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Let ${\mathrm{Sym}}_n$ be the vector space of real symmetric matrices and considerAt $R\in O(n)$,This derivative is surjective: for $S\in {\mathrm{Sym}}_n$, choosing $H=RS/2$ gives $D_R\Phi (H)=S$. Thus $I$ is a regular value and the orthogonal group $O(n)={\Phi }^{-1}(I)$ is a submanifold of codimensionThe determinant takes only the values $\pm 1$ on $O(n)$, so the special orthogonal group $SO(n)=\{R\in O(n):\det R=1\}$ is an open-and-closed submanifold of $O(n)$. Therefore

The tangent space is the kernel of $D_R\Phi$:where the Special orthogonal Lie algebra $\mathfrak{so}(n)$ is the vector space of skew-symmetric matrices. This proves the special orthogonal group as a submanifold description.

Given a solution $u(t)$, construct $a_j(t)=\frac{1}{2}{e}^{u_j(t)/2}$ and hence the smooth matrices $L(t)$ and $B(t)$. Let $U(t)$ solve the linear ordinary differential equationBecause $B(t)$ is skew-symmetric,so $U(t)$ is an orthogonal matrix and is invertible. Moreover,It follows thatThus the Lax pair flow is an Isospectral Lax equation: every $L(t)$ is similar to $L(0)$ and has the same eigenvalues and characteristic polynomial.

Every symmetric function of these constant eigenvalues is a first integral. In particular, the trace invariants of a Lax equationare constant because the cyclic property of the trace givesFor example,is an explicit first integral of equation $(†)$.

The special orthogonal group in three dimensions isDifferentiating $A(t{)}^{T}A(t)=I$ at the identity shows that its Lie algebra is the space of skew-symmetric matrices. A convenient basis iswith $[J_a,J_b]={ϵ}_{abc}{J}_c$.

The infinitesimal action of $\exp (t{J}_a)$ on a point $x\in {\mathbb{R}}^3$ is $J_{a}x=e_a\times x$. It therefore generates the vector fieldFundamental vector fields for this left action form an antihomomorphism with the stated convention, and direct differentiation givesso their span is closed under the Lie bracket of vector fields.

The brackets $\{x,y\}=z$, $\{y,z\}=x$, and $\{z,x\}=y$ define the rotational Lie-Poisson structure on R3. With the convention that a Hamiltonian vector field acts by $X_H(f)=\{f,H\}$,Hence the required Hamiltonians are simplyThe quadratic functionsatisfies $\{F,x_a\}=0$ for all $a$, so it is a Casimir function of a Poisson manifold. Its nonzero regular level sets $F=R^2$ are spheres. The Poisson tensor has rank two there and is tangent to each level set, so it inverts to a symplectic form; each sphere is a symplectic leaf. Rotations preserve both $F$ and the alternating tensor ${ϵ}_{abc}$, hence preserve the restricted Poisson tensor and its inverse symplectic form. The $SO(3)$ action therefore restricts to a symplectic action on every sphere $S_R^2\subset {\mathbb{R}}^3$.

Over $\mathbb{C}$, the real skew-symmetric matrix $A$ is normal because $A^{\ast }=-A$. Part (a) gives an orthonormal complex eigenbasis, and the nonzero eigenvalues are purely imaginary pairs $\pm i\lambda$.

For each $\lambda >0$, choose a unit real vector $x$ in the kernel of $A^2+{\lambda }^{2}I$ and putThen $x\perp y$, $\Vert y\Vert =1$, andThus $A$ has matrixon the orthonormal basis $(x,y)$. Distinct such invariant planes are orthogonal; complete them by an orthonormal basis of $\ker A$. Taking these basis vectors as the columns of an orthogonal matrix $R$ gives

If $X$ belongs to the Special orthogonal Lie algebra, which is also the Lie algebra of $O(3)$, then $e^{tX}\in O(3)$ near $t=0$. Differentiatingat zero gives $X^{\mathsf{T}}+X=0$. Conversely, the matrix exponential of every real skew-symmetric matrix is orthogonal. Hence

Differentiate the squared Euclidean norm and use the skew-symmetric matrix identity $A^T=-A$:Thus $\Vert \mathbf{y}(t){\Vert }_2^2$ is constant, and continuity of the nonnegative square root gives

Set $\overline{\mathbf{y}}=({\mathbf{y}}_n+{\mathbf{y}}_{n+1})/2$. The implicit midpoint rule readsTaking the inner product with ${\mathbf{y}}_{n+1}+{\mathbf{y}}_n=2\overline{\mathbf{y}}$ givesThe last equality again follows from skew-symmetry. Mathematical induction therefore yields

Write $A_n=A({\mathbf{y}}_n)$. The trapezoidal rule isTaking the inner product with ${\mathbf{y}}_{n+1}+{\mathbf{y}}_n$ givesThe two quadratic terms vanish because each $A_j$ is skew-symmetric. Moreover ${\mathbf{y}}_{n+1}^T{A}_n{\mathbf{y}}_n=-{\mathbf{y}}_n^T{A}_n{\mathbf{y}}_{n+1}$. HenceUnlike the implicit midpoint rule, the trapezoidal rule evaluates $A$ at two different states, so the mixed terms need not cancel.

Take the real $L^2$ inner product of the equation with $u$. The homogeneous Dirichlet boundary conditions and integration by parts giveThus $\Vert u(t){\Vert }_{L^2}\le \Vert u_0{\Vert }_{L^2}$, and applying the same estimate to the difference of two solutions gives continuous dependence on the initial data. The constant-advection term is skew-symmetric under these boundary conditions and contributes no energy. Hence

Write the semidiscrete system as ${\mathbf{u}}^{\prime }=L_h\mathbf{u}+\alpha D_h\mathbf{u}$. With zero boundary values, the centered second-difference matrix $L_h$ is symmetric negative definite and the centered first-difference matrix $D_h$ is skew-symmetric. ThereforeThis is a mesh-uniform stability estimate for the centered convection-diffusion semidiscretization, valid for every real $\alpha$.

Both centered differences have local spatial error $O(h^2)$ for a sufficiently smooth solution. Stability plus consistency gives convergence, equivalently by the semidiscrete form of the Lax equivalence theorem. Thus

Choose a basis $u_1,\dots ,u_r$ of $U$, extend it to a basis $u_1,\dots ,u_r,w_1,\dots ,w_s$ of $W$, and extend again to a basis of $V$ by adding $v_1,\dots ,v_t$. DefineEvery member of the full basis lies in $W$ or $W^{\prime }$, so the sum of vector subspaces is $W+W^{\prime }=V$. If $x\in W\cap W^{\prime }$, express it in both displayed bases. Uniqueness of the coordinates in the full basis forces its $w_j$ and $v_j$ coefficients to vanish. Thus $x\in U$. The reverse inclusion is built into the construction, proving $W\cap W^{\prime }=U$. This is a subspace supplement with a prescribed intersection.

For the matrix example, letThe skew-symmetric matrices are exactly $U=\mathrm{span}\{J\}$, and the traceless matrices form $W$. A suitable answer isThe matrix trace of $aI+bJ$ is $2a$, so $W\cap W^{\prime }=\mathrm{span}\{J\}=U$. Every $2\times 2$ real matrix has the decompositionwhose first term lies in $W$ and second in $W^{\prime }$. Therefore $W+W^{\prime }=M_2(\mathbb{R})$, as required.

The transpose is a linear map, so $T(A)=A+A^T$ is a linear map on the $n^2$-dimensional vector space $M_{n\times n}(F)$. Let $E_{ij}$ denote a matrix unit. The answer depends on the characteristic of a field.

If $\mathrm{char}F\ne 2$, then $T(A)=0$ means $A^T=-A$. Such skew-symmetric matrices have zero diagonal, and their entries above the diagonal can be chosen freely. A basis of the kernel of a linear map is $E_{ij}-E_{ji}$ for $i<j$, giving nullity $n(n-1)/2$. The image of a linear map consists of all symmetric matrices: $T(A)$ is symmetric, and every symmetric matrix $S$ is $T(S/2)$. Its basis comprises $E_{ii}$ and $E_{ij}+E_{ji}$ for $i<j$, giving rank of a linear map $n(n+1)/2$.

If $\mathrm{char}F=2$, then $-A=A$, so $\ker T$ consists of all symmetric matrices, with dimension $n(n+1)/2$. Every $T(A)$ is symmetric with zero diagonal, and every symmetric zero-diagonal matrix $S$ arises by taking $A_{ij}=S_{ij}$ for $i<j$ and all other entries zero. The image of a linear map therefore has basis $E_{ij}+E_{ji}$ for $i<j$, and dimension $n(n-1)/2$. This is the symmetrization map in characteristic two.

In both cases the rank-nullity theorem is verified explicitly:

Let $A$ be a skew-symmetric matrix. For every real vector $x$, the real scalar $x^{T}Ax$ equals its own transpose, butHence $x^{T}Ax=0$. If $Av=\lambda v$ is a real eigenvector equation with $v\ne 0$, it follows thatso $\lambda =0$. If $A$ is a diagonalisable matrix over $\mathbb{R}$, it has a real basis of eigenvectors, all with eigenvalue zero. Thus $A$ vanishes on a basis and $\boxed{A=0_n}$.

The field matters: a nonzero real skew-symmetric matrix can be a diagonalisable matrix over the complex numbers, for example $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ with eigenvalues $i,-i$.

Take the vector subspace of skew-symmetric matrices,It is closed under vector addition and scalar multiplication, and a basis isIndeed, its diagonal entries vanish, and its entries below the diagonal are forced by those above it. Hence $\boxed{\dim K=\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$. Part (a)(ii) proves that the only diagonalisable matrix in $K$ is $0_n$.

Yes, in even dimension. For example, choose the skew-symmetric matrixHere $B^2=-I$, so $(I+B{)}^{-1}=(I-B)/2$ and $C=(I-B{)}^2/2=-B$. It is an orthogonal skew-symmetric matrix. More generally the skew-symmetry condition $C=-C^{-1}$ gives $(I-B{)}^2=-(I+B{)}^2$, or $B^2=-I$; no real odd-dimensional matrix can satisfy that condition.

The symmetric part of a matrix isIt satisfies $S^{\mathsf{T}}=S$. The complementary term $(A-A^{\mathsf{T}})/2$ is a skew-symmetric matrix.

The original PDF requires $B$ itself to be an orthogonal skew-symmetric matrix; it does not require $C$ to be diagonal. A real two-dimensional skew-symmetric matrix has the form $B=\left(\begin{array}{cc}0& u\\ -u& 0\end{array}\right)$. The orthogonal matrix condition gives $u^2=1$, so choose $u=1$. Then $C=B^{\mathsf{T}}AB$ is forced:Direct multiplication verifies $B^{\mathsf{T}}=-B$, $B^{\mathsf{T}}B=I$ and $BC{B}^{\mathsf{T}}=A$. Choosing $-B$ gives the same $C$. Replacing $B$ by a matrix of eigenvectors to make $C$ diagonal would generally fail the printed antisymmetry requirement.

Commuting inverse. Assume $I-M$ is invertible, as required by the displayed inverse. Multiplication givesMultiply on both sides by $(I-M{)}^{-1}$ to obtainMore generally, a matrix commuting with an invertible matrix also commutes with its inverse.

Orthogonality. For a real skew-symmetric matrix $A$, $x^{T}Ax=0$. If $(I-A)x=0$ for a real vector $x$, multiply by $x^T$ to obtain $\Vert x{\Vert }^2=0$. Thus $I-A$ is invertible; the same argument applies to $I+A$. For the Cayley transform of a real skew-symmetric matrix,The factors commute by the first result, soTherefore $\boxed{N\text{ is an orthogonal matrix}.}$

Eigenvector mapping. If $Av=av$, then $(I-A{)}^{-1}v=(1-a{)}^{-1}v$. ConsequentlyThe denominator cannot vanish. In fact the eigenvalues of a real skew-symmetric matrix are purely imaginary: $v^{†}Av$ is the negative of its complex conjugate, so $a=-\overline{a}$. For $a=i\beta$ the corresponding eigenvalue of $N$ is $(1+i\beta )/(1-i\beta )=e^{2i\arctan \beta }$, consistent with orthogonality.

Spectrum in three dimensions. Let $Q$ be any real three-by-three orthogonal matrix. For a complex eigenvector $v$, $\Vert Qv\Vert =\Vert v\Vert$ implies $|\lambda |=1$. A real cubic characteristic polynomial has a real root, necessarily $+1$ or $-1$. Any nonreal roots form a complex-conjugate pair of unit modulus. If all three roots are real, two of the three signs agree and can be grouped as a pair with angle zero or $\pi$. Thus in every case the eigenvalues can be writtenTheir product is $\epsilon =\det Q$.

Determinant and fixed axis for the Cayley transform. Since $A^T=-A$,so $\boxed{\det N=1}$. This determinant argument works in every dimension. In dimension three the preceding spectrum result then guarantees an eigenvector with $Nx=x$. Set $z=(I-A{)}^{-1}x$. The fixed-vector equation says $(I+A)z=(I-A)z$, so $Az=0$ and $x=(I-A)z=z$. HenceThis is the axis fixed by the three-dimensional rotation. Equivalently, a real odd-dimensional skew-symmetric matrix has zero determinant, because $\det A=\det A^T=\det (-A)=-\det A$.

For a skew-symmetric matrix, $M^T=-M$. Invariance of the determinant under transpose givesFor odd $n$, $2\det M=0$, so $\det M=0$ over the real or complex numbers.

Collect the paired orthonormal bases from part (iv), followed by an orthonormal basis of $\ker M^2=\ker M$. The matrix $Q$ of these columns is an orthogonal matrix, andIndeed, each pair obeys $M{e}_j={\mu }_j{f}_j$, $M{f}_j=-{\mu }_j{e}_j$, and all kernel vectors are annihilated. This proves the real block form of a skew-symmetric matrix directly, including repeated block parameters. In particular its rank is $2r$, always even.