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For a possibly complex eigenvector $v\ne 0$ with $Av=\lambda v$, invertibility of $I-A$ gives $(I-A{)}^{-1}v=(1-\lambda {)}^{-1}v$. ThereforeThus every eigenvector of $A$ remains an eigenvector of $N$. For a real skew-symmetric matrix, its eigenvalues have the form $\lambda =it$ with $t$ real, so $\nu =e^{2i\arctan t}$ lies on the unit circle, consistently with orthogonality.

First, $I-A$ is invertible. If $(I-A)x=0$ for a real vector, then $Ax=x$, whereas the skew-symmetric matrix identity gives $x^{T}Ax=0$; hence $x^{T}x=0$ and $x=0$. Thus its real kernel is trivial, which suffices for invertibility. The same argument applies to $I+A$.

Using $A^T=-A$ and the rule for the transpose of a product,The factors $I+A$ and $I-A$ commute, as do the required inverses. ConsequentlyThus $\boxed{N\text{ is orthogonal}}$. This is the Cayley transform of a real skew-symmetric matrix.

The Hermitian conjugate of a real matrix is its transpose, soTherefore $\boxed{M\text{ is anti-Hermitian for all }\theta ,\varphi .}$ It is also a real skew-symmetric matrix. It is Hermitian as well only in the exceptional case $M=0$, precisely when $R_1$ is symmetric. This includes the identity and half-turn cases from part (d), so the zero-matrix overlap should not be excluded.

Since both real matrices are antisymmetric, $A^{†}=-A$ and $B^{†}=-B$. Thus $U^{†}=-A+iB$. Expand the two unitary identities:Subtracting and adding give $\boxed{AB=BA,A^2+B^2=-I}$. Because $A$ has simple eigenvalues, each of its complex eigenspaces is one-dimensional. Commutation makes each one invariant under $B$: if $Av=av$, then $A(Bv)=a(Bv)$. Hence $Bv=bv$ for some scalar $b$. The complete eigenbasis of $A$ is therefore a common eigenbasis, which also diagonalizes $U$.

The skew-Hermitian decomposition of a unitary matrix is explicit:If $Uv=uv$, unitarity gives $U^{†}v=u^{-1}v=\overline{u}v$. ConsequentlyEquivalently, for $u=e^{i\theta }$ the paired eigenvalues are $\boxed{a=i\sin \theta ,b=-i\cos \theta }$. The hypotheses can only occur in even dimension: $U^T=-U$ and $U$ is invertible, whereas an odd-dimensional skew-symmetric matrix has zero determinant.

The required actions have already been identified:In three dimensions choose a unit vector $w$ perpendicular to both. Its one-dimensional orthogonal complement to the pair is invariant under $M$. On a one-dimensional real space, a skew-symmetric matrix must act as zero: $Mw=\eta w$ would imply $\eta =w^{T}Mw=0$. Thus $Mw=0$.

The columns of $W=(u_1{v}_{1}w)$ form an orthonormal basis, so $W$ is an orthogonal matrix. The displayed actions giveand thereforeIf $M=0$, take ${\mu }_1=0$ and any orthogonal matrix $W$. Otherwise its nonzero two-dimensional block supplies the pair used above.

For the instantaneous Lorenz error growth at a zero-y crossing, direct differentiation of the Lorenz system givesAt a crossing of $y=0$ its symmetric part is particularly simple:Thus a perturbation initially in the $z$ direction has norm-squared rate $-2b$ and initially decays in norm at rate $b$. The $x$-dependent couplings between $y$ and $z$ are skew-symmetric at the crossing and do not contribute to the instantaneous total norm derivative. They can rotate the perturbation later: a pure $\delta z$ immediately induces $\dot{\delta y}=-x\delta z$. It is therefore incorrect to claim that such an error remains an independently decaying $z$ mode along the complete trajectory.

For the $(x,y)$ block, the trace is $-2(\sigma +1)<0$ and the determinant is $4\sigma -(\sigma +r-z{)}^2$. A negative determinant gives one positive and one negative eigenvalue. Equivalently,In particular the requested sufficient lower-height condition isThere is also a second growth range $z>\sigma +r+2\sqrt{\sigma }$; it must not be lost by removing the absolute value. At either endpoint the largest rate is zero. Between them both block eigenvalues are negative. At a crossing in either growing range, some planar error directions have positive instantaneous exponential rate. This statement is local in time: a single instantaneous positive eigenvalue of $S$ does not prove sustained exponential error growth over an evolving Lorenz trajectory.

If $Q(t)$ is a smooth curve of orthogonal matrices with $Q(0)=I$, differentiating $Q(t{)}^{T}Q(t)=I$ gives $Q^{\prime }(0{)}^T+Q^{\prime }(0)=0$. Thus each tangent velocity is a skew-symmetric matrix.

Conversely, if $A^T=-A$, the curve $Q(t)=\exp (tA)$ satisfiesusing commutation of $-tA$ and $tA$. By part (a), $Q^{\prime }(0)=A$. Both inclusions prove

The matrix exponential is the absolutely convergent series $e^{tA}={\sum }_{n=0}^{\infty }{t}^n{A}^n/n!$, with derivative $A{e}^{tA}$ and inverse $e^{-tA}$. If $K^T=-K$, then $(e^{tK}{)}^T{e}^{tK}=e^{-tK}{e}^{tK}=I$. Its determinant is continuous in $t$, is never zero, and starts at one, so it is one throughout: $e^{tK}\in SO(3)$. Conversely, differentiating $(e^{tK}{)}^T{e}^{tK}=I$ at zero gives $K^T+K=0$. ThusThe quantifier matters: membership at one isolated time alone would not imply skewness.

Write $P(F)=DW(F)$ and $J=\det F>0$. For any skew matrix $K$, differentiate frame indifference along $e^{tK}F$:A matrix orthogonal to every skew matrix is symmetric. Therefore $P(F)F^T$ and the Cauchy stress tensor $T=J^{-1}P(F)F^T$ are symmetric.

Let $\mathcal{C}=D^{2}W(I)$ and assume $P(I)=0$. Differentiating symmetry of $P(F)F^T$ at $I$ in direction $H$ shows $\mathcal{C}[H]$ is symmetric for every $H$. Hence $c_{ijkl}=c_{jikl}$. As a Hessian, $\mathcal{C}$ also has pair-interchange symmetry $c_{ijkl}=c_{klij}$. Combining the two gives $c_{ijkl}=c_{ijlk}$. Equivalently $\mathcal{C}$ annihilates infinitesimal rotations. This proves both requested minor symmetries and shows exactly where the stress-free assumption enters. It is stress-free linearization of an objective energy.

For the particular energy, ${\sum }_i{v}_i^2=\mathrm{tr}(F^{T}F)$ and ${\prod }_i{v}_i=J$, soAt $I$ this is $[4\alpha +h^{\prime }(1)]I$. Therefore the reference is stress-free exactly when $\boxed{h^{\prime }(1)=-4\alpha }$. To differentiate again, use $\mathrm{cof}F=J{F}^{-T}$. The determinant and inverse derivatives at $I$ givewhich also proves the supplied hint. It follows thatFor a symmetric $e$ this becomesNo additional positivity of the bulk modulus follows from the stated smoothness of $h$ alone; this calculation identifies the moduli rather than imposing extra material assumptions.

For a skew-symmetric matrix, $A^T=-A$. By parts (c) and (d),If $n$ is odd, $\det (A)=-\det (A)$, hence $2\det (A)=0$ over $\mathbb{R}$. Therefore $\det (A)=0$, so $A$ is a singular matrix. The conclusion depends on odd dimension: in dimension two, $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ is skew-symmetric and has determinant one.

True. Using the same transpose identity and $X^T=-X$ givesThe final equality follows by multiplying $-X$ and $-X^{-1}$ in either order and obtaining the identity. Thus the inverse is a skew-symmetric matrix as well. The invertibility hypothesis is retained throughout.

Let $E_{ij}$ denote a matrix unit, with its only nonzero entry equal to one in row $i$, column $j$. A basis for the vector space of real skew-symmetric matrices isEach upper-triangular entry is free, its reflected lower-triangular entry is its negative, and every diagonal entry is zero. This is the basis of the space of real skew-symmetric matrices.

Using linearity and involutivity of the matrix transpose,Therefore $B-B^{\top }$ is a skew-symmetric matrix.

Split $B$ into its symmetric part and skew-symmetric part:Taking the transpose gives $S^{\top }=S$ and $K^{\top }=-K$. Thus $S$ is a symmetric matrix and $K$ a skew-symmetric matrix. These formulas also show uniqueness, since a real matrix that is both symmetric and skew-symmetric is zero.

If $A$ is a skew-symmetric matrix, the scalar $x^{\top }Ax$ equals its own transpose, soand it is zero over the real numbers.

Conversely, suppose the quadratic form $x^{\top }Ax$ vanishes for every real vector. Substituting $x=e_i$ gives $a_{ii}=0$. Substituting $x=e_i+e_j$ gives $a_{ij}+a_{ji}=0$ for $i\ne j$. Therefore $A^{\top }=-A$. This proves the quadratic-form criterion for a skew-symmetric matrix:

A symmetric operator $L$ on a real Hilbert space is strictly positive definite when $⟨Lv,v⟩>0$ for every nonzero $v$ in its operator domain. Strict positivity proves uniqueness of a solution of $Lu=f$, but in infinite dimension it need not prove existence for every $f$. A uniformly positive definite symmetric operator additionally has a positive lower bound independent of $v$. Symmetry must be stated separately when positivity is defined only by a real quadratic expression: adding a skew-symmetric matrix changes the operator without changing that expression.

Proof. For a skew-symmetric matrix, the scalar $x^{\top }Ax$ equals its transpose $x^{\top }{A}^{\top }x=-x^{\top }Ax$, so it vanishes. Conversely, substituting standard basis vectors $e_i$ into the vanishing quadratic form gives $a_{ii}=0$. Substituting $e_i+e_j$ gives $a_{ij}+a_{ji}=0$. Thus $A^{\top }=-A$. In particular, a real quadratic form depends only on a matrix's symmetric part.

For stage states $Y_i=y_n+h{\sum }_j{a}_{ij}{k}_j$ and a vector field with $Y_i^T{k}_i=0$, expansion of the new squared norm gives $\Vert y_{n+1}{\Vert }^2-\Vert y_n{\Vert }^2=-h^2{\sum }_{i,j}{M}_{ij}{k}_i^T{k}_j$, where $M_{ij}=b_i{a}_{ij}+b_j{a}_{ji}-b_i{b}_j$. Hence $M=0$ preserves the quadratic form exactly for every well-defined step. The same argument with a fixed symmetric matrix in the inner product preserves any quadratic invariant of the vector field. In particular it preserves the Euclidean norm for a skew-symmetric matrix differential equation.

The displayed parts of a unitary matrix are skew-Hermitian matrices. Expanding $U{U}^{†}=U^{†}U=I$ gives $AB=BA$ and $A^2+B^2=-I$. They admit a common orthonormal eigenbasis. For a unitary eigenvalue $u$, the corresponding part eigenvalues are $i\mathrm{Im}u$ and $-i\mathrm{Re}u$. If $U^T=-U$, then $U^{†}=-\overline{U}$, so these are exactly its real and imaginary parts and both are real skew-symmetric matrices. Such an invertible antisymmetric matrix has even dimension, by its determinant. If one part has distinct eigenvalues, commutation alone shows its one-dimensional eigenspaces are preserved by the other part.

For a nonzero real three-dimensional skew-symmetric matrix, choose real orthogonal vectors $u,v$ of equal length with $Au=\theta v$, $Av=-\theta u$ and $\theta >0$, and a real unit vector $e_3$ in its kernel. The matrix exponential acts by $e^{A}u=u\cos \theta +v\sin \theta$, $e^{A}v=v\cos \theta -u\sin \theta$ and $e^A{e}_3=e_3$. Its matrix in this orthonormal basis is a planar rotation matrix with an unchanged axis. A purely imaginary nonzero eigenvalue and its complex eigenvector give $u,v$ by taking real and imaginary parts; antisymmetry proves their equal norms and orthogonality.

An orthogonal similarity preserves skew symmetry, so upper-Hessenberg reduction of a real skew-symmetric matrix is automatically tridiagonal with zero diagonal. For a Householder reflection, the scalar $v^{T}Av$ vanishes, eliminating the fourth term of the general similarity update. Storing only one triangle and using the displayed skew rank-two update costs $2n^3/3+O(n^2)$ scalar products, the same leading product count as optimized symmetric tridiagonalization; the simplification saves lower-order work.