Are all the eigenvalues of Antisymmetric matrix must be imaginary (ti)?

Question

(The field is $R$ - reals) Antisymmetric matrix ($A^t=-A$) is a normal matrix ($A^tA=AA^t$) because $A^tA=-AA=-A^2$. $$AA^t=A(-A)=-AA=-A^2\ .$$

So, are all the eigenvalues of $A$ have to be imaginary ($t=a+ib$ where $a=0$) ? It makes sense but I don't know how to prove it.

Thanks

Answer 1

You can consider the same matrix in the complex space; this does not change the eigenvalues. Now if $Au=\lambda u$, then $\lambda||u||^2=(u,\lambda u)=(u,Au)=-(Au,u)=-(\lambda u,u)=-\bar\lambda||u||^2$. Thus, $\lambda=-\bar\lambda$ and hence $\lambda$ is pure imaginary. However, as @Surb rightly says, if you exclude zero from imaginary numbers, the statement will not be true.

Answer 2

This is not true in general. The $0$ matrix has only real eigenvalues (0) and is anitsymmetric. You can also consider $$A=\begin{pmatrix} 0 & 1& 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},$$ then $A\neq 0$ is antisymetric and has $0$ as eingevalue since it is not invertible. But as shown by Vladimir, the only possible real eigenvalue of an antisymmetric matrix is $0$.