IMC 2025 · Day 2 · P8

If $\lambda = 0$, the claim holds as $0 \in \mathbb{R}$. Assume $\lambda \ne 0$ is an eigenvalue of $A$ with a corresponding eigenvector $x \in \mathbb{C}^n \setminus \{0\}$.

We will first express the operation $A \mapsto A^R$ algebraically. The element at position $(i, j)$ in $A$ ends up at position $(n + 1 - j, i)$ in $A^R$. Thus, the rotation is defined by the relation $(A^R)_{i,j} = A_{j, n+1-i}$.

Let $J$ be the matrix where $J_{i,j} = \delta_i^{n+1-j}$. The operation of transposing $A$ and then reversing the rows gives the matrix $J A^\top$. The $(i,j)$-th element of this matrix is $$(J A^\top)_{i,j} = \sum_{k=1}^{n} J_{i,k} (A^\top)_{k,j} = (A^\top)_{n+1-i, j} = A_{j, n+1-i}.$$ This matches the definition of $A^R$, so we get the identity $A^R = J A^\top$. Note that the matrix $J$ is symmetric ($J = J^\top$) and it is its own inverse ($J^2 = I$).

The given condition $A = A^R$ thus means $A = J A^\top$. Left-multiplying by $J$ yields $$\tag{$*$} J A = J (J A^\top) = (J^2) A^\top = A^\top.$$ Now, consider the standard Hermitian inner product $(u, v) = v^* u$ on $\mathbb{C}^n$. We evaluate $(Ax, Ax)$ in two ways. First, using our choice of $x$ as an eigenvector corresponding to $\lambda$: $$(Ax, Ax) = (\lambda x, \lambda x) = |\lambda|^2 \|x\|^2.$$ Second, using the adjoint property and $(*)$: $$(Ax, Ax) = (A^* A x, x) = (A^\top A x, x) = (J A (\lambda x), x) = \lambda (J A x, x) = \lambda^2 (J x, x).$$ Together, these give us $|\lambda|^2 \|x\|^2 = \lambda^2 (Jx, x)$.

The term $(Jx, x)$ is real, since $(Jx, x)^* = (J^* x, x) = (Jx, x)$ because $J$ is real and symmetric. Since $\lambda \ne 0$ and $x \ne 0$, the left side $|\lambda|^2 \|x\|^2$ is a positive real number. This implies that $\lambda^2 (x, Jx)$ must also be a positive real number. And as $(x, Jx)$ is real, so is $\lambda^2$.

Thus, either $\lambda$ is real (if $\lambda^2 > 0$) or its real part is 0 (if $\lambda^2 < 0$). This completes the proof.