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No. A pure product state has a pure reduced density matrix, whereas part (b) gives $I/2$.

With $p_{-1}=1,p_0=a_0$, $q_{-1}=0,q_0=1$ and $p_n=a_n{p}_{n-1}+p_{n-2}$, $q_n=a_n{q}_{n-1}+q_{n-2}$, induction gives $p_n{q}_{n-1}-p_{n-1}{q}_n=(-1{)}^{n+1}$. Matrix multiplication gives $[a_0,\dots ,a_n,\beta ]=(p_n\beta +p_{n-1})/(q_n\beta +q_{n-1})$; monotonicity in $\beta >0$ places it strictly between the adjacent convergents.

For a two-qubit density operator, the real three-by-three matrix $T$ encodes the spin-product expectation values: unit measurement axes $\mathbf{a},\mathbf{b}$ have $E(\mathbf{a},\mathbf{b})={\mathbf{a}}^{T}T\mathbf{b}$. The tensor does not include local Bloch-vector data and alone need not specify the complete density operator. The CHSH optimum from the correlation tensor depends on the two largest eigenvalues of $T^{T}T$.

The strictly positive case states that a real square matrix $W$ with $W_{ij}>0$ has a positive simple eigenvalue $\lambda$ and positive left and right eigenvectors, with all other eigenvalues strictly smaller in modulus. The Brouwer fixed-point theorem applied to $v\mapsto Wv/(1^{T}Wv)$ on the nonnegative unit simplex gives $v>0$ and $Wv=\lambda v$. For another eigenvector $z$, maximize $|z_i|/v_i$; the triangle inequality gives $|\mu |\le \lambda$. Equality forces every modulus ratio and complex argument to agree because every entry is positive, so $z$ is proportional to $v$ and $\mu =\lambda$. Apply the same argument to $W^T$ for a positive left eigenvector; its positive pairing with $v$ rules out a generalized eigenvector at $\lambda$, establishing algebraic simplicity. For merely nonnegative matrices a nonnegative leading eigenvector exists. An irreducible nonnegative matrix has a positive leading eigenvector and simple Perron eigenvalue; a primitive nonnegative matrix has the strict modulus gap.

A convex cone containing no nonzero line through the origin, equivalently $K\cap (-K)=\{0\}$. The copositive cone is pointed: a matrix in both it and its negative has zero quadratic form on the nonnegative orthant; testing $e_i$ and $e_i+e_j$ kills all entries.

If a real matrix $B$ has a positive diagonal matrix $W$ such that $WB$ is a symmetric matrix, then $H=W^{1/2}B{W}^{-1/2}$ is a symmetric matrix related to $B$ by a similarity transformation: $B^{T}W=WB$ gives $H^T=H$. Thus $B$ is a diagonalizable matrix with real eigenvalues. If ${\mathbf{z}}^{T}WB\mathbf{z}\le 0$ for all real vectors $\mathbf{z}$, its eigenvalues are nonpositive. This converts symmetry in a weighted inner product into Euclidean symmetry.

For real positive semidefinite matrices $A,B$, their product need not be symmetric, but its matrix trace is nonnegative:The equality uses the cyclic property of the matrix trace, and the last matrix is a positive semidefinite matrix. Consequently the Loewner order inequality $A⪯C$ implies $\mathrm{tr}(AB)\le \mathrm{tr}(CB)$ whenever $B⪰0$.

For a two-level Fourier recurrence ${\hat{u}}^{n+1}=a(\theta ){\hat{u}}^n+b(\theta ){\hat{u}}^{n-1}$, the amplification matrix isStability requires its powers to be uniformly bounded in the time index and the mesh frequencies. If its two amplification roots have modulus at most one and their separation has a positive mesh-independent lower bound, the eigenvectors $({\xi }_j,1{)}^T$ give a uniformly bounded diagonalization of a matrix, proving stability. A repeated unit-modulus root of this companion matrix instead produces a Jordan block and linear growth in time. Thus checking only the moduli of the roots is insufficient.

For a nonnegative bounded weak subsolution and $\beta >1$, test with ${\eta }^2(u+\epsilon {)}^{\beta -1}$. Uniform ellipticity and Cauchy-Schwarz inequality give $\int |Du{|}^2{u}^{\beta -2}{\eta }^2\le 4{\Lambda }_{\ast }^2/[{\lambda }^2(\beta -1{)}^2]\int u^{\beta }|D\eta {|}^2$, with ${\Lambda }_{\ast }$ an operator norm bound for the coefficient matrix. The Sobolev chain rule and $\epsilon \downarrow 0$ justify fractional powers and zero values. An entrywise bound $\Lambda$ yields ${\Lambda }_{\ast }\le n\Lambda$.

For a nonzero positive-weight eigenform at a good prime, the two oldforms $f(\tau ),f(p\tau )$ are independent. At level multiplied by $p$, the Hecke operator is the bad-prime operator, with matrix $\left(\begin{array}{cc}\lambda & 1\\ -\chi (p)p^{k-1}& 0\end{array}\right)$. Distinct roots $\alpha ,\beta$ of its characteristic polynomial give eigenforms $f-\beta f(p\tau )$ and $f-\alpha f(p\tau )$. Weight-zero constants are an exception to the independence assertion.

A square nonnegative matrix $A$ is primitive if some positive integer power $A^k$ has strictly positive entries. It is therefore an irreducible nonnegative matrix. The Perron–Frobenius theorem gives a leading eigenvalue whose modulus is strictly larger than that of every other eigenvalue. Every strictly positive matrix is primitive; the two-cycle permutation matrix is irreducible but not primitive.

A principal minor is the determinant of a square submatrix selected using the same index set for rows and columns. For a three-by-three matrix, the coefficient of $\lambda$ in its monic characteristic polynomial is the sum of its three principal minors of order two.

Optimization of a quadratic form or quadratic-plus-linear objective over specified constraints. Convexity depends on the matrix and feasible domain; discrete restrictions lead to binary quadratic optimization.

Suppose the columns $a_i$ of a matrix over the rational numbers satisfy the columns condition with blocks $B_1,\dots ,B_s$. Choose rational coefficients ${\lambda }_{ij}$ such thatChoose a positive integer $c$ clearing all denominators and $p\ge \max |c{\lambda }_{ij}|$. For generators of any full positive M-p-c set, the coordinateslie in that set and satisfy $Ax=0$. Thus the monochromatic m-p-c set theorem supplies the sufficiency direction of Rado's theorem.

A random matrix is a matrix whose entries are random variables. Its eigenvalues form a random configuration, and its empirical spectral measure records their distribution with equal mass on each eigenvalue. Entry assumptions and normalization determine the spectral regime.

For any finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ and rational positive-determinant $\gamma$, the intersection $S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}\Gamma \gamma$ has finite index. Multiply $\gamma$ by a positive rational scalar to obtain an integral matrix $A$ with positive integer determinant $D$. For $h=I+DB\in \Gamma (D)$, $Ah{A}^{-1}=I+AB\mathrm{adj}(A)$ is integral, so $\Gamma (D)$ lies in $S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}S{L}_2(\mathbb{Z})\gamma$. This intersection has finite index. Its further intersection with ${\gamma }^{-1}\Gamma \gamma$ has relative index at most $[S{L}_2(\mathbb{Z}):\Gamma ]$. No assumption that $\Gamma$ itself is a congruence subgroup is needed.

A rational matrix is a matrix whose entries are rational numbers. Multiplying by a common denominator produces an integer matrix and preserves its homogeneous solution set.