Let \(A\) be an \(n \times n\) matrix with complex entries such that \(A^* = -A\), and let \(B = e^A\). Show that
\(\det B = e^{\operatorname{tr} A}\);
\(B^* = e^{-A}\);
\(B\) is unitary.
Solution
1. A skew-adjoint matrix is unitarily diagonalizable with purely imaginary characteristic values.
Since \[ A^*=-A, \] one has \[ AA^*=A(-A)=-A^2=A^*A, \] so \(A\) is normal. By the spectral theorem there is a unitary matrix \(U\) and scalars \(\lambda_1,\ldots,\lambda_n\) such that \[ A=U\operatorname{diag}(\lambda_1,\ldots,\lambda_n)U^*. \] If \(Av=\lambda v\) with \(v\ne0\), then \[ \lambda\|v\|^2=(Av\mid v) =-(v\mid Av) =-\overline\lambda\|v\|^2, \] so \(\lambda=-\overline\lambda\). Thus every \(\lambda_j\) is purely imaginary.
2. Prove \(\det(e^A)=e^{\operatorname{tr}A}\).
The matrix exponential respects similarity, and for a diagonal matrix it is computed entrywise. Hence step 1 gives \[ e^A =U\operatorname{diag}(e^{\lambda_1},\ldots,e^{\lambda_n})U^*. \] Therefore \[ \det(e^A) =\prod_{j=1}^n e^{\lambda_j} =e^{\sum_j\lambda_j}. \] But \[ \operatorname{tr}A=\sum_j\lambda_j, \] so \[ \boxed{\det B=e^{\operatorname{tr}A}}. \]
3. Prove \(B^*=e^{-A}\).
From the absolutely convergent power series \[ e^A=\sum_{k=0}^\infty\frac{A^k}{k!} \] and continuity of adjoint in finite dimensions, \[ \begin{aligned} B^* &=\left(\sum_{k=0}^\infty\frac{A^k}{k!}\right)^*\\ &=\sum_{k=0}^\infty\frac{(A^*)^k}{k!}\\ &=\sum_{k=0}^\infty\frac{(-A)^k}{k!}\\ &=e^{-A}. \end{aligned} \] Thus \[ \boxed{B^*=e^{-A}}. \]
4. Conclude that \(B\) is unitary.
Because \(A\) commutes with \(-A\), \[ e^{-A}e^A=e^0=I. \] Using step 3, \[ B^*B=e^{-A}e^A=I. \] Hence \[ \boxed{B\text{ is unitary}.} \]