My textbook defines a skew-Hermitian and skew-Symmetric operator as $(T(x),x) = -(x,T(x))$. It is Hermitian when the inner-product is complex, Symmetric when inner-product is real. And it has nice proofs that eigenvalue $ \lambda = \frac{(T(x),x)}{(x,x)} , \bar\lambda = \frac{(x,T(x))}{(x,x)}$.
So, in case of skew-Hermitian $\lambda = -\bar\lambda,\lambda $ is purely imaginary. But for skew-Symmetric, since the euclidean space is real, and $\lambda = -\bar\lambda$, the eigenvalue(s) are both real and imaginary, i.e. all of them are zero.
So far all good, but this is not matching with a simple skew-symmetric example I am checking. $$ A = \begin{pmatrix} 0 & a\\-a&0\end{pmatrix} , $$ $$(\lambda I-A)=0 \Rightarrow \begin{pmatrix} \lambda-0 & -a\\a&\lambda-0 \end{pmatrix} = 0 \Rightarrow \lambda^2 = -a^2 \therefore \lambda = +ia,-ia$$ As $a$ is any random real number, the eigenvalues for the two dimensional skew-symmetric matrix are pure imaginary, contradictory to what is written in my textbook. Can someone please point out where is the fallacy in my understanding? If $A$ does not belong to skew-symmetric space (with real inner product), (why??) then can you give an example for a skew-symmetric matrix with real inner product (and zero eigenvalues only)?