Let A be a real skew-symmetric matrix and ($\lambda,v)$ eigenpair. Since A is skew-symmetric, the associated opearator is also skew-symmetric and we may write:
$\langle\lambda v, v\rangle = \langle T(v),v\rangle=-\langle v, T(v)\rangle = -\bar{\lambda}\langle v,v\rangle$
Thus, we have that $\lambda = -\bar{\lambda}$, showing that $\lambda$ is purely imaginary.
Ok. My question is: we are considering a matrix A over the reals. Since its eigenvalues are all complex, is it still diagonalizable?
Best Answer
For skew-symmetrix matrices, first consider $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$. It's a rotation by 90 degrees in $\mathbb{R}^2$, so over $\mathbb{R}$, there is no eigenspace, and the matrix is not diagonalizable. It is of course, diagonalizable over $\mathbb{C}$ though.
See here for the corresponding statement about complex skew-symmetric matrices using unitary matrices instead of orthogonal ones. Note that the blocks in the matrix $\Sigma$ at this link are themselves diagonalizable, so $\Sigma$ is diagonalizable. Is $\Sigma$ diagonalizable with unitary matrices?