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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?

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  • $\begingroup$ The Spectral Theorem guarantees that if you have a matrix that is skew-symmetric, then the eigenvalues are purely imaginary. Where did you hear that the eigenvalues are zero? $\endgroup$ – EpicMochi Nov 3 '14 at 5:34
  • $\begingroup$ T is skew-symmetry if <T(x).x>= - <x.(Tx)> . Since inner product is real number, it has to be zero? $\endgroup$ – dkim Nov 3 '14 at 5:37
  • $\begingroup$ Thats what my book said $\endgroup$ – dkim Nov 3 '14 at 5:37
  • $\begingroup$ Inner product need not be a real number, depending on what space we're talking about. Take $$A = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)$$ and $$v = (1,i)^t$$ Observe that $<Av,v> = i\sqrt{2}$. Remember that the inner product over complex spaces requires us to take a complex conjugate of the second vector in our inner product. $\endgroup$ – EpicMochi Nov 3 '14 at 6:10
  • $\begingroup$ When your book said that the eigenvalues are 0, what field was it working with? $\endgroup$ – EpicMochi Nov 3 '14 at 6:12

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