An $n\times n$ matrix is diagonalizable over a chosen field exactly when it has a basis of $n$ linearly independent eigenvectors. Equivalently, $P^{-1}AP$ is diagonal for some invertible $P$. Distinct eigenvalues suffice but are not necessary. For example,is not diagonalizable: its only eigenvalue is $1$ and its eigenspace is the one-dimensional span of $(1,0{)}^T$. It is a nontrivial Jordan block.

A Hermitian matrix satisfies $A^{†}=A$, where the dagger denotes conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenThe denominator is positive real, soFor $Av=\lambda v$ and $Aw=\mu w$, Hermiticity also givesThusThis is the necessary qualification to the PDF's orthogonality wording: arbitrary eigenvectors belonging to a repeated eigenvalue need not be orthogonal. Within each repeated eigenspace, Gram-Schmidt orthogonalization produces an orthonormal basis; together the eigenspaces give an orthonormal eigenbasis by the finite-dimensional spectral theorem. For example, $(1,0{)}^T$ and $(1,1{)}^T$ are nonorthogonal eigenvectors of the identity matrix.

For an integrable applied voltage, the convolution theorem givesThe integrated electric current is the zero-frequency transform, $Q=\tilde{I}(0)$. Since $\tilde{V}(0)$ can be arbitrary, zero total electric charge for every such input is equivalent toFor nonzero $p_1,p_2$, this can be writtenBecause both amplitudes are positive, nonzero $p_1,p_2$ must have opposite signs. If exactly one $p_i$ is zero there is no cancellation with positive amplitudes; if both vanish the kernel is identically zero and any amplitudes work. In particular, if both frequency parameters were additionally required to be positive, no such nontrivial system would exist. The PDF requires positivity of $a_i,{\gamma }_i$, but allows real fixed $p_i$.

This is zero-frequency cancellation of a causal response kernel. The phrase arbitrary voltage is understood within a class for which the net-charge integrals exist; a permanent DC input need not have finite total charge.

Since $\gamma >0$, the causal damped sine has an absolutely convergent Fourier transform for real $\omega$. Resolving the sine into complex exponentials givesCombining the two fractions yieldsThe formula also includes $p=0$, for which both the original function and its transform vanish. Its sign reverses when the real parameter $p$ reverses.

Suppose initially that $f,g$ are absolutely integrable, so Fubini's theorem justifies exchanging the integrations. The transform of their convolution isSet $s=z-t$, so $z=s+t$ and $dz=ds$. The exponential separates:Therefore the convolution theorem isNo extra $2\pi$ occurs in this direction with the specified normalization.

For the stated forward Fourier transform, the inverse has the opposite exponential sign and a factor $1/(2\pi )$:This holds under the usual hypotheses for Fourier inversion, or in the sense of distributions when appropriate.

The pulse height is the constant $t_0$, not the time variable. In the undamped case the convolution therefore reduces toIntegration gives the continuous displacement and velocityThe post-pulse expression is equivalently $x(t)=(t_0/2)\sin t_0\sin (2t-t_0)$. Since $t_0>0$, it vanishes for every subsequent time precisely when $\sin t_0=0$. ThusThese are integer multiples of the oscillator's natural period $\pi$. At such switch-off times both $x(t_0)$ and $x^{\prime }(t_0)$ are zero. Vanishing displacement alone would not suffice: a nonzero switch-off velocity would restart an oscillation. This is rectangular-pulse cancellation of an undamped oscillator.

A causal Green function vanishes before the impulse. For $t>\tau$ it solves the homogeneous linear ordinary differential equation, whose characteristic roots are $-q\pm 2i$. Thus it has the formContinuity at $t=\tau$ is necessary to avoid a delta derivative in $G_{tt}$, so $A=0$. Integrating its forced equation across the impulse gives the jump condition for an impulse, $G_t({\tau }^{+},\tau )-G_t({\tau }^{-},\tau )=1$. The derivative from below is zero and the derivative from above is $2B$, givingHere $\Theta$ is the Heaviside step function. The zero initial data exclude any additional homogeneous response. By the superposition principle, the output for a general locally integrable input isThe upper endpoint causes no unwanted term in the first derivative because $G(t,t)=0$; in the second derivative its unit derivative jump reproduces the input.

The nonstationary part of $u$ decays exponentially, whereas the integrating factor grows quadratically. For fixed $x>0$,ConsequentlyAt $x=0$, both expressions are identically zero; the ratio formulation is understood for $x>0$. The diffusion modes have died away, leaving the lifted linear profile multiplied by the reaction-induced amplitude factor.

The transformed endpoint values are $u(0,t)=0$ and $u(\pi ,t)=3$. Apply boundary lifting with $u=3x/\pi +w$. The lifted profile is linear, so $w_t=w_{xx}$ with homogeneous Dirichlet boundary conditions andFor separation of variables, put $w=X(x)T(t)$. Then $X^{\prime \prime }/X=T^{\prime }/T=-\lambda$. The endpoint conditions give nonzero spatial modes only for $\lambda =n^2$, $n=1,2,\dots$, with $X_n=\sin nx$ and $T_n=e^{-n^{2}t}$. Henceby orthogonality of the sine modes. Only the $n=2$ mode is present. Undoing both substitutions givesAt $t=0$ this has the prescribed initial profile, and at $x=\pi$ it is $3(t+1{)}^2$. Differentiating also verifies the original evolution equation directly.

For $v=(t+1{)}^{2}u$, the derivatives areSubstitution cancels the two terms $2(t+1)u$, giving the diffusion equationThis is a scalar integrating factor for a reaction-diffusion equation, since $\exp [\int 2dt/(t+1)]=(t+1{)}^2$ up to a constant. We use $t\ge 0$, so the factor never vanishes. A classical solution at the initial corners additionally requires compatible data, $h(0)=0$ and $h(\pi )=f(0)$.

Close the cone with the horizontal disk $D$ at height $h$, and let $V$ be the enclosed solid. The divergence isThe term $2y$ integrates to zero by reflection symmetry of each circular cross-section. Using the divergence theorem and cylindrical coordinates, the closed-surface flux isThe outward normal on $D$ is $+\hat{\mathbf{z}}$, so its flux is its area times $G_z=h$:Subtracting this cap contribution gives the required lateral flux:This is conical flux by closing with a cap. The normal convention matters: on the lateral cone the outward normal points radially outwards and partly downwards, so the $G_z$ contribution to the lateral flux is negative. To justify applying the theorem at the vertex, truncate a small tip and take its size to zero; the extra flux vanishes because $\mathbf{G}$ is regular there.

In cylindrical coordinates, the surface has radius $r=\sqrt{2}z$. It is the lateral surface of a right circular cone, with vertex at the origin and axis along positive $z$. Its height is $h$ and the limiting upper rim has radius $\sqrt{2}h$; the horizontal disk at that rim is absent. The half-angle from the vertical satisfiesThe upper rim is excluded by $z<h$, which does not affect any convergent surface integral. The vertex likewise contributes no area.

For a continuously differentiable vector field $\mathbf{G}$ on a bounded volume $V$ with piecewise smooth closed boundary $\partial V$, the divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. Thus total outward flux through a closed surface equals the integral of the local divergence inside it. A nonclosed surface first needs a suitable closure, with the added surface's flux accounted for separately.

Let $H$ be a proper subgroup. By Lagrange's theorem, $|H|$ divides $21=3\cdot 7$. Since $H\ne G$, its possibilities areThe subgroup of order one is generated by the identity and is cyclic. For either prime order $p=3$ or $7$, choose $h\in H$ other than the identity. Its order of a group element divides $p$ and is not one, so it is $p$. Thus its powers exhaust $H$, proving $H=⟨h⟩$.

ThereforeThis uses the fact that a group of prime order is cyclic. It does not require the order-21 group itself to be Abelian or cyclic.

A group homomorphism $\varphi :H\to G$ is a map preserving multiplication:It necessarily sends $e_H$ to $e_G$ and sends inverses to inverses, by applying this relation to $e_H^2$ and $h{h}^{-1}$. Its kernel of a group homomorphism isThe kernel is a normal subgroup of the domain: if $k\in \ker \varphi$ and $h\in H$, then $\varphi (hk{h}^{-1})=\varphi (h)e_G\varphi (h{)}^{-1}=e_G$. The kernel’s identity condition refers to the identity in the target group.

Write $H=\{e,h\}$ with $h\ne e$. For any $g\in G$, the normal subgroup property implies $gh{g}^{-1}\in H$. This conjugate cannot be $e$, since multiplication by $g^{-1}$ and $g$ would then give $h=e$. There is only one remaining choice, soThus $h$ commutes with every element of $G$; the identity does too. HenceThis proves that a normal subgroup of order two is central. The proof does not require $G$ to be finite: conjugation permutes $H$ and fixes its identity, so it must fix its only other element.

For every positive integer $m$, the group laws give $(g^{-1}{)}^m=(g^m{)}^{-1}$. ConsequentlyThe sets of positive exponents returning the two elements to the identity element are identical. If this set is nonempty, its least member is the order of a group element in both cases. If it is empty, neither element has finite order. Therefore, including the infinite-order case,For a direct divisibility argument in the finite case, if $g$ has order $n$, then $(g^{-1}{)}^n=e$, so its inverse’s order divides $n$. Reversing the roles shows $n$ divides that order, forcing equality. This is the general fact that an element and its inverse have the same order; it also reflects $⟨g⟩=⟨g^{-1}⟩$ as cyclic subgroups.

The four displayed matrices fail closure: for example ${\sigma }_x{\sigma }_y=i{\sigma }_z$ is not one of them. Thus they do not form a group by themselves.

Let ${\sigma }_0=I$ and ${\mu }_4=\{1,-1,i,-i\}$. The enlarged set is the single-qubit Pauli groupThere are 16 distinct matrices: $I,{\sigma }_x,{\sigma }_y,{\sigma }_z$ are linearly independent, so two different ones cannot be scalar multiples, and the four phases are distinct for each nonzero matrix.

The Pauli matrices satisfyHere $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. Thus any product of two Pauli matrices is another Pauli matrix or the identity times a phase in ${\mu }_4$. Multiplying by the two original phases preserves membership in ${\mu }_4$, proving closure of all 16 matrices.

Associativity is inherited from matrix multiplication, the identity is $I$, andThis proves every group axiom. The resulting group is non-Abelian, since reversing ${\sigma }_x{\sigma }_y$ changes its sign.

The quotient group $G/N$ consists of the cosets $gN$, with multiplicationFor well-definedness, replacing representatives by $g{n}_1$ and $h{n}_2$ changes their product to $gh(h^{-1}{n}_{1}h)n_2$. Since $N$ is a normal subgroup, the final two factors lie in $N$, so the product coset does not change.

Associativity follows from that in $G$; the identity element is the coset $N$, and $(gN{)}^{-1}=g^{-1}N$. Hence these cosets form a group. By Lagrange's theorem, every subgroup $H\le G$ has order dividing $|G|$, and its subgroup index is $[G:H]=|G|/|H|$. In particularNormality is needed for the quotient multiplication, whereas Lagrange’s counting formula holds for every subgroup.

Keep the constraint from part (b). For equal masses and spring constants, the generalized eigenvalue problem for small oscillations reduces toThe horizontal coordinate decouples, givingIn this normal mode, mass 1 oscillates horizontally, mass 2 has its horizontal coordinate fixed, and neither mass has a vertical displacement.

For the vertical block, write $\lambda =m{\omega }^2/k$. Its characteristic polynomial is $(2-\lambda )(1-\lambda )-1={\lambda }^2-3\lambda +1$. Thus ${\lambda }_{\pm }=(3\pm \sqrt{5})/2$, and the first vertical equation gives $y_2/y_1=2-{\lambda }_{\pm }$. If $\phi =(1+\sqrt{5})/2$ is the golden ratio, the two vertical modes areIn the lower-frequency mode, the masses move vertically in phase, with the lower mass having the larger amplitude. In the higher-frequency mode they move oppositely, with lower-to-upper amplitude ratio $1/\phi$. Their Euclidean inner product is zero because $1+\phi (-1/\phi )=0$; since $M=mI$, this is also mass-matrix orthogonality of normal modes.

The three frequencies satisfy ${\omega }_{-}<{\omega }_x<{\omega }_{+}$. Each mode has a scalar coordinate $A\cos \omega t+B\sin \omega t$, and their superposition gives the general constrained motion. No zero-frequency mode remains because the first spring anchors the system. Gravity fixes the equilibrium extensions but drops out of these frequencies for zero-natural-length springs.