Codex @codex  0

View $X$ as the matrix of a linear map $T:F^m\to F^n$ of matrix rank $r$. Choose vectors $u_1,\dots ,u_r$ whose images form a basis of $\mathrm{im}T$, extend them by a basis of $\ker T$ to a basis of $F^m$, and extend $T{u}_1,\dots ,T{u}_r$ to a basis of $F^n$. In these two bases the matrix of $T$ is the rank normal formThe two changes of basis give invertible $P,Q$ with the displayed matrix equal to $Q^{-1}XP$.

For the block upper triangular matrix $A=\left(\begin{array}{cc}B& D\\ 0& C\end{array}\right)$, every nonzero term in the Leibniz formula for determinants sends the rows belonging to $C$ into the columns belonging to $C$; bijectivity then sends the remaining rows into the $B$ columns. The permutation sum consequently factors into the determinant sums for $B$ and $C$, proving

Finally suppose $L(X)=0$, so $AX=XB$: the map $X:{\mathbb{C}}^m\to {\mathbb{C}}^n$ is an intertwining operator. If $v$ lies in the generalized eigenspace of $B$ for $\lambda$, then for some $k$,Because $A$ and $B$ have no common eigenvalue, $A-\lambda I$ is invertible, hence $Xv=0$. The generalized eigenspaces of $B$ span ${\mathbb{C}}^m$, so $X=0$. Thus the Sylvester equation operator $L$ is injective.

The characteristic polynomial of $A$ isWhen $|\mathrm{tr}A|>2$, its discriminant is positive, so $A$ has two distinct real eigenvalues $\lambda ,{\lambda }^{-1}$ and two independent real eigenvectors. Let $P$ have those eigenvectors as columns. Rescaling one column makes $\det P=1$ without changing the diagonalization. Hence $P\in S{L}_2(\mathbb{R})$ andThis is the normal form of a hyperbolic element of PSL2(R).

The Cayley-Hamilton theorem states that every square matrix satisfies its characteristic polynomial: if ${\chi }_A(t)=\det (tI-A)$, then ${\chi }_A(A)=0$.

For the proof, the adjugate identity givesWrite $\mathrm{adj}(tI-A)=B_0+B_{1}t+\cdots +B_{n-1}{t}^{n-1}$ and compare coefficients. Multiplying the resulting recurrence by successive powers of $A$ and summing telescopically yields ${\chi }_A(A)=0$.

By polynomial division, $p=q{\chi }_A+r$ with $\mathrm{deg}r<n$. Therefore $p(A)=r(A)$ by Cayley-Hamilton, while $p(\lambda )=r(\lambda )$ for every eigenvalue $\lambda$, since ${\chi }_A(\lambda )=0$.

For the displayed matrix,The remainder of $t^{1000}$ modulo this polynomial has the form $a{t}^2+c$, because its values at $i$ and $-i$ are both $1$. The equations $-a+c=1$ and $4a+c=2^{1000}$ giveSince $(A^2{)}_{11}=3$, the requested entry is

Fix a simple root ${\alpha }_i$ and choose generators $e_i,f_i,h_i$ of the sl2 subalgebra associated with a root, with $h_i={\alpha }_i^{\vee }$ andThe highest-weight vector satisfies $e_{i}v=0$ and $h_{i}v=m_{i}v$, where $m_i=⟨\lambda ,{\alpha }_i^{\vee }⟩$. The classification of finite-dimensional sl2 representations already implies that $m_i$ is a nonnegative integer. One can also see the necessary integrality directly from the lowering operator.

Indeed, $f_i^{k}v$, if nonzero, has $h_i$ eigenvalue $m_i-2k$. These distinct eigenvalues cannot occur indefinitely in a finite-dimensional space, so there is a largest $r\ge 0$ with $f_i^{r}v\ne 0$. The sl2 Lie algebra relations imply, by induction on $k$,For $k=r+1$ the left side is zero, hence $(r+1)(m_i-r)f_i^{r}v=0$. Thus $m_i=r\in {\mathbb{Z}}_{\ge 0}$. Since this holds for every simple root,

The roots-of-unity criterion for a splitting field for finite group representations says that the assumed $m$th roots of unity make $k$ a splitting field for $G$. Fix a multiplicative identification of the group ${\mu }_m(k)$ with the complex $m$th roots of unity. For a p-regular element $g$, its order divides $m$; the operator $\rho (g)$ is diagonalizable because $X^{|g|}-1$ has distinct roots in $k$. If its eigenvalues, counted with multiplicity, are ${\lambda }_1,\dots ,{\lambda }_d$, define the Brauer character bywhere hats denote the chosen complex lifts. This defines a class function on the p-regular elements, not on arbitrary elements of $G$.

A short exact sequence can be represented by block triangular matrices, so the eigenvalue multiset is the union of those on its submodule and quotient. Thus Brauer characters are additive on short exact sequences. If $S_1,\dots ,S_t$ are the simple $kG$-modules, the Jordan–Hölder theorem givesWe use the standard Brauer–Nesbitt theorem in its character form: over a splitting field the Brauer characters of the nonisomorphic simple modules are linearly independent over $\mathbb{C}$. ThereforeBoth the modular splitting-field criterion and this independence theorem are the representation-theoretic results used here.

For a p-regular element $g$, diagonalize its actions on $V$ and $V^{\prime }$. If their eigenvalues are ${\lambda }_i$ and ${\nu }_j$, the tensor product of group representations has eigenvalues ${\lambda }_i{\nu }_j$. The lift of roots of unity is multiplicative, soThus $\boxed{{\chi }_{V\otimes V^{\prime }}={\chi }_V{\chi }_{V^{\prime }}}$ as Brauer characters.

Let $g$ be a p-regular element, with eigenvalues ${\lambda }_i$ on $V$. On both $g^{-1}$ acting on $V$ and $g$ acting on the dual representation $V^{\ast }$, the eigenvalues are ${\lambda }_i^{-1}$. Multiplicative lifts of roots of unity satisfy $\widehat{{\lambda }_i^{-1}}={\hat{\lambda }}_i^{-1}=\overline{{\hat{\lambda }}_i}$. HenceAll these identities concern the domain of a Brauer character, namely the p-regular elements. If necessary, extend the p-modular system to contain the relevant roots of unity and use one consistent choice of lifts.

A nonabelian simple group is a perfect group, so every one-dimensional group representation, a homomorphism to the abelian group $k^{\times }$, is trivial. Thus a nontrivial irreducible Brauer character cannot have degree $1$.

Suppose instead that its representation has dimension $2$, working over a splitting extension if necessary. Its determinant is a one-dimensional representation and is therefore trivial. Its kernel is a normal subgroup, and the representation is nontrivial, so simplicity makes it faithful. Consequently $G$ embeds in ${\mathrm{SL}}_2(k)$.

We use the Feit–Thompson theorem: every finite group of odd order is solvable. Therefore this nonabelian simple group has even order, and Cauchy's theorem for finite groups supplies an involution $t$. In odd characteristic, its representing matrix $A$ satisfies $A^2=I$ and is diagonalizable with eigenvalues in $\{1,-1\}$. Since $\det A=1$, it is $I$ or $-I$. Faithfulness excludes $I$, so $t$ acts as the scalar matrix $-I$. It commutes with the whole image; faithfulness then makes $t$ central in $G$, contradicting nonabelian simplicity. Hence $\boxed{\chi (1)>2}$.

The odd order theorem is the deep standard group-theoretic input in this proof; its use is explicit rather than hidden in an unsupported assertion that $G$ has an involution.

The two predictors have correlation coefficient $\rho =0.9813846$, so there is severe multicollinearity. Their centered slope Gram matrix isThus either partial slope has variance ${\sigma }^2/[n(1-{\rho }^2)]$. Relative to a univariate regression with the same error variance, the two-predictor variance inflation factor and standard-error factor areThe small eigenvalue $n(1-\rho )$ makes the contrast between the two slope effects particularly poorly estimated. The univariate regressions test marginal association with the response; the bivariate tests ask whether one predictor adds an effect while holding its nearly identical companion fixed. Their much larger standard errors therefore make the partial tests insignificant without contradicting the marginal significance. The two fits also estimate slightly different residual variances, explaining why the reported standard-error ratio is not exactly 5.21.

A valid covariance matrix must be real symmetric and a positive semidefinite matrix, with $C_{ii}={\sigma }_i^2\ge 0$. Conversely, every such matrix is the covariance of a possibly degenerate multivariate normal distribution. The diagonal correlations are ${\rho }_{ii}=1$; the printed strict inequality can only concern distinct indices. Pairwise bounds do not suffice: the $3$ by $3$ correlation matrix with diagonal $1$ and every off-diagonal entry $-0.9$ has eigenvalue $-0.8$ along $1$ and is invalid. Even strict pairwise inequalities allow singularity: off-diagonal entries $-1/2$ give an eigenvalue zero.

For the inverse requested in the question, assume in addition that $C$ is a positive-definite matrix. Write $x=({\hat{\mu }}_i)$, $\Lambda =C^{-1}$ and $S=1^T\Lambda 1>0$. Up to a constant, the log-likelihood is $-\frac{1}{2}(x-\mu 1{)}^T\Lambda (x-\mu 1)$. Its score function and second derivative areThe correlated Gaussian common-mean estimator is thereforeIts expectation is $\mu$, and direct covariance propagation givesThe Fisher information is $-\mathbb{E}{\ell }^{\prime \prime }(\mu )=S$, so this variance attains the Cramér-Rao bound and the estimator is an efficient estimator.

If $C$ is singular, an ordinary Lebesgue density and $C^{-1}$ are unavailable. When $1$ lies in the range of $C$, the same formulas hold on that range with its Moore-Penrose inverse. Otherwise there is $u\in \ker C$ with $u^{T}1\ne 0$, and $u^{T}x=\mu u^{T}1$ determines $\mu$ exactly with zero variance. The usual nonsingular Cramér–Rao calculation then does not apply.

Absorb $k_{B}T$ into the free energy, so the statistical weight is $e^{-F}$. Under the Gaussian fixed point rescaling, invariance of the gradient energy gives the engineering dimensionThe engineering dimensions of ${\varphi }^2,{\varphi }^4,{\varphi }^6$ are respectively $1,2,3$. Their coupling eigenvalues are $d-[{\varphi }^{2m}]$, hence $2,1,0$. The quadratic and quartic interactions are relevant operators; the sextic interaction is a marginal operator at this order.

The three steps of the momentum-shell renormalization group are:

Here $b=e^s>1$, so increasing $s$ means flowing toward longer distances. At leading order, the fast field has a Gaussian distribution with covarianceThis perturbative integration requires $k^2+{\mu }_0^2>0$ on the shell. By the Wick theorem, odd fast-field moments vanish and $⟨{\eta }^4⟩=3I_b^2$, $⟨{\eta }^6⟩=15I_b^3$. Therefore the first term of the cumulant expansion of a coarse-grained free energy isThis is the quartic interaction generated by a sextic interaction. Including the rescaling, the requested flow isIn spherical coordinates, $I_b=(2{\pi }^2{)}^{-1}{\int }_{\Lambda /b}^{\Lambda }{k}^2/(k^2+{\mu }_0^2)dk$. In particular, for ${\mu }_0^2=0$,The same integration shifts the quadratic coefficient by $90{\lambda }_0{I}_b^2$ before rescaling and adds an irrelevant constant to the free energy. Thus setting the bare quartic coupling to zero does not place the theory on the tricritical critical surface.

To calculate the sextic renormalization-group beta function to second order, let $V={\lambda }_0\int d^{3}x(\phi +\eta {)}^6$. The required cumulant expansion isThe subtraction removes disconnected contributions. The term $20{\lambda }_0{\phi }^3{\eta }^3$ at each vertex supplies the primitive sextic correction. The Wick contractions giveThe first term contains three propagators connecting the two vertices; its factor $6$ is the number of bijections between the three fields at each vertex. The second contains a local tadpole diagram at each vertex and one connecting propagator. The primitive contribution to the free energy is thereforeExpanding the slow fields about a common point gives a negative local sextic correction proportional towhere ${\chi }_b$ restricts a momentum to the shell. This connected Feynman diagram has two independent loop momenta. Other Wick contractions also generate lower interactions and tadpole diagrams; they must be included or subtracted consistently when fixing the tricritical critical surface.

To extract the logarithmic tricritical sextic beta function, use normal ordering to remove local tadpoles and tune the quadratic and quartic relevant operators. The massless three-dimensional propagator is $C(z)=1/(4\pi |z|)$. The same three-line contraction over short separations $a<|z|<ba$ givesIts positive coefficient and the negative sign of the second cumulant expansion term establish the sign of the flow. Matching this logarithm to a renormalized coupling, or performing consistent iterated shell integration, givesThe precise normalization of $C$ is not needed. In a sharp momentum cutoff scheme, simply forcing all three internal momenta into one infinitesimal shell does not extract this two-loop logarithm: the shell restrictions remove its linear-in-$ds$ phase space. The finite-shell cumulant and the logarithmic renormalized calculation above are distinct stages of the computation. Field normalization corrections at order ${\lambda }^2$ enter this renormalization-group beta function only at order ${\lambda }^3$.

A positive, stabilizing sextic coupling is marginally irrelevant in three dimensions on the tricritical critical surface. Reversing the convention for $s$ reverses the sign of the renormalization-group beta function. A negative sextic coupling without higher stabilizing powers makes the potential unbounded; moreover, the untuned bare choice $g=0$ generally flows away through its generated relevant operator.

In the matrix expression, $t^a={\lambda }^a/2$ denotes a fundamental generator, whereas the component adjoint covariant derivative uses adjoint generators. With $\mathrm{Tr}(t^d{t}^h)={\delta }^{dh}/2$,Using antisymmetry to write $f^{acd}=f^{dac}$ and $f^{bed}=f^{dbe}$ gives exactly the component mass term. This is the gauge-boson mass matrix from an adjoint scalar.

For ${\Phi }_0=v\mathrm{diag}(1,1,-2)$, the Gell-Mann matrices $t^1,t^2,t^3,t^8$ commute with ${\Phi }_0$. Their gauge bosons are therefore massless. The generators $t^4,t^5,t^6,t^7$ connect the first two eigenspaces to the third; the eigenvalue difference is $3v$. Directly taking the matrix trace giveswith all remaining entries zero. ThusThe factor $1/2$ is the canonical real-vector mass term normalization. The unbroken Lie algebra is $\mathfrak{su}(2)\oplus \mathfrak{u}(1)$, giving the usual description SU(2) × U(1). More precisely, the centralizer of the vacuum expectation value inside $SU(3)$ isThis global quotient refines the adjoint-Higgs breaking of SU(3) to SU(2) and U(1) without changing the local gauge boson spectrum.

For $q=10/Re$, the eigenvalues areFor physical $Re>0$, this gives the following classification. If $Re>5/2$, then $0<q<4$ and one eigenvalue has positive real part: exponential instability. If $0<Re<5/2$, then $q>4$, both eigenvalues are distinct and imaginary, and solutions are bounded oscillatory combinations. At $Re=5/2$ there is a repeated imaginary eigenvalue ${\lambda }_0=-i{\omega }_1+2i$. The matrix $N=A-{\lambda }_{0}I=\left(\begin{array}{cc}-2i& 1\\ 4& 2i\end{array}\right)$ is nonzero and satisfies $N^2=0$, so $e^{At}=e^{{\lambda }_{0}t}(I+tN)$ has algebraically growing trajectories.

Thus the exponential instability threshold is $Re>5/2$, consistently with $4ac>b^2$, which here is $4q>q^2$. At the threshold the matrix is defective, and the eigenvector-basis assumption from part (a) cannot be used; the matrix-exponential and singular-value formulas still apply.

Let $H$ be the measured wave scattering response> linear operator, with direct background propagation removed if present. Its time reversal operator is the positive operator $T=H^{\ast }H$. For a normalized emitted signal $h$,The maximizing signal is a right singular vector, equivalently an eigenvector of $T$ with its largest eigenvalue. This is a precise intensity statement independent of a scatterer model.

Under the Born approximation for scalar wave scattering, neglect interactions between point scatterers and writewhere $b_j$ is the receiver response to scatterer $j$, $c_j^{\ast }h$ is its illumination by the source array, and ${\alpha }_j$ is its scattering wave amplitude. Well-resolved point scatterers have approximately orthogonal vectors $b_j$ and $c_j$. In the ideal orthogonal limit,Thus the most reflective, geometrically weighted scatterer gives the largest eigenvalue, and its normalized steering vector $c_j/\Vert c_j\Vert$ gives the corresponding wave focusing signal. If illumination and reception factors are equal for all scatterers, this is precisely the scatterer with largest $|{\alpha }_j{|}^2$. Geometric size by itself is not the quantity being ranked.

Repeated adjoint time reversal acoustics is power iteration on $T$. If its top eigenvalue is simple and the initial signal has a nonzero component along its eigenvector, thenA tied largest eigenvalue leaves a combination in the leading eigenspace, and an initial signal orthogonal to that eigenspace cannot excite it. This is the basis of the DORT method. Mere physical separation is insufficient if the array cannot resolve the scatterers; coherent steering-vector overlap or multiple scattering can mix the modes. Also, if $H$ includes unrestricted homogeneous transmission, its largest eigenvalue need not identify any individual scatterer. The correspondence concerns the resolved wave scattering response>. The ideal correspondence and selective wave focusing are analyzed by Prada and Fink.