Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 6

A skew-Hermitian matrix satisfies $A^{†}=-A$, where $†$ denotes the conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenSince $v^{†}v>0$, this gives $\boxed{\overline{\lambda }=-\lambda ,\text{ so }\lambda \in i\mathbb{R}}$.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. It preserves the Hermitian inner product and norm. Thus if $Uv=\mu v$,and therefore $\boxed{|\mu |=1}$. Zero is allowed among the purely imaginary eigenvalues in the first statement, but not among the unitary eigenvalues.

Let $A{v}_i={\lambda }_i{v}_i$ and $A{v}_j={\lambda }_j{v}_j$. The skew-Hermitian matrix property giveswhere the last equality uses part (a). Distinct eigenvalues therefore imply $\boxed{v_i^{†}{v}_j=0\text{ for }i\ne j}$. Normalize each eigenvector to length one. Since there are $n$ distinct eigenvalues, these vectors form an orthonormal basis.

Put them in the columns of $U$. Then $U$ is a unitary matrix and $AU=UD$, so

To avoid confusing the diagonalizing matrix with the exponential, write $A=WD{W}^{†}$ from part (b). The convergent matrix exponential series givesEach ${\lambda }_j$ is purely imaginary, so $|e^{{\lambda }_j}|=1$. The diagonal exponential is therefore a unitary matrix, andHence the matrix called $U=\exp (A)$ in this part is unitary. The series identity $(e^A{)}^{†}=e^{A^{†}}=e^{-A}$ gives the same conclusion directly and also covers repeated eigenvalues.

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.