Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 5

Let $A{v}_j=a_j{v}_j$. Commutation gives $A(B{v}_j)=B(A{v}_j)=a_{j}B{v}_j$, so $B$ preserves each eigenspace of $A$. Since the eigenvalues of $A$ are nondegenerate, each eigenspace is one-dimensional and $B{v}_j=b_j{v}_j$. The spectral theorem for a Hermitian matrix lets us choose these vectors as an orthonormal eigenbasis. With $U=(v_1,\dots ,v_n)$,This is simple spectrum gives simultaneous diagonalization. The common eigenvectors can be chosen up to their phases and ordering. In fact degeneracy does not prevent a common eigenbasis for commuting Hermitian matrices, but the one-dimensional argument already suffices here.

The convergent matrix exponential series respects similarity, so $e^A=U{e}^{{\Lambda }_A}{U}^{†}$ and likewise for $B$. Scalar exponentials multiply on each diagonal entry, givingThe essential hypothesis is commutation; Hermiticity supplies this convenient diagonalization proof.

Expand both sides through second order as $ϵ\to 0$, holding $X,Y$ fixed. The product isPut $Z=ϵ(X+Y)+{ϵ}^2[X,Y]/2$. Its exponential isThusThis is the quadratic part of the Baker--Campbell--Hausdorff formula. The remainder estimate is in any fixed finite-dimensional matrix norm, and is an asymptotic statement rather than an exact formula for arbitrary finite $ϵ$.

Use the lower antisymmetric block shown in the original PDF. Write $M=\mathrm{diag}(aK,bJ)$ withPowers preserve the block structure. Splitting the matrix exponential series into even and odd powers gives $e^{aK}=I\cosh a+K\sinh a$. For the other block, the matrix exponential when the square is minus the identity gives $e^{bJ}=I\cos b+J\sin b$. ThereforeThe upper block is hyperbolic and the lower block is a rotation. The transcription's altered last row and hyperbolic functions in the lower block would describe a different problem.