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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 $[1-{\lambda }^{60}]/m$.

The parity-check matrix has rank $3$, so its kernel has dimension $7-3=4$ and size $16$; see linear code. Its columns are precisely the seven distinct nonzero vectors of ${\mathbb{F}}_2^3$. Thus no weight-one or weight-two word is in the kernel, while three dependent columns give a weight-three word. Hence $d=3$.

The normal linear model is $Y=X\beta +\epsilon$ with $\epsilon \sim N(0,{\sigma }^2{I}_n)$. Full column rank givesWith hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$, the $i$th regression leverage is $h_{ii}$. Since $e=(I-H)Y=(I-H)\epsilon$ and $I-H$ is an orthogonal projection,High-leverage observations can move their own fitted values strongly and may exert disproportionate influence; residual size alone can therefore conceal them.

The displayed orthogonal matrix is a rotation through $2\alpha$, where ${\sin }^2\alpha =M/N$.