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With $p_i=\mathbb{P}({\mathrm{dis}}_i=1)$, the fitted logistic model isRisk category 1 is the reference level, so its effect is included in the intercept; adding a separate coefficient for both levels would make the design matrix linearly dependent.

For $q=k^2$, the mode matrix is $J-q\mathrm{diag}(D_1,D_2)$. Its trace remains negative. Its determinant isIt is negative for some $q>0$ exactly whenTogether with homogeneous stability, the spatial-instability region isFor fixed $\beta$, this is the region below the last curve, above $\alpha =\beta$, and left of $\lambda =1/\beta$.

If $D_1=D_2=D$, the linearized matrix for wavenumber $k$ isIts eigenvalues are those of the homogeneous Jacobian shifted left by $D{k}^2$, so a stable homogeneous equilibrium remains stable for every spatial mode. Thus equal diffusivities cannot produce a Turing instability. In part (b), equal diffusivities mean $\lambda =1$; because homogeneous stability requires $\beta >1$, the necessary condition $1-\lambda \beta >0$ fails, consistently ruling out instability.

A strongly regular graph with parameters $(k,a,b)$ is $k$-regular, with every adjacent pair having $a$ common neighbours and every distinct nonadjacent pair having $b$ common neighbours. Its adjacency matrix satisfiesOn the orthogonal complement of the all-one vector, the two possible eigenvalues areUsing $1+m_r+m_s=n$ and $k+m_{r}r+m_{s}s=0$ gives exactly the two displayed expressions for $m_r,m_s$. They are eigenspace dimensions and hence integers, proving the rationality condition.

The Petersen graph has parameters $(3,0,1)$, so its spectrum isIf three Petersen graphs partitioned $E(K_{10})$, their adjacency matrices would satisfy $A+B+C=J-I$. The five-dimensional eigenvalue-one spaces of $A$ and $B$ inside the nine-dimensional space $1^{\perp }$ intersect nontrivially. For a nonzero common vector $v$, $Av=Bv=v$ and thereforecontradicting the Petersen spectrum. No such partition exists.

For a nonzero ideal $I\subset O_K$, its norm isFor $d=[K:\mathbb{Q}]$ and embeddings ${\sigma }_1,\dots ,{\sigma }_d$, defineThe field discriminant is the discriminant of any integral basis of $O_K$. If ${\alpha }_1,\dots ,{\alpha }_d$ is a $\mathbb{Z}$-basis of $I$, then

Multiplication by $\alpha$ on an integral basis has integer matrix $M_{\alpha }$. Its image lattice is $\alpha O_K$, soThe determinant of this $\mathbb{Q}$-linear map is the field norm $N_{K/\mathbb{Q}}(\alpha )$, proving

The matrixis the orthogonal projection onto the row space of $X$. Since $XP\beta =X\beta$, the empirical risks of $h_{\beta }$ and $h_{P\beta }$ are equal. Orthogonality givesso $q(\beta )\ge q(P\beta )$, with strict inequality unless $(I-P)\beta =0$. Therefore the minimizer lies in the row space:Because $X^T$ is injective when $X{X}^T$ is invertible, minimizing over that row space is equivalent to minimizing $r(\alpha )=q(X^T\alpha )$.

Since $A_k=Q_k{R}_k$,Thus every step is an orthogonal similarity, so it preserves eigenvalues and symmetry.

For a symmetric $r$-banded matrix, its subdiagonal entries can be eliminated by Givens rotations ordered along the band. Each rotation creates only the next local bulge; multiplication in reverse order chases that bulge out without creating entries beyond the original upper band. Symmetry supplies the corresponding lower band. This is symmetric bandwidth preservation under QR iteration, and proves that $A_{k+1}$ is again $r$-banded.

The unshifted QR algorithm computes the eigenvalues of $A$. For a real symmetric matrix it drives $A_k$ toward a diagonal or block-diagonal matrix; the accumulated orthogonal factors simultaneously approximate its eigenvectors.

A complex matrix is diagonalisable when it is similar to a diagonal matrix, equivalently when the vector space has a basis of its eigenvectors.

Set $y=tz$ and $u=(x,z{)}^T$. The system becomesThe matrix has eigenpairs $1,(2,1{)}^T$ and $-2,(1,1{)}^T$. A particular solution is $(-t/3-2,-1{)}^T$. Therefore

The eigenvalues of the recurrence matrix have modulus $e^{-\kappa T}<1$, so transients decay and the coefficients approach the fixed pointThusFor $T=k\pi$ and $\kappa k\ll 1$, this isEven multiples deliver impulses in phase and produce resonance; odd multiples alternate the phase and largely cancel.

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.