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For $f(z)=(az+b)/(cz+d)$, finite fixed points obeywith the point at infinity included in the usual way when appropriate. The fundamental theorem of algebra on the Riemann sphere gives at least one fixed point. Unless $f$ is the identity, the equation is nonzero of degree at most two, so there are one or two distinct fixed points.

Let $\zeta$ be a primitive $m$th root of unity. For every $u\in \mathbb{C}$,has order $m$, and these transformations are distinct as $u$ varies.

Projectivising matrices is a homomorphism. Thus $B=CA{C}^{-1}$ in $S{L}_2(\mathbb{C})$ implies $g=[C]f[C{]}^{-1}$ in the Möbius group.

The converse as stated is false because matrix representatives may be rescaled: $A=I$ and $B=2I$ define the same Möbius transformation, while neither $B$ nor $-B$ is conjugate to $A$ (their traces differ).

Invertible matrices are closed under multiplication, contain $I$, have associative multiplication, and have inverses by the adjugate formula over the field ${\mathbb{F}}_p$. The first column can be any nonzero vector and the second any vector outside its span, so

For $p=2$, the matrix $\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$ has order three. There is no element of order six: $G{L}_2({\mathbb{F}}_2)\cong S_3$, whose element orders are $1,2,3$.

For $p>2$, $S{L}_2({\mathbb{F}}_p)=\ker \det$ is a proper normal subgroup. For $p=2$, the order-three subgroup in the copy of $S_3$ is proper and normal.

In $G{L}_2({\mathbb{F}}_{11})$, takewhere $A\le {\mathbb{F}}_{11}^{\times }$ is the subgroup of order five. This semidirect product has order $55$ and is nonabelian because a nontrivial diagonal element does not commute with translations.

The Leibniz definition isThe adjugate is the transpose of the cofactor matrix. Cofactor expansion givesFor the displayed tridiagonal matrix, expansion along the last row gives $D_n=2D_{n-1}-D_{n-2}$, with $D_1=2,D_2=3$. Induction yields

A Jordan block is $J_m(\alpha )=\alpha I+N$, where $N$ has ones on the superdiagonal and zeros elsewhere. Since $N^2=0$ for $m=2$, Taylor expansion givesThe Jordan normal form is the block-diagonal matrix, unique up to block order, to which $A$ is similar and whose blocks are Jordan blocks.

A block of size at least two for eigenvalue $\lambda$ contributes a generalized eigenvector in $\ker (A-\lambda I{)}^2\setminus \ker (A-\lambda I)$. Thus the displayed equality for every $\lambda$ is equivalent to every block having size one, which is diagonalizability.

The transpose of each Jordan block is similar to it by reversing its basis. Hence $A^T$ has the same Jordan form as $A$.

Write $A=PJ{P}^{-1}$. Let $R$ be block diagonal with the reversal matrix on each Jordan block. Both $R$ and $RJ$ are symmetric and $R$ is invertible. Thereforewhere $C,D$ are symmetric and $C$ is nonsingular.

The matrix is antisymmetric, so the value is zero. For a common candidate $p=(a,b,c,d{)}^T$, the column-player inequalities are $Ap\le 0$:The first three imply $a+d\le b\le c\le a-2d$, hence $d=0$ and $a=b=c$. Normalization givesIndeed $Ap=(0,0,0,-1{)}^T/3\le 0$, and antisymmetry gives the corresponding row inequalities.

If $V=V_1\oplus V_2$ were a nontrivial invariant decomposition, each restricted minimal polynomial would be a power of $x-\alpha$ of degree at most $\dim V_i<n$. Their least common multiple is the minimal polynomial of $\varphi$, contradicting its degree $n$.

Since $m_{\varphi }$ has degree $n$, $\varphi$ has a cyclic vector $b_1$. Put $b_k={\varphi }^{k-1}{b}_1$. Then $\varphi (b_k)=b_{k+1}$ for $k<n$. In this basis the matrix is the companion matrix with ones immediately below the diagonal and last columnobtained by expanding $(x-\alpha {)}^n$.

On the closed class, the transition matrix isIts stationary distribution is $(3/11,8/11)$. The class is irreducible and aperiodic because of its self-loops, while the other class is transient and enters it almost surely. Hence

The symmetric bilinear form associated with a real quadratic form $Q$ is obtained by polarization identity:In coordinates $Q(x)=x^{T}Ax$, replacing $A$ by its symmetric part does not change $Q$, and the formula gives $phi(u,v)=u^{T}Av$; this proves existence. A symmetric form is positive semidefinite when $phi(v,v)\ge 0$ for every $v$, and positive definite when the inequality is strict for every $v\ne 0$.

The diagonalization theorem for real quadratic forms says that some basis puts any symmetric form intowith $r$ further zero coordinates. Sylvester's law of inertia says that $(p,q,r)$ is independent of the diagonalizing basis. To prove this, let $P$ be the span of the positive coordinate vectors and let $N\oplus Z$ be the span of the negative and zero vectors. If $P^{\prime }$ is positive definite for another diagonalization and $\dim P^{\prime }>p$, thenso the two spaces intersect nontrivially. A vector in the intersection would have both positive and nonpositive square, a contradiction. Thus $p^{\prime }\le p$; symmetry gives $p=p^{\prime }$. Applying the same argument to $-\varphi$ gives $q=q^{\prime }$, and then $r=r^{\prime }$.

For the nondegenerate form on $V$, write its inertia as $(p,q,0)$, so $p+q=2n$. The restriction vanishes identically on $E$: polarization gives $phi(u,v)=0$ for $u,v\in E$. Projection of $E$ to the positive coordinate space is injective, since a vector with zero positive projection cannot be isotropic unless it is zero. Hence $k\le p$; projection to the negative space likewise gives $k\le q$. Therefore $k\le \min (p,q)\le n$.

The matrix of $l^2$ is $l{l}^T$. If $l\ne 0$, it has rank one and inertia $(1,0,n-1)$, hence signature one; if $l=0$, its rank and signature are zero. The coefficientwise product $(l^2,s^2)$ has matrix $t{t}^T$, where $t_i=l_i{s}_i$, so it has the same rank-one conclusion when $t\ne 0$ and is zero otherwise.

Finally, diagonalization writes every positive semidefinite form as a sum of squares, say $f={\sum }_a{l}_a^2$ and $g={\sum }_b{s}_b^2$. Bilinearity of coefficientwise multiplication givesa sum of positive semidefinite rank-at-most-one forms. Thus $(f,g)$ is positive semidefinite.

Write $Y_i=X_i^T\beta +{\epsilon }_i$, where conditionally $\mathbb{E}({\epsilon }_i|X_i)=0$ and $\mathrm{Var}({\epsilon }_i|X_i)={\sigma }^{2}v(X_i)$. ThenThe law of large numbers sends the matrix towhile the vector tends to zero, proving consistency. The central limit theorem and Slutsky's theorem give

Similarly, with $B=\mathbb{E}(X{X}^T)$ and $C=\mathbb{E}(v(X)X{X}^T)$,These conclusions require the displayed matrices to be finite and nonsingular.

Assume first that $D=b^T{V}^{-1}b>0$. Since $X_M={\theta }_M^{T}Z$,and for $Y={\varphi }^{T}Z$,The least-squares affine-regression coefficients are thereforeThe minimum iswhereThis matrix is symmetric and positive semidefinite. If $b=0$, then $X_M=0$, one may take $\alpha =0$ and arbitrary $\beta$, and the minimum is ${\varphi }^{T}V\varphi$, corresponding to $Q=V$.

For a positive semidefinite matrix, the sum of the eigenvalues is its trace. ThusIf $M,N\in S$ and $0\le t\le 1$, then $tM+(1-t)N⪰0$ andHence $S$ is convex.

Equality of the mixed derivatives givesFor a fundamental matrix solution this is equivalent to the zero-curvature condition

Within each degenerate unperturbed eigenspace of fixed $n_2$, the perturbation has matrixIts eigenvectors are $|+{⟩}_1$ and $|-{⟩}_1$, with first-order shifts $+g$ and $-g$. Thus degenerate perturbation theory gives exactly the states and energies found in part (i); again, all higher corrections vanish because the perturbation commutes with the second-oscillator Hamiltonian and is exactly diagonalized within each degenerate block.

SetThe integral equation becomesEvaluating it at the three delta functions givesThus

For $x>a$, all three Green-function terms are proportional to $e^{ikx}$. Hence

Now define $\gamma =ik{\hslash }^2/(\lambda m)$, so that $q=-1/\gamma$. If $M$ denotes the matrix multiplying $\mathbf{\Psi }$, direct evaluation givesThe displayed algebraic equation is therefore exactly the condition that the finite-dimensional scattering system become singular. Its solutions are poles of the analytically continued scattering amplitude.

To see the imaginary-axis poles directly, write $k=i\kappa$ with $\kappa >0$. When $a\to 0$, the equation reduces toand its nonzero root is $\gamma =-3$, orThis is the bound state of the coincident potential $-3\lambda \delta (x)$. When $a\to \infty$, the three centres decouple and the roots approachwith exponentially small splitting. A pole at $k=i\kappa$ has energy $E=-{\hslash }^2{\kappa }^2/(2m)$ and an exponentially decaying wavefunction, so these upper imaginary-axis singularities represent bound states.

Starting with $A_0=A$, suppose columns $1,\dots ,j-1$ have already been reduced to tridiagonal form. Let $x$ be the part of column $j$ in rows $j+1,\dots ,n$. A Householder reflection $P_j$ on these coordinates can map $x$ to $\pm \Vert x\Vert e_1$. Extend it by the identity on the first $j$ coordinates and call the resulting orthogonal matrix $U_j$.

The updatezeros every entry below row $j+1$ in column $j$. Because it is an orthogonal similarity, it preserves eigenvalues; because the same transformation is applied on both sides, it preserves symmetry and zeros the corresponding row entries without disturbing earlier columns. After $n-2$ such steps,is symmetric and tridiagonal and has the same eigenvalues as $A$. Every reflector is obtained from finitely many matrix entries using finitely many arithmetic operations and a square root, so this is a finite construction.

In the $A$ basis, the transformations whose columns are the new basis vectors areThus $U_C$ is a rotation through $-\pi /3$, while $U_B$ is a real orthogonal reflection, equivalently a rotation with one basis-vector phase reversed.

For any real orthogonal matrix $U$ with columns $|u_j⟩$,Applying this to $U_B$ and $U_C$ and normalizing proves

$A=\left(\begin{array}{cc}{x}_3& x_1-i{x}_2\\ x_1+i{x}_2& -x_3\end{array}\right)$, so $\mathrm{tr}A=0$ and $\det A=-|x{|}^2$. Unitary conjugation preserves Hermiticity, trace, and determinant; every traceless Hermitian $2\times 2$ matrix is $x\prime \cdot \sigma$ with real $x\prime$, and $|x\prime |=|x|$.

A Jordan form is block diagonal with blocks $J_k(\lambda )$ having $\lambda$ on the diagonal and ones on the superdiagonal. Over $\mathbb{C}$ every square matrix is similar to one, uniquely up to permutation of blocks.

The shift matrix obeys $A^4=I$ and $B=aI+bA+c{A}^2+d{A}^3$. Evaluating $g(t)=a+bt+c{t}^2+d{t}^3$ at $1,i,-1,-i$ gives

The transition matrix is symmetric with stationary projector $J/m$ and one other eigenvalue $\lambda =1-\alpha m/(m-1)$. Therefore, starting at BBC1, the probability of BBC2 after $60$ steps is $[1-{\lambda }^{60}]/m$.