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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?

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  • $\begingroup$ Well, it obviously is not diagonalisable over $\Bbb R$ (if $A\ne0$): its characteristic polynomial has some non-real roots. $\endgroup$ – user228113 Dec 8 '16 at 10:51
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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?

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If $A$ is a skew-sym. matrix with real entries, then there are 3 possibilities:

  1. $A=0$. Then a is diagonalizable.

  2. if $A$ is regarded as a member of $M_n(\mathbb R)$ and $A \ne 0$, then $A$ is not diagonalizable.

  3. if $A$ is regarded as a member of $M_n(\mathbb C)$ , then $A$ is diagonalizable, since $A$ is normal.

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