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The map $\theta :\mathbb{C}[x]\to V$, $p\mapsto p(\varphi )a_1$, is surjective because $1,x,\dots ,x^{n-1}$ map to the given basis. Its kernel is generated by the minimal polynomial and the quotient has dimension $n$, so $\mathrm{deg}{m}_{\varphi }=n$. The relation for $\varphi (a_n)$ gives

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 $(x-\alpha {)}^n$.

The northwest-corner rule constructs an initial basic feasible solution: fill the current cell with the smaller remaining supply and demand, delete the exhausted row or column, and continue.

For a current spanning-tree basis, solve $u_i+v_j=c_{ij}$ on its occupied cells, fixing one potential to zero. The reduced cost of an unoccupied cell isIf all reduced costs are nonnegative, part (a) proves optimality. Otherwise choose a cell with negative reduced cost. Adding its edge to the tree creates a unique even cycle. Mark its cells alternately $+$ and $-$, starting with $+$ at the entering cell, and setAdd $\theta$ on the plus cells and subtract it on the minus cells. Row and column totals are unchanged, the entering cell becomes positive, and a minimizing minus cell leaves the basis. In the nondegenerate case the objective decreases strictly. There are finitely many bases, so no basis repeats and the algorithm terminates at a basis with no negative reduced cost, which is optimal.

The dual space $V^{\ast }$ is the vector space of linear maps $V\to \mathbb{F}$. If $B=(v_1,\dots ,v_n)$, its dual family satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Every functional $f$ obeysso the family spans; evaluation on each $v_j$ proves linear independence. Thus it is a basis without any prior dimension argument.

For $p=a+bt+c{t}^2$,The basis dual to $(f_0,f_1,f_2)$ isas direct substitution gives $f_i(p_j)={\delta }_{ij}$. Hence the three functionals form a basis of $P_2^{\ast }$.

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.

In one dimension the defining relations forcegiving four characters.

Every irreducible two-dimensional representation is, after a choice of basis,where $\lambda \in {\mathbb{C}}^{\times }\setminus \{1,-1\}$. It is irreducible because the two distinct $r$-eigenlines are interchanged by $t$. Conversely, part (a) gives this basis from $v,tv$. If the only eigenvalue of $r$ were $\pm 1$, its eigenspace would be $t$-invariant and contain a $t$-eigenvector, contradicting irreducibility. Parameters $\lambda$ and ${\lambda }^{-1}$ give isomorphic representations, and there are no other identifications.

Since $17$ is square-free and $17\equiv 1(\mathrm{mod}4)$, the standard quadratic-field integral-basis theorem givesIndeed, the displayed generator has monic polynomial $x^2-x-4$, and the discriminant of this basis is $17$, the field discriminant; no larger order is possible.

In the ordered basis $(|0⟩,|1⟩)$,

For $H_0=E_2{B}_2^{†}{B}_2$, the eigenspaces of energies $0$ and $E_2$ are each two-dimensional because $n_1$ is free. Degenerate perturbation theory requires diagonalizing $E_1{B}_1^{†}{B}_1$ within each space. It is already diagonal in the product basis, with eigenvalues $0,E_1$. Therefore the first-order eigenvectors remainand their energies are $0,E_1,E_2,E_2+E_1$, respectively. These are the exact answers, so all higher corrections vanish.

In the ordered basis $(|00⟩,|01⟩,|10⟩,|11⟩)$,$B_1^{†}$ and $B_2^{†}$ are their transposes, and

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

Let $P$ have the old domain basis as columns, $C$ the old codomain basis, and $D$ the new codomain basis. Change of coordinates gives $A\prime =D^{-1}CA{P}^{-1}$, hence