Set $M=A-\lambda I$. Its conjugate transpose is $M^{†}=A^{†}-\overline{\lambda }I$, soSince $A$ is a normal matrix, the two products coincide. Therefore $A-\lambda I$ is normal for every complex $\lambda$.

For the standard complex inner product, $\Vert y{\Vert }^2=y^{†}y$. Using the conjugate transpose and the defining equality for a normal matrix,Both norms are nonnegative, so $\Vert Ax\Vert =\Vert A^{†}x\Vert$ for every $x$.

Starting from $Y_{n-2}=0$ forces $X_{n-2}=0$. The first nonzero value is consequently $X_{n-1}=-1$. Remaining nonzero at the next step forces $X_n=1$, because $p_{-1,-1}=0$. Hence, on an event of positive probability,The two values are unequal by assumption. They expose memory of the hidden sign: immediately after leaving zero the sign is negative, whereas after two consecutive nonzero observations it is positive. Whenever the relevant observed histories coexist with positive probability, the Markov property for $Y$ fails, since its next-step law is not determined by its present value alone. In particular, the absolute-value projection does not preserve the Markov property for all initial laws; the strong lumpability condition would require $p_{-1,0}=p_{1,0}$.

There is a small qualification to a blanket conclusion about an unspecified initial law. The printed assumptions do not ensure that either conditioning event is possible, and a degenerate initial law may never reach the hidden states that cause the failure. For example, in the order $-1,0,1$, takeAll printed constraints hold, yet $Y_n=0$ always and is a Markov chain. Thus the correct general conclusion is the transformed process is not necessarily Markov; the two computed probabilities give the intended failure when their histories are available.

The authoritative definition here is $Y_n=|X_n|$. On the conditioning event, $X_{n-1}=0$. Since $p_{0,1}=0$, the additional condition $Y_n=1$ then forces $X_n=-1$. The Markov property therefore givesAs usual, this conditional probability is asserted when the conditioning event has positive probability.

Take $G=S_3$ and its normal subgroup $N=A_3=⟨r⟩$, where $r=(123)$. Normality follows either from the sign of a permutation or directly from conjugation sending $r$ to $r$ or $r^{-1}$. The quotient group has order two, and $t=(12)$ provides a section $i:C_2\to S_3$ because $t^2=1$ and $tN$ is its nonidentity coset.

However, the map $\Phi (q,n)=i(q)n$ from the direct product of groups is not a group homomorphism. In its domain $(1,r)$ commutes with $(tN,1)$, whereas their images $r$ and $t$ do not commute: $tr{t}^{-1}=r^{-1}\ne r$. This is a split group extension giving a genuine semidirect product instead of a direct product. The pair $S_3,A_3$ supplies the required split but non-direct example.

For the general bijectivity assertion, let $\pi :G\to G/N$ be the quotient projection and assume $\pi \circ i$ is the identity. Given $g\in G$, put $q=\pi (g)$ and $n=i(q{)}^{-1}g$. Then $\pi (n)=q^{-1}q=1$, so $n\in N$ and $g=i(q)n$. This proves surjectivity of $\Phi$. If $i(q)n=i(q^{\prime })n^{\prime }$, applying $\pi$ gives $q=q^{\prime }$, after which cancellation gives $n=n^{\prime }$. ThusThis section normal form for a split group extension does not require finiteness. The obstruction to its being a homomorphism is the nontrivial conjugation action of the section on $N$.

Use additive notation and take $G=\mathbb{Z}/4\mathbb{Z}$, $N=\{0,2\}$. Its quotient group is $C_2$. A group homomorphism from $C_2$ sends its generator to an element annihilated by $2$, so the image must be either $0$ or $2$. Both project to the identity coset $N$, and neither can project to the nonidentity generator of $G/N$. Therefore this quotient projection has no homomorphic section. It is a nonsplit group extension, despite both $G$ and $N$ being nontrivial finite groups.

Take $G=C_2\times C_2$ and $N=C_2\times \{1\}$. Both are nontrivial, and $N$ is normal because $G$ is abelian. The second-coordinate projection identifies the quotient group $G/N$ with $C_2$. Swapping the two coordinates identifies $(G/N)\times N$ with $G$, soThis is a direct product of groups with its first factor as the specified normal subgroup.

In three dimensions the curl can be written with the Levi-Civita symbol:Both terms contract an antisymmetric array with a symmetric one, and therefore vanish. Every differentiable radial field of this form has zero curl away from the origin. This is the three-dimensional part of the divergence and curl of a radial vector field.

Apply the product rule and the preceding gradient formula to the vector field $F_i=f(r)x_i$. Summing over the repeated index givesHere ${\delta }_{ii}=n$ for the Kronecker delta. Thus the divergence and curl of a radial vector field give, in any dimension,

The antisymmetric part need not be a tensor. Its contraction with every symmetric test is zero, so the premise imposes no transformation law on it. This is the blindness of symmetric contraction tests to antisymmetric arrays, now with a third free index.

For a concrete example, prescribe the same numerical array in every orthonormal frame, with onlynonzero, and take $d^s=0$. Every contraction $d_{ijk}^a{s}_{ij}$ vanishes, and the zero result is a genuine vector in every frame. But under $R=\mathrm{diag}(-1,-1,1)$ the rank-three tensor transformation law would send the $121$ component to $-1$, whereas our framewise prescription leaves it equal to $1$. Thus the array is not a tensor despite satisfying every stipulated contraction test. An antisymmetric part could be a tensor if an additional transformation law were supplied; it is simply not forced to be one.

Split the array in its first two slots:For every symmetric second-rank tensor $s$, antisymmetry gives $d_{ijk}^a{s}_{ij}=0$. Hence the stipulated vector equals $d_{ijk}^s{s}_{ij}$.

Let $R$ change the orthonormal frame, so $s_{ab}^{\prime }=R_{ai}{R}_{bj}{s}_{ij}$. The vector transformation law givesfor every symmetric test $s$. Both coefficient arrays in $i,j$ are symmetric, so equality against every symmetric matrix forces equality of the coefficients; one may test the individual diagonal entries and the symmetric off-diagonal basis matrices. Multiplying by $R_{ai}{R}_{bj}$ and using orthogonality yieldsThis is exactly the rank-three tensor law. It is the symmetric-test criterion for a third-rank Cartesian tensor, a form of the quotient theorem for Cartesian tensors. Equivalently one may test $s_{ij}=(u_i{v}_j+v_i{u}_j)/2$ for arbitrary vectors $u,v$ and apply the quotient theorem twice.

Let $R$ be rotation through $2\pi /3$ about $z$, and let $S$ be rotation through $\pi$ about $x$. These generate the order-six dihedral group with relationsnamely $\{I,R,R^2,S,RS,R^{2}S\}$. We specify its order explicitly because dihedral notation has two conventions. It is a proper finite subgroup of the full axial rotation group.

To see that it is sufficient, write a general matrix in blocks $T=\left(\begin{array}{cc}B& u\\ v^T& c\end{array}\right)$. Commutation with $R$ forces $u=v=0$, since a nontrivial planar $120$-degree rotation fixes no nonzero planar vector. Its planar part is $\cos (2\pi /3)I+\sin (2\pi /3)J$, with nonzero sine, so $B$ must commute with $J$ and has the form $aI+bJ$. Commutation with $S$ again kills $b$. Hence this six-element subgroup has precisely the required invariant tensors.

It is also smallest by order. Groups of orders $2,3,5$ are cyclic: a nonidentity element has order dividing the prime group order by Lagrange theorem. For a cyclic rotation subgroup, the nonzero antisymmetric tensor representing cross product with its rotation-axis direction is invariant, so it cannot force the symmetric form above. The trivial subgroup plainly imposes no constraint. A group of order four is either cyclic or has three nonidentity elements of order two; in the latter case they commute, since $(AB{)}^{-1}=AB$ and also equals $BA$. Commuting distinct half-turns in three dimensions have perpendicular axes: conjugation by one half-turn must preserve the other's axis, and a distinct preserved axis lies in its perpendicular plane. In the specified axial group, their three axes are consequently $z$ and two perpendicular horizontal directions. In coordinates along them every diagonal matrix is invariant, including one with unequal horizontal entries. Such a group again does not force the required form.

Thus the smallest subgroup has six elements, generated by the axial $120$-degree rotation and one horizontal half-turn. It is the smallest axial rotation group forcing second-rank transverse isotropy; conjugating the horizontal axis gives equally valid choices.

Under a proper rotation matrix $R$, invariance of a Cartesian second-rank tensor reads $T=RT{R}^T$, equivalently $TR=RT$. The half-turn $R_z(\pi )=\mathrm{diag}(-1,-1,1)$ forces all entries mixing the $z$ direction with the $xy$ plane to vanish. Thus $T$ consists of a planar $2\times 2$ block $B$ and the entry $T_{33}=c$.

Commutation with the planar quarter-turn $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ gives $B=aI+bJ$. This block already commutes with every planar rotation. Now use the half-turn $R_x(\pi )=\mathrm{diag}(1,-1,-1)$: on the planar block its conjugation fixes $I$ and sends $J$ to $-J$, so invariance forces $b=0$. ThereforeThe formula uses the Kronecker delta and is sufficient as well as necessary: rotations preserving the unoriented $z$-axis fix both $I$ and $e_3{e}_3^T$. This is an axially invariant second-rank tensor with the planar antisymmetric part removed by the horizontal half-turn.

Write a nonzero complex number as $w=r{e}^{i\theta }$. Matching modulus and complex argument gives $r^4=2\sqrt{2}$ and $4\theta =\pi /4+2\pi k$. Thus the four distinct fourth roots areThese angles differ by $\pi /2$, so the values are distinct. Every further integer $k$ repeats one of them; this is multiplication by the fourth roots of unity.

Squaring first avoids unnecessary expansion: $(2+2i{)}^2=8i$, and squaring again gives the fourth power $\boxed{z^4=-64}$. In the polar form of a complex number, $z=2\sqrt{2}{e}^{i\pi /4}$; multiplying its complex argument by four gives the same negative-real result.

For circular orbits in an exponential central potential, a constant positive radius requires $V_{\mathrm{eff}}^{\prime }(r)=0$, orLet $h(r)=r^3{e}^{-r}$. Its derivative is $h^{\prime }(r)=r^2{e}^{-r}(3-r)$: it rises from zero to $h(3)=27/e^3$, then falls back to zero. HenceThe strict lower bound matters: the printed upper bound also allows $l=0$, but then there is no finite-radius circular orbit. Nonzero angular momentum is an implicit hypothesis of the two-orbit claim.

At a circular orbit, use $l^2=r^3{e}^{-r}$ to obtainA small radial displacement $\eta$ satisfies $\ddot{\eta }=-V_{\mathrm{eff}}^{\prime \prime }(r)\eta$. Therefore the inner circular orbit is radially stable at fixed angular momentum, while the outer orbit is unstable. The inner perturbation oscillates with squared frequency $e^{-r_1}(3/r_1-1)$; the outer perturbation has an exponentially growing solution. At the limiting value $l^2=27/e^3$, the radii coalesce at $r=3$ and the linear restoring coefficient vanishes.

The central force has no transverse component. Its angular equation is thereforeThus the specific angular momentum is the constantThe radial equation isSubstitute $\dot{\theta }=l/r^2$:Consequently the radial motion for $r>0$ is exactly a unit-mass one-dimensional motion in the effective potentialIts conserved energy is $\frac{1}{2}{\dot{r}}^2+V_{\mathrm{eff}}(r)$; the second term in the effective potential accounts for the angular kinetic energy.

Choose $x\notin H$, possible because $H$ is proper. For any $h\in H$, the element $xh$ is also outside $H$, since otherwise multiplying by $h^{-1}$ would put $x$ inside $H$. Thus both $x$ and $xh$ belong to $K=⟨G\setminus H⟩$, and $h=x^{-1}(xh)\in K$. So $K$ contains $H$ as well as its entire complement, provingThe complement of a proper subgroup generates the group argument does not actually require finiteness.

If one subgroup contains the other, their union is the larger subgroup. Conversely, if neither contains the other, choose $h\in H\setminus K$ and $k\in K\setminus H$. If the union were a subgroup it would contain $hk$. If $hk\in H$, then $k=h^{-1}(hk)\in H$, a contradiction; if $hk\in K$, then $h=(hk)k^{-1}\in K$, again a contradiction. ThusThe union of two subgroups cannot generally be treated like their generated subgroup.

For $U_t=\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$ and $D_r=\mathrm{diag}(r,r^{-1})$, direct multiplication gives $D_r{U}_1{D}_r^{-1}=U_{r^2}$. In ${\mathbb{F}}_{11}$, $5^2=3$ and $5^{-1}=9$, soTo rule out all conjugators when $p=5$, suppose $PA{P}^{-1}=B$ and write $P=\left(\begin{array}{cc}r& s\\ t& u\end{array}\right)$. Comparing $PA=BP$ gives $t=0$ and $r=3u$. Its determinant condition is $ru=1$, so $r^2=3$. The nonzero squares modulo five are $1$ and $4$, and therefore no such $P$ exists. The two matrices are not conjugate in ${\mathrm{SL}}_2(5)$. This is the square-class obstruction in unipotent conjugacy in SL2 over a finite field; checking only diagonal conjugators would not by itself exclude every possible conjugator.