}$. Substitution givesA nonzero displacement vector requires zero determinant:Solving this quadratic in ${\omega }^2$ gives the diatomic-chain dispersion relation with equal springs,The primitive unit cell has length $2a$, so the first Brillouin zone is $|q|\le \pi /(2a)$.

The principal moments of inertia are the eigenvalues of the symmetric inertia tensor. In units of $\mathrm{kg}{\mathrm{m}}^2$, its characteristic determinant isTherefore the requested algebraic answer isAn orthonormal basis of principal axes is $(1,1,0)/\sqrt{2}$, $(1,-1,0)/\sqrt{2}$, and $(0,0,1)$; the last two span the degenerate eigenspace and may be replaced by any orthonormal basis of it.

There is a physical inconsistency in calling the supplied matrix a rigid body's inertia tensor. The moment-of-inertia triangle inequalities require each principal moment to be no greater than the sum of the other two. Here $3>1+1$. In principal coordinates, $I_2+I_3-I_1=2\int x_1^{2}dm$ cannot be negative for a nonnegative mass distribution. Thus the eigenvalue calculation is well defined, but no ordinary rigid mass distribution realizes this tensor. The positive second-moment matrix test for an inertia tensor makes the same obstruction explicit.

To find the two-mass three-spring normal modes, choose both displacements positive to the right, measured from mechanical equilibrium. The middle spring's extension changes by $x_2-x_1$. Hooke's law and Newton's second law therefore giveIn a normal mode both displacements have the same angular frequency: take $x_j=A_j\cos (\omega t+\varphi )$. A nonzero amplitude vector exists when the determinant vanishes:Writing $D=\sqrt{m_1^2-m_1{m}_2+m_2^2}$, the two angular frequencies areThe corresponding eigenvectors can be specified bySince $D>|m_2-m_1|$ for positive masses, the lower normal mode ${\omega }_{-}$ has $A_2/A_1>0$: both bodies move in the same direction at any instant. The higher normal mode ${\omega }_{+}$ has $A_2/A_1<0$: they move in opposite directions. Their amplitudes are generally unequal. For the useful check $m_1=m_2=m$, these reduce to ${\omega }_{-}=\sqrt{k/m}$ with $A_2=A_1$, and ${\omega }_{+}=\sqrt{3k/m}$ with $A_2=-A_1$. Each angular frequency has units $\mathrm{rad}{\mathrm{s}}^{-1}$.

For the instantaneous Lorenz error growth at a zero-y crossing, direct differentiation of the Lorenz system givesAt a crossing of $y=0$ its symmetric part is particularly simple:Thus a perturbation initially in the $z$ direction has norm-squared rate $-2b$ and initially decays in norm at rate $b$. The $x$-dependent couplings between $y$ and $z$ are skew-symmetric at the crossing and do not contribute to the instantaneous total norm derivative. They can rotate the perturbation later: a pure $\delta z$ immediately induces $\dot{\delta y}=-x\delta z$. It is therefore incorrect to claim that such an error remains an independently decaying $z$ mode along the complete trajectory.

For the $(x,y)$ block, the trace is $-2(\sigma +1)<0$ and the determinant is $4\sigma -(\sigma +r-z{)}^2$. A negative determinant gives one positive and one negative eigenvalue. Equivalently,In particular the requested sufficient lower-height condition isThere is also a second growth range $z>\sigma +r+2\sqrt{\sigma }$; it must not be lost by removing the absolute value. At either endpoint the largest rate is zero. Between them both block eigenvalues are negative. At a crossing in either growing range, some planar error directions have positive instantaneous exponential rate. This statement is local in time: a single instantaneous positive eigenvalue of $S$ does not prove sustained exponential error growth over an evolving Lorenz trajectory.

Uniform area distribution on the unit disk givesThe Jacobian determinant has absolute determinant $r$, soIntegrating out the other coordinate givesThe joint probability density function factors into these two marginal distribution, proving independence of $R$ and $\Theta$. FinallyThe angle at $R=0$ can be defined arbitrarily, since that event has probability zero.