A matrix is a diagonalizable matrix if there is an invertible matrix $P$ for which $P^{-1}MP$ is diagonal. Its columns then form a basis of eigenvectors, and the diagonal entries are their corresponding eigenvalues.

Here the given eigenvectors form an orthonormal basis. Put them into the columns of $Q$:Define $\Lambda =Q^{T}MQ$. Its entries areThis proves the claimed diagonalization of a matrix, rather than merely guessing it:The trace follows directly from this factorization:because each column of $Q$ has unit length. Trace is the sum of the eigenvalues, with multiplicity.

For the particular matrix, $(M^2{)}_{ij}={\sum }_k{M}_{ik}{M}_{kj}$. If $i=j$, every $k\ne i$ contributes one and $k=i$ contributes zero. If $i\ne j$, precisely $k=i$ and $k=j$ contribute zero; the other $n-2$ terms contribute one. ThusApplying this identity to a nonzero eigenvector givesEvery eigenvalue is therefore $n-1$ or $-1$.

If $r$ is the multiplicity of $n-1$, the trace formula for a diagonalizable matrix and the zero diagonal giveso $r=1$. Up to ordering, the required diagonal form isFor $n=1$ this means just $(0)$. For $n\ge 2$, the eigendirections are also transparent: the all-ones vector has eigenvalue $n-1$, and the $(n-1)$-dimensional subspace whose coordinate sum is zero has eigenvalue $-1$, since $(M\mathbf{x}{)}_i={\sum }_j{x}_j-x_i$. The spectrum consists of one collective mode and $n-1$ zero-sum modes.

For the reality of the eigenvalues, one may initially allow a nonzero complex eigenvector $\mathbf{v}$ with $M\mathbf{v}=\lambda \mathbf{v}$. Since $M$ is real symmetric, it is a Hermitian matrix, andis real: the numerator equals its own complex conjugate and the denominator is positive. Once $\lambda$ is real, a nonzero real or imaginary part of $\mathbf{v}$ supplies a real eigenvector. We work with the real unit eigenvectors of the problem.

For two such eigenvectors, symmetry givesDistinct eigenvalues therefore give zero dot product. Together with the prescribed unit lengths,These $n$ vectors form an orthonormal basis of ${\mathbb{R}}^n$.

Expand a real unit vector as $\mathbf{x}={\sum }_a{c}_a{\mathbf{e}}_a$. Then ${\sum }_a{c}_a^2=1$, and its Rayleigh quotient isMoreover,Every coefficient on the right is strictly positive, so equality forces $c_a=0$ for $a\ge 2$. Equality occurs precisely along the top eigendirection:

To obtain a unit vector in $S$ using just ${\mathbf{e}}_1,{\mathbf{e}}_2$, let $u=({\mathbf{e}}_1{)}_1$ and $v=({\mathbf{e}}_2{)}_1$ denote their first coordinates. If $u^2+v^2>0$, takeThen ${\alpha }_{1}u+{\alpha }_{2}v=0$ and ${\alpha }_1^2+{\alpha }_2^2=1$. Hence ${\alpha }_1{\mathbf{e}}_1+{\alpha }_2{\mathbf{e}}_2$ has first coordinate zero and unit length. If $u=v=0$, simply take ${\alpha }_1=1,{\alpha }_2=0$. In either case the Rayleigh quotient of the constructed vector isThe continuous quadratic form attains its maximum on the compact space $S$, so

For $\mathbf{x}=(0,\mathbf{y}{)}^T$, deletion of the first row and column gives ${\mathbf{x}}^{T}M\mathbf{x}={\mathbf{y}}^{T}A\mathbf{y}$, while $|\mathbf{x}|=|\mathbf{y}|$. The spectral theorem for real symmetric matrices applied to $A$ identifies its largest eigenvalue with this maximum:Combining the upper bound for every unit vector with the constructed lower bound provesThis is largest-eigenvalue interlacing for a principal submatrix: restricting a quadratic form to a coordinate hyperplane cannot exceed its old maximum, but leaves some direction in the span of the top two eigenvectors.

Every counted point has integer coordinates and lies in the square $[-n,n{]}^2$, while the origin is always counted. HenceNo precise circle-area estimate is needed; this elementary bound controls the growth of the power series coefficients.

The comparison test for series states that if $0\le u_n\le v_n$, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of the smaller nonnegative series forces divergence of the larger one.

For $|z|=r<1$,For $0<r<1$, the ratio of successive terms on the right tends to $r<1$, so the ratio test and the comparison test for series show absolute convergence. The case $r=0$ is immediate.

If $|z|\ge 1$, then $|a_n{z}^n|\ge 1$, so the terms do not tend to zero. The term test for divergence therefore rules out convergence, including every point on the boundary $|z|=1$. Thus the radius of convergence isThe series converges exactly in the open unit disc. This is an example of polynomial coefficient growth with unit power-series radius: the lattice-count bound is polynomial rather than exponential.

A sequence $(x_n)$ is a convergent sequence with limit $x$ ifThe index $N$ may depend on $\epsilon$, but must work for every later term.

Fix $\epsilon >0$. Choose $N$ such that $|x_i-x|<\epsilon /2$ for $i\ge N$, and putThis is a fixed finite number. For $n\ge N$, the triangle inequality givesChoose $n$ sufficiently large that $C/n<\epsilon /2$ as well. The error is then less than $\epsilon$, provingTaking successive arithmetic averages preserves the limit. This is the Cesaro theorem for convergent sequences: any troublesome initial terms contribute only a fixed numerator divided by $n$.

A counterclockwise planar rotation matrix has the formTo recover this form from preservation of the Euclidean norm, let $\mathbf{u},\mathbf{v}$ be the columns of $R$. Applying the norm condition to the two coordinate unit vectors gives $|\mathbf{u}|=|\mathbf{v}|=1$. Applying it to their sum givesso $\mathbf{u}\cdot \mathbf{v}=0$. The columns are an orthonormal basis, and $R^{T}R=I$: $R$ is an orthogonal matrix.

Write $\mathbf{u}=(\cos \theta ,\sin \theta {)}^T$. There are two perpendicular unit vectors available for $\mathbf{v}$, namely $\pm (-\sin \theta ,\cos \theta {)}^T$. The first gives determinant $1$, the second $-1$. The positive determinant hypothesis selects the first. A norm-preserving planar map with positive determinant is a rotation.

For the commutation condition, writeDirect matrix multiplication givesEquality forces $c=-b$ and $d=a$, so, with $u=a$ and $v=-b$,If $u=v=0$, take $\lambda =0$ and $R=I$. Otherwise set $\lambda =\sqrt{u^2+v^2}>0$ and choose $\cos \theta =u/\lambda$, $\sin \theta =v/\lambda$. ThenThis describes the centralizer of a planar quarter-turn. Its nonzero elements act as a rotation followed by a uniform scaling, just as multiplication by a nonzero complex number does.

The stated bound is Lipschitz continuity with constant $1$. For every $x$ and every $\epsilon >0$, taking $\delta =\epsilon$ ensures $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$. The same $\delta$ works at every point, so $f$ is both continuous and uniformly continuous.

For an integer $N\ge 1$, partition $[0,1]$ into $N$ equal intervals and define a step functionAt the remaining endpoint set $g_N(1)=f((N-1)/N)$, so the final interval may be taken closed on the right. The Lipschitz bound givesChoosing $N\ge 1/\epsilon$ proves the requested uniform step approximation on a compact interval:A sufficiently fine partition gives a piecewise constant approximation uniformly over the whole interval.

The continuous function $f$ and every finite step function $g_N$ are Riemann integrable. Put $c_j=f(j/N)$. Integrating each constant piece explicitly givesSince the sine differences have absolute value at most $2$, the triangle inequality and uniform approximation giveTo prove convergence, fix $\epsilon >0$ and first choose $N$ with $1/N<\epsilon /2$. With this $N$ held fixed, the finite number ${\sum }_j|c_j|$ is independent of $n$, so the second term is also less than $\epsilon /2$ for sufficiently large $n$. ThereforeFirst make the approximation error small, then let the oscillation frequency grow. This is the step-function proof of the Riemann-Lebesgue lemma; it does not require assuming that a Lipschitz function has an everywhere-defined derivative.

The Rolle theorem states that if a real function is continuous on $[u,v]$, differentiable on $(u,v)$, and has equal endpoint values, then its derivative vanishes somewhere in $(u,v)$. Order the $k$ distinct real roots as $r_1<\cdots <r_k$. The theorem gives a root of $f^{\prime }$ in each interval $(r_j,r_{j+1})$. These intervals are disjoint, so the obtained roots are distinct:

We prove the requested bound for positive roots of a polynomial by mathematical induction on the number of nonzero terms. In fact, the proof also works with roots counted according to their multiplicity. Write $V(p)$ for the number of sign changes and $N_{+}(p)$ for the positive-root count.

A one-term polynomial $a_1{x}^{d_1}$ has no positive roots and no sign changes. For the induction step, factor out the smallest power of $x$:For $x>0$ this factor is nonzero, so $N_{+}(q)=N_{+}(p)$, including multiplicities, and $V(q)=V(p)$. Differentiation removes exactly the constant term:The exponents remain strictly increasing, and the positive multipliers $e_j$ do not change coefficient signs. The induction hypothesis therefore gives

First suppose $a_1{a}_2<0$. Removing $a_1$ removes exactly one sign change, so $V(q)=V(q^{\prime })+1$. If $q$ has positive roots $r_1<\cdots <r_k$ of multiplicities $m_1,\dots ,m_k$, its derivative has multiplicity $m_j-1$ at $r_j$ and, by the Rolle theorem, an additional root between each pair. Thus the Rolle root count with multiplicities giveswhen there is a positive root. ConsequentlyIf there is no positive root, the desired bound is automatic.

Now suppose $a_1{a}_2>0$, so $V(q)=V(q^{\prime })$. Again the zero-root case is immediate. Otherwise, multiplying $q$ by $-1$ if necessary lets us assume $a_1>0$ and $a_2>0$ without changing roots or sign changes. For sufficiently small $x>0$,since $q(0)=a_1$ and the lowest-power term of $q^{\prime }$ has positive coefficient. Choose such a point $u$ before the first positive root $r_1$. The mean value theorem on $[u,r_1]$ gives a point $v\in (u,r_1)$ withBut $q^{\prime }(u)>0$. The intermediate value theorem applied to the polynomial $q^{\prime }$ therefore gives an extra root in $(u,v)$, before $r_1$. It is distinct from the roots at $r_j$ and those between consecutive roots. HenceBoth sign cases complete the induction:An initial sign change allows one extra positive root; without that sign change, the derivative must already turn before the first positive root. This is the upper-bound part of the Descartes rule of signs. The multiplicity claim used above follows by differentiating $q(x)=(x-r{)}^{m}h(x)$ with $h(r)\ne 0$: the derivative has a factor $(x-r{)}^{m-1}$ and its remaining factor is nonzero at $r$.

The effective number of relativistic energy degrees of freedom count two photon polarizations, four electron/positron spin states, and six neutrino/antineutrino states, with the Fermi-Dirac distribution factor $7/8$ for fermions. Treating these species as relativistic and at a common temperature gives $\boxed{g_{\ast }=2+(7/8)(4+6)=10.75}$.

The mass-equivalent radiation mass density is $\rho =({\pi }^2/30)g_{\ast }(k_{B}T{)}^4/({\hslash }^3{c}^5)$. Therefore the Friedmann equation givesAt $T_0=10^{10}\mathrm{K}$, $H_0\simeq 0.503{\mathrm{s}}^{-1}$. Equating the specified weak interaction rate to $H_0(T/T_0{)}^2$ gives $(T_{\ast }/T_0{)}^3=H_0/(1{\mathrm{s}}^{-1})$ and henceThe approximate equilibrium neutron-to-proton ratio, neglecting chemical potentials, is $r_{\ast }\simeq \exp [-(m_n-m_p)c^2/(k_B{T}_{\ast })]$, about $0.152$ using the mass difference $1.293\mathrm{MeV}$. This is a freeze-out estimate rather than a detailed reaction-network computation.

After weak cosmological weak freeze-out, free neutrons undergo beta-minus decay; neutrinos decouple, and electron-positron annihilation heats the photons relative to the neutrinos. Initially the abundant energetic photons destroy deuterium almost as soon as it forms. Once the deuterium bottleneck ends, nuclear reactions rapidly pass through deuterium, tritium and helium-three to helium-4. Reactions eventually become too slow compared with cosmic expansion; nearly all surviving neutrons are bound in helium-four. Its mass fraction is approximately $Y_4=2r/(1+r)$, where $r$ is evaluated when nuclear assembly begins.

Increasing $g_{\ast }$ increases the Hubble parameter, raises $T_{\ast }$, increases the freeze-out neutron-to-proton ratio, and leaves less time for free-neutron decay after freeze-out. Both effects increase $Y_4$; decreasing $g_{\ast }$ reverses them. Most deuterium is only an intermediate product and is burned into helium, so its residual abundance and its absolute change are tiny compared with the helium mass fraction. This does not imply a negligible fractional change in residual deuterium. Faster expansion also shortens the time available to burn it, and its relative abundance can be appreciably sensitive to extra radiation. The distinction matters when interpreting the final wording of the question; joint deuterium and helium abundances constrain extra relativistic species, as discussed in the Particle Data Group nucleosynthesis review.