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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

$Pu=0$, hence $HPu=0$. In the diagonal basis, $f(\lambda )=1-{\sum }_j{u}_j^2{\mu }_j/({\mu }_j-\lambda )=-\lambda {\sum }_j{u}_j^2/({\mu }_j-\lambda )$. Between consecutive ${\mu }_j$ its nonzero factor is strictly monotone from one infinite sign to the other, giving one root; zero weights give equality at endpoints. Hence ${\mu }_1\le {\sigma }_1\le {\mu }_2\le \cdots \le {\sigma }_{n-1}\le {\mu }_n$.

A rank-$k$ Cartesian tensor has components $T_{i_1\dots i_k}$ transforming under an orthogonal change of basis $R$ as $T{\prime }_{i_1\dots i_k}=R_{i_1{j}_1}\cdots R_{i_k{j}_k}{T}_{j_1\dots j_k}$.

Gram–Schmidt applied to $1,t$ gives the orthonormal basis $1$ and $\sqrt{12}(t-1/2)$.

Replace the quadratic constraint by the two linear inequalities $-2\le x_1-x_2+x_3\le 2$. Simplex pivots lead to the active constraints $x_1=0$, $-x_2+x_3=2$, and $3x_2+x_3=4$. Thus $x^{\ast }=(0,1/2,5/2)$ and the value is $3$. With right-hand side $4+\epsilon$, the same basis remains optimal for small nonzero $\epsilon$, giving $x^{\ast }=(0,1/2+\epsilon /4,5/2+\epsilon /4)$.

At the maximal ideal $\mathfrak{m}=(2,X)$, the images of $4,2X,X^2$ form a three-dimensional basis of $I/\mathfrak{m}I$ over ${\mathbb{F}}_2$, so two generators are impossible. More generally $(2^{n-1},2^{n-2}X,\dots ,2X^{n-2},X^{n-1})$ needs exactly $n$ generators by the same argument.

Generators define a surjection $R^k\to M$, and conversely images of the standard basis generate any quotient. Writing $\varphi (m_i)={\sum }_j{a}_{ij}{m}_j$ with $a_{ij}\in I$ and applying the adjugate to $XI-A$ gives a monic annihilating polynomial whose lower coefficients lie in $I$. Taking $\varphi =1$ when $IM=M$ yields $xM=0$ with $x-1\in I$; for $M=\mathbb{Z}/3$ and $I=(2)$ take $x=3$. The Jacobson radical criterion follows by placing a nonunit $1-rs$ in a maximal ideal. It yields Nakayama’s lemma. Applying the determinant trick to preimages of a finite generating set constructs a polynomial right inverse to any surjective endomorphism, proving injectivity.

$A={\mathbb{Z}}^2/⟨(6,9)⟩$. The class of $(2,3)$ has order $3$, so $A$ has torsion and is not free. A unimodular change of basis (Smith normal form, using $\gcd (6,9)=3$) gives $A\cong \mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}$.

$V^{\ast }$ is the space of linear functionals and the dual basis satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Nondegeneracy and equal finite dimensions make $u\mapsto ⟨u,\cdot ⟩$ an isomorphism; evaluation similarly identifies $V$ with $V^{\ast \ast }$. Riesz representation applied to $v\mapsto ⟨Tv,u⟩$ gives the unique adjoint. If $T$ is diagonalizable, declare an eigenbasis orthonormal to make it self-adjoint. Conversely, for self-adjoint $T$ and invariant $W$, $⟨Tv,w⟩=⟨v,Tw⟩=0$ shows $W^{\perp }$ invariant.

$|V|=p^n$. An ordered basis is chosen successively in $(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})$ ways, which also counts bijective endomorphisms. Counting ordered independent $k$-tuples and dividing by the number of bases of ${\mathbb{F}}_p^k$ gives the Gaussian binomial ${\prod }_{i=0}^{k-1}(p^n-p^i)/(p^k-p^i)$.

A fundamental system of a root system is a subset $\Delta \subset \Phi$ which is a basis of $V$ and for which every $\beta \in \Phi$ has an expansionwhose coefficients are either all nonnegative or all nonpositive. Its associated positive system of a root system isThus $\Phi =\Pi \bigsqcup (-\Pi )$, and the elements of $\Delta$ are the simple roots.

The Weyl reflection in $\alpha$ satisfiesNow take $\beta \in \Pi \setminus \{\alpha \}$ and expand it in the basis $\Delta$. At least one coefficient belonging to a simple root other than $\alpha$ is positive. Sincethe reflection changes only the coefficient of $\alpha$. Every root has simple-root coefficients of one sign, so the unchanged positive coefficient prevents $s_{\alpha }(\beta )$ from being negative. Hence $s_{\alpha }(\beta )\in \Pi$. Because $s_{\alpha }$ is an involution, it permutes $\Pi \setminus \{\alpha \}$, while it exchanges $\alpha$ and $-\alpha$. Therefore

In the orthonormal basis $(|G⟩,|B⟩)$, direct multiplication gives(up to the equivalent opposite orientation convention).

Conditional on Alice and Bob using the same basis, Eve chooses the wrong basis with probability $2/3$, then Bob disagrees with probability $1/2$, so the conditional error is $1/3$. Alice and Bob match with probability $1/3$, hence the per-transmission detection probability is $1/9$.

Applying $I\otimes I,X,Z,$ or $XZ$ to one qubit permutes the Bell basis up to phases. Composing two such Pauli operations maps any chosen Bell state to any other, proving local equivalence.

Charlie measures in the $X$ basis and sends one bit to Alice or Bob; the remaining pair becomes $|{\Phi }^{+}⟩$ or $|{\Phi }^{-}⟩$, corrected by a local $Z$. Alice then performs standard teleportation: CNOT from $|\xi ⟩$ to her GHZ qubit, Hadamard, two computational measurements, and sends the two outcomes to Bob, who applies the corresponding Pauli correction.

Let ${\ell }_j$ be the Lagrange basis polynomials for the nodes. Necessity and existence give uniquely $A_j={\int }_{-1}^1{\ell }_j(t)dt$, because every polynomial of degree at most $n-1$ is ${\sum }_{j}p(t_j){\ell }_j(t)$.

The alternative complete commuting set is $H_0,L^2,S^2,J^2,J_z$, since these operators mutually commute. Clebsch-Gordan coefficients give the unitary change of basis.