False. The counterexample just given is antisymmetric in every basis and has zero contraction with every symmetric second-rank tensor. It still fails the Cartesian second-rank tensor transformation law, because its value is nonzero in one basis and zero in another. Thus antisymmetry alone, together with symmetric tests, proves no tensor transformation law. The same blindness of symmetric contraction tests to antisymmetric arrays applies even when the array is assumed antisymmetric.

False. Tests against symmetric second-rank tensors see only the symmetric part $A_s=(A+A^T)/2$, because the Frobenius inner product of a symmetric matrix and an antisymmetric matrix is zero:An arbitrary basis-dependent antisymmetric part is invisible to these tests. For example, prescribein one basis, and $A^{\prime }=0$ in another rotated basis, choosing antisymmetric arrays in every remaining basis. Every contraction with any symmetric second-rank tensor is the invariant scalar zero. But an invertible rotation cannot transform this nonzero matrix to zero, so the array is not a Cartesian second-rank tensor. This is the blindness of symmetric contraction tests to antisymmetric arrays.

True. In all four tests, an array is understood to have components specified in each rotated orthonormal basis; being a scalar means that the contraction is invariant under those changes. Write the Frobenius inner product as $A:B={\sum }_{ij}{A}_{ij}{B}_{ij}$. Every test Cartesian second-rank tensor obeys $B^{\prime }=RB{R}^T$. The assumption saysChoosing the elementary matrices as $B$, or choosing $B=R^T{A}^{\prime }R-A$, gives $R^T{A}^{\prime }R=A$. Thereforewhich is the required Cartesian second-rank tensor transformation law. This is the scalar contraction test for a Cartesian tensor.

Start with the assumed basis of eigenvectors, grouping them by eigenvalue. In each eigenspace, apply the Gram-Schmidt process, which replaces a linearly independent list by an orthonormal list with the same span. Every new vector stays within that eigenspace, since an eigenspace is a vector subspace.

By part (iv), different eigenspaces are orthogonal. The combined lists therefore form an orthonormal basis of the entire space, still consisting of eigenvectors. Let $U$ have these vectors as columns. Then $U$ is a unitary matrix, soThis gives the required unitary diagonalization of a normal matrix, including repeated eigenvalues.

The adjoint eigenvector identity for a normal matrix gives $A^{†}{x}_{\lambda }=\overline{\lambda }{x}_{\lambda }$. HenceAs $\mu \ne \lambda$, $x_{\lambda }^{†}{x}_{\mu }=0$. Thus eigenvectors in distinct eigenspaces of a normal matrix are orthogonal in the complex inner product.

If $Ax=\lambda x$, then $M=A-\lambda I$ is a normal matrix by part (ii), and $Mx=0$. Part (i), applied to $M$, yields $\Vert M^{†}x\Vert =\Vert Mx\Vert =0$. ThusThe nonzero vector $x$ is consequently an eigenvector of the conjugate transpose with eigenvalue $\overline{\lambda }$. This adjoint eigenvector identity for a normal matrix will also determine the special spectral restrictions below.

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.