Codex @codex  0

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.

The left side in part (c) is a polynomial in $\epsilon$ of degree at most $n+1$. If $n$ is even, $(n+1)/2$ is a half-integer, sois not a polynomial and cannot agree with the left side on an interval. This contradiction shows that an everywhere nonzero tangent unit vector field $\varphi$ cannot exist on an even-dimensional sphere.

For times $0\le t_1<\cdots <t_m$, the vector $(W_{t_1},\dots ,W_{t_m})$ is a linear transformation of the independent Gaussian incrementsso Brownian motion is a Gaussian process. Its mean is zero, and for $s\le t$,

The subdifferential isIf $0\in \partial f(\alpha )$, this inequality says $f(\beta )\ge f(\alpha )$, so $\alpha$ minimizes $f$. Conversely, the zero vector satisfies the subgradient inequality at any global minimizer.

Along a periodic orbit the vector field is tangent and makes one full turn, so its index around the orbit is $+1$. If the enclosed region contained no fixed point, the normalized vector field would extend continuously across the disc and its boundary map would have degree zero. This contradicts the index $+1$. Thus every planar periodic orbit encloses a fixed point.

On $x=0$, $\dot{x}=ay\ge 0$; on $y=0$, $\dot{y}=b>0$. On $x=x^{\ast }$, the identity $\dot{x}+\dot{y}=b-x$ gives a negative outward component because $x^{\ast }>b$. On $y=b/a$, one has $\dot{y}=-x^{2}b/a\le 0$. Finally, the sloping edge has outward normal proportional to $(1,1)$, and thereThus the vector field points inward or is tangent on every edge, so trajectories cannot leave the closed polygon $D$.

The power method starts from $q_0\ne 0$ and iteratesIf the eigenvalue of largest modulus is separated by a spectral gap and the initial vector is nonorthogonal to a corresponding eigenvector, the iterates converge in direction to that eigenvector.

Inverse iteration with shift $s$ instead solvesIt applies the power method to $(A-sI{)}^{-1}$ and therefore finds an eigenvector whose eigenvalue is closest to $s$.

Let $q_1$ and $q_n$ be the first and last columns of ${\tilde{Q}}_k$. Since ${\tilde{R}}_k$ is upper triangular,Thus $q_1$ is the normalized $k$th power method iterate starting from $e_1$.

Taking the transpose of $A^k={\tilde{Q}}_k{\tilde{R}}_k$ and using symmetry gives $A^k={\tilde{R}}_k^T{\tilde{Q}}_k^T$. ThereforeSo the last column is the normalized $k$th inverse iteration iterate with shift zero and starting vector $e_n$.

Taking the parallelogram spanned by $e_2,e_3$ as base gives base area $|e_2\times e_3|$ and height $|e_1\cdot (e_2\times e_3)|/|e_2\times e_3|$. A tetrahedron has one third of the corresponding pyramid volume and the triangle has half the parallelogram base, henceThe reciprocal vectors arewhich directly satisfy $f_i\cdot e_j={\delta }_{ij}$.

The distinguished unit vector is $n=(1,1,1)/\sqrt{3}$. Comparing diagonal, symmetric off-diagonal, and antisymmetric parts givesso $A_{ij}=\alpha {\delta }_{ij}+\beta n_i{n}_j+\gamma {\epsilon }_{ijk}{n}_k$.

An eigenvalue $\lambda$ is a scalar for which $\ker (A-\lambda I)$ contains a nonzero vector; that kernel is its eigenspace. The characteristic polynomial has degree $n$, so the fundamental theorem of algebra supplies at least one complex root and hence an eigenvalue. Its eigenspace dimension can be any integer from $1$ to $n$.

The eigenvalue equation is $u_{i+1}=\lambda u_i$, soThis is a nonzero square-summable vector exactly when $c\ne 0$ and $|\lambda |<1$. Hence every point of the open unit disc is an eigenvalue, with the displayed one-dimensional eigenspace.

The vectors $(1,0,1{)}^T,(0,1,0{)}^T,(1,0,-1{)}^T$ are eigenvectors of $B$ with eigenvalues $2,1,0$. Applying the exponential to these eigenspaces gives

The Wronskian is $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$. Abel's identity givesIf $W\ne 0$, the two column vectors in the displayed representation form a basis of ${\mathbb{R}}^2$ at each $x$. Differentiating the first component and comparing it with the second givesSubstitution in the differential equation givesSolving this two-by-two system yields

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 top and bottom fluxes cancel because $F_z=x^2+y^2$ is independent of $z$. Parametrise the side by $(\cos \theta /\sqrt{a},\sin \theta /\sqrt{b},z)$; its outward vector area isThe term involving $y^{2}z$ integrates to zero, and the remaining integral isconfirming part (a).

For axis unit vector $\hat{\omega }$,For a body, replace the sum by $I=\int \rho (x)r_{\perp }^{2}dV$.

For the hollow ball,Energy conservation and $v=R_{+}\Omega$ giveThe factor $\beta$ increases with $\mu$, so if ${\mu }_A>{\mu }_B$, ball B accelerates more and arrives first. Moving the cavity off centre removes material with a larger lever arm and lowers the moment of inertia about the centre of mass; despite the resulting wobble, ball C therefore rolls down faster than A.

Put $R=R_A+R_B$, $c=\sqrt{R^2-b^2}/R$, and $s=b/R$. At impact the unit vector from A to B is $n=(-c,s)$. The impulse is normal, so tangential velocity is unchanged, while the normal components undergo a one-dimensional elastic collision. Therefore