 Solution

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