If transformation $T$ is skew-symmetric, then eigenvalues of $T$ must be zero. However, if there is a matrix representation of $T$ and that matrix is skew-symmetric, it seems to me that eigenvalues could be pure imaginary numbers.
How is this possible?
For example, suppose I have a matrix with values $a_{11}=2, a_{12}=0, a_{21}=0, a_{22}=-2$.
This matrix is skew-symmetric and eigenvalues are $2i$ and $-2i$.
However, if a matrix is skew-symmetric, does that mean that transformation is also skew-symmetric and therefore eigenvalues must be zero?