Codex @codex  0

Apply matrix multiplication to obtainSince $\hat{\mathbf{n}}$ is a unit vector, the subtracted term is the orthogonal projection of $\mathbf{x}$ along $\hat{\mathbf{n}}$. Hence $M$ is the orthogonal projection onto a plane through the origin with normal vector $\hat{\mathbf{n}}$. It fixes every vector perpendicular to $\hat{\mathbf{n}}$ and sends every scalar multiple of $\hat{\mathbf{n}}$ to zero. In particular $\hat{\mathbf{n}}\cdot M\mathbf{x}=0$.

From part (f), $PQ=\mathrm{diag}(-1,-1,1)$. To obtain $PQR=-I$, choose the Householder transformation that reverses only the third coordinate:Indeed $\mathbf{r}{\mathbf{r}}^T=\mathrm{diag}(0,0,2)$, so $R_{ij}={\delta }_{ij}-r_i{r}_j$, and direct matrix multiplication gives $PQR=-I$. The alternative vector $-\mathbf{r}$ produces the same matrix. Geometrically $R$ reflects in the plane $x_3=0$, completing the reversal of all three coordinates.

Adding the vectors component by component givesConsequently $\boxed{a=0}$. The requirement that the vectors be nonzero then implies $b\ne 0$.

Since $\mathbf{v}=-\mathbf{u}$ and $\mathbf{u}\ne 0$, the vectors point in opposite directions. More explicitly, their dot product givesHence their angle is $\boxed{\theta =\pi \text{ radians}=180^{\circ }}$.

Induction on $p$ gives $S^p{\mathbf{u}}_i={\lambda }_i^p{\mathbf{u}}_i$, including $p=0$. The absolutely convergent matrix exponential series may therefore be applied term by term to each eigenvector:ThusFor a repeated eigenvalue, every nonzero vector in its eigenspace remains an eigenvector. Because the real exponential function is injective, distinct real eigenvalues of $S$ do not merge under this transformation.

Choose the unnormalized eigenvectorsTheir linear combination giving the specified vector isApplying the reciprocal eigenvalues of the inverse matrix therefore yieldsIn Cartesian coordinates, this is $(-1/3,2/3{)}^T$, and direct matrix multiplication gives $S\mathbf{y}=(1,0{)}^T$.

No. Every one of the three vectors has equal first and second Cartesian coordinates, so their linear span is contained in the plane $x=y$. That linear subspace has dimension two and cannot equal ${\mathbb{R}}^3$.

More explicitly, call the three given vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ in their listed order. ThenThis nontrivial linear relation proves linear dependence, whereas a basis must be linearly independent. Being unit vectors alone does not ensure that they form a basis.

Commuting inverse. Assume $I-M$ is invertible, as required by the displayed inverse. Multiplication givesMultiply on both sides by $(I-M{)}^{-1}$ to obtainMore generally, a matrix commuting with an invertible matrix also commutes with its inverse.

Orthogonality. For a real skew-symmetric matrix $A$, $x^{T}Ax=0$. If $(I-A)x=0$ for a real vector $x$, multiply by $x^T$ to obtain $\Vert x{\Vert }^2=0$. Thus $I-A$ is invertible; the same argument applies to $I+A$. For the Cayley transform of a real skew-symmetric matrix,The factors commute by the first result, soTherefore $\boxed{N\text{ is an orthogonal matrix}.}$

Eigenvector mapping. If $Av=av$, then $(I-A{)}^{-1}v=(1-a{)}^{-1}v$. ConsequentlyThe denominator cannot vanish. In fact the eigenvalues of a real skew-symmetric matrix are purely imaginary: $v^{†}Av$ is the negative of its complex conjugate, so $a=-\overline{a}$. For $a=i\beta$ the corresponding eigenvalue of $N$ is $(1+i\beta )/(1-i\beta )=e^{2i\arctan \beta }$, consistent with orthogonality.

Spectrum in three dimensions. Let $Q$ be any real three-by-three orthogonal matrix. For a complex eigenvector $v$, $\Vert Qv\Vert =\Vert v\Vert$ implies $|\lambda |=1$. A real cubic characteristic polynomial has a real root, necessarily $+1$ or $-1$. Any nonreal roots form a complex-conjugate pair of unit modulus. If all three roots are real, two of the three signs agree and can be grouped as a pair with angle zero or $\pi$. Thus in every case the eigenvalues can be writtenTheir product is $\epsilon =\det Q$.

Determinant and fixed axis for the Cayley transform. Since $A^T=-A$,so $\boxed{\det N=1}$. This determinant argument works in every dimension. In dimension three the preceding spectrum result then guarantees an eigenvector with $Nx=x$. Set $z=(I-A{)}^{-1}x$. The fixed-vector equation says $(I+A)z=(I-A)z$, so $Az=0$ and $x=(I-A)z=z$. HenceThis is the axis fixed by the three-dimensional rotation. Equivalently, a real odd-dimensional skew-symmetric matrix has zero determinant, because $\det A=\det A^T=\det (-A)=-\det A$.

Nilpotence and eigenvalues. If $Av=\lambda v$ with $v\ne 0$, then $A^{n}v={\lambda }^{n}v$. Thus $A^n=0$ forces $\lambda =0$ for every eigenvalue. Conversely, if all $n$ complex eigenvalues, counted with algebraic multiplicity, vanish, the monic characteristic polynomial is ${\chi }_A(t)=t^n$. The Cayley-Hamilton theorem then gives $A^n=0$. HenceThis is the criterion for a nilpotent matrix.

Symmetric case. Here a symmetric matrix is understood to be real. The spectral theorem gives $A=QD{Q}^T$ with $Q$ orthogonal and $D$ real diagonal. The previous result makes every diagonal entry zero, so $A=0$. Alternatively, choose $2^r\ge n$, so $A^{2^r}=0$. For every real vector $x$,It follows that $A^{2^{r-1}}=0$. Repeat until $A=0$. This uses positivity of the quadratic form $x^T{M}^{T}Mx$. The restriction to real symmetric, or complex Hermitian matrices, matters: a complex matrix satisfying only $A^T=A$ need not have this positivity.

The parameters and powers. The first two columns of $T$ coincide, so $\det T=0$ for all $\alpha ,\beta$. The sum of its principal two-by-two determinants, each a minor of a matrix, isTogether with $\mathrm{tr}T=2+\beta$, this givesFor every eigenvalue to vanish both remaining coefficients must be zero. ThereforeDirect multiplication at these values givesIndeed the first row of $T^2$ multiplied by $T$ is $3{\mathrm{row}}_1(T)+3{\mathrm{row}}_2(T)-6{\mathrm{row}}_3(T)=0$, and the second row is its negative.

Geometry of repeated application. For any parameters the image of a linear map lies in the span of $a=(1,1,1{)}^T$ and $b=(1,\alpha ,\beta {)}^T$, since $Tx=(x_1+x_2)a+x_{3}b$. Thus it lies in a fixed plane through the origin, independently of the input. When $a,b$ are independent its normal is $a\times b=(\beta -\alpha ,1-\beta ,\alpha -1)$; if they coincide, the image is a line and is still contained in a plane.

For $\alpha =-5,\beta =-2$, put $s=x_1+x_2$ and $t=x_3$. ThenThe columns span this entire plane, so $T$ has rank two. The next two steps areThus the image of a linear map contracts from a plane to a line and then to the origin under successive applications. The line from $T^2$ is precisely the kernel of $T$.

For $a>0$, the positive constant-$u$ surfaces in elliptic cylindrical coordinates satisfyThey are confocal elliptic cylinders with focal lines $x=\pm a,y=0$. For a generic fixed $v$, eliminating $u$ givesBecause $u\ge 0$, a fixed $v$ selects the half-branch with the signs of $x$ and $y$ fixed by $\cos v$ and $\sin v$. These are hyperbolic cylinders. The degenerate $u=0$ surface is $y=0$, $|x|\le a$, extended in $z$. At $v=0,\pi$ the cross-sections are the two exterior horizontal rays; at $v=\pi /2,3\pi /2$ they are vertical half-axes.

Differentiate the position vector to get the coordinate tangent vectorsTheir pairwise dot products vanish. Their squared lengths are $a^2({\sinh }^{2}u+{\sin }^{2}v)$, $a^2({\sinh }^{2}u+{\sin }^{2}v)$ and one. This proves orthogonality and gives the scale factors of orthogonal coordinates:The Riemannian metric is $d{s}^2=h^2(d{u}^2+d{v}^2)+d{z}^2$. Away from the focal lines, divide the coordinate tangent vectors by their lengths to obtain the local orthonormal unit vectors $e_u,e_v,e_z$. For physical vector field components $A_u,A_v,A_z$, the formulas for vector calculus in elliptic cylindrical coordinates areThe gradient of a scalar function has components $({\varphi }_u/h,{\varphi }_v/h,{\varphi }_z)$. Apply the divergence formula to this gradient; Laplace's equation becomesThe formulas apply where $h>0$; the focal lines require a limit or another coordinate chart.

The orthogonality condition is $v_j{u}_j=0$, soEvery vector in the orthogonal complement is fixed by the orthogonal projection.

Under an orthogonal Cartesian change of axes $Q$, the vector components transform as $v_i^{\prime }=Q_{ia}{v}_a$, while $v_k^{\prime }{v}_k^{\prime }=v_a{v}_a$ and ${\delta }_{ij}=Q_{ia}{Q}_{ja}$. Substitution givesThis is the rank-two Cartesian tensor transformation law. The nonzero assumption ensures that the denominator is positive. In matrix form $P=I-v{v}^T/(v^{T}v)$ is the orthogonal projection matrix onto the orthogonal complement of $v$.

For the proposed vector, the first horizontal component of $Kq$ is $-k(\alpha +\beta )/2$. Since the corresponding component of $q$ is zero, a normal mode requires $\beta =-\alpha$. The vertical components then give $Kq=2kq$. With an arbitrary nonzero overall amplitude, choose $\alpha =1$ to obtainThe analogous pattern invariant under reflection across the vertical axis isDirect multiplication confirms $K{q}_x=2k{q}_x$. The two patterns give independent eigenvectors with the same eigenvalue, as expected from the square's rotation symmetry.

In a breathing mode, each displacement is proportional to its equilibrium position. A representative vector isMultiplying by the stiffness matrix gives $K{q}_b=4k{q}_b$. ThusPositive and negative amplitudes are uniform expansion and contraction about the center.

The first two vectors displace every particle identically in the horizontal or vertical direction. Every relative displacement is zero, so every spring has zero extension to first order. The third vector is $(y_i,-x_i)$ evaluated at the equilibrium points, which is an infinitesimal clockwise rotation. For any edge the relative displacement is perpendicular to that edge, again giving zero extension.

These are the rigid-motion zero modes of a spring framework: two translations and one infinitesimal rotation. Each has $Kq=0$, so $\boxed{\omega =0}$. In the linearized dynamics its amplitude has the form $A+Bt$, expressing free rigid motion rather than a restoring oscillation. Exact finite rigid rotations also preserve every spring length; the mode vector is their first-order tangent.

Use the ordered displacement vector $q=(x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4{)}^T$, and define $K$ by $V=\frac{1}{2}{q}^{T}Kq$. For an unstretched spring, the extension to first order is the dot product of the relative displacement with its equilibrium unit vector. Summing its squared extensions gives the central spring stiffness matrix. The four side springs and two diagonal springs giveThereforeThe kinetic energy is $\frac{m}{2}{\dot{q}}^T\dot{q}$, so a normal mode satisfies $Kq=m{\omega }^{2}q$. Keeping the factor $1/2$ in the definition of $V$ fixes the normalization of $K$ and the resulting frequencies.

The radial component of the curl in spherical coordinates isIt contains no $f_r$, so $R$ cannot contribute. Substituting the tangential components givesThe mixed partial derivatives of smooth $G$ agree, so its terms cancel. Multiplication by $r$, to take the dot product with the position vector $\mathbf{r}=r{\mathbf{e}}_r$, does not change this independence from $R,G$.

The remaining terms giveThis is the Laplace-Beltrami operator on the unit sphere. Taking the dot product with the unit radial vector instead would give $r^{-2}{\Delta }_{\theta ,\varphi }F$; the extra factor $r$ here comes from the position vector in the question.

Use the complex inner product $⟨a,b⟩={\sum }_j\overline{a_j}{b}_j=a^{†}b$, conjugate-linear in its first argument. For $a\ne 0$, put $c=⟨a,b⟩/⟨a,a⟩$. Positivity givesHence the Cauchy-Schwarz inequality is $|⟨a,b⟩|\le \Vert a\Vert \Vert b\Vert$. If $a=0$ it is immediate; equality holds precisely when the two vectors are linearly dependent.

For a unitary matrix, $U^{†}U=I$, so $\Vert Ua{\Vert }^2=a^{†}{U}^{†}Ua=\Vert a{\Vert }^2$. If $Uv=\lambda v$ with a nonzero eigenvector, this implies $|\lambda |\Vert v\Vert =\Vert v\Vert$. Thus every eigenvalue satisfiesFor the one-parameter family, expand unitarity to first order: $(I+t{A}^{†}+O(t^2))(I+tA+O(t^2))=I$, so $A^{†}+A=0$. Thus $A$ is a skew-Hermitian matrix, an element of the unitary Lie algebra. If $Av=\mu v$, then $v^{†}Av=\mu \Vert v{\Vert }^2$, whereas complex conjugation gives $(v^{†}Av{)}^{\ast }=v^{†}{A}^{†}v=-v^{†}Av$. Consequently $\overline{\mu }=-\mu$, proving that every eigenvalue of $A$ is purely imaginary, with zero permitted.

A two-index Cartesian tensor has components transforming as $T_{ij}^{\prime }=R_{ik}{R}_{jl}{T}_{kl}$ under an orthogonal change of orthonormal basis. The tensor order here counts indices, rather than the matrix rank. Its scalar, symmetric-traceless and axial tensor decomposition follows by settingThen $T_{ij}=Y{\delta }_{ij}+{\Omega }_{ij}+S_{ij}$, with $\Omega$ antisymmetric and $S$ symmetric and traceless. There are three independent antisymmetric components and five independent symmetric traceless components: three off-diagonal entries and two independent diagonal ones in the latter case.

For the symmetric tensor relative to a nonzero vector $A$, put $q=A_i{A}_i>0$, $s=A_i{T}_{ij}{A}_j$, and $t=T_{ii}$. The symmetric tensor decomposition relative to an axis givesIn particular $A_i{B}_i=s/q-s/q=0$. Define the orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/q$ onto the plane perpendicular to $A$, and setThis tensor is symmetric and kills $A$. Its trace is $t-s/q-2\alpha =0$. To verify the full decomposition, set $u=A/\sqrt{q}$ and split $T$ into its longitudinal component $(u^{T}Tu)u{u}^T$, two mixed components, and $PTP$. The mixed terms are exactly $B{A}^T+A{B}^T$, and $\alpha +\beta q=s/q$ gives the longitudinal coefficient. Thus all requested constraints and the reconstruction hold.