Real Skew Symmetric $3\times 3$ matrix has one eigen value $2i$ Real Skew Symmetric $3\times 3$ matrix has one eigen value $2i$ so another eigen value is $-2i$ another eigen value must be $0$ right? as we know these kind of matrix has eigen values $0$ or purely imaginary.
 A: In general, a real skew-symmetric $3 \times 3$ matrix $K$ looks like
$K = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}, \tag{1}$
so that the characteristic polynomial is
$\det(\lambda - K) = \lambda^3 + (a^2 + b^2 + c^2) \lambda, \tag{2}$
which is easily seen to have roots
$\lambda = 0, \pm i\sqrt{a^2 + b^2 + c^2}, \tag{3}$
which follow the general pattern the OP suggested; apparently the matrix she/he was thinking of satisfies $\sqrt{a^2 + b^2 + c^2} = 2$; but $0$ will always be an eigenvalue for such $K$ in any case.
Hope this helps.  Cheers.
A: Since a real matrix $A$ is skew if, and only if, $A=-A^T$, since $\text{spec} (A)=\text{spec}(A^T)=-\text{spec}(-A^T)$ and since the characteristic polynomial has degree $3$ and non-real roots come in pairs, it follows that $0\in\text{spec}(A)\cap \text{spec}(A^T)$.
A: Yes, the characteristic polynomial of a real $3\times 3$ matrix will be a real cubic polynomial.  If not identically zero, the real skew matrix will have one pair of conjugate imaginary eigenvalues and a zero eigenvalue, each of multiplicity one.
A: We know that a skew symmetric matrix $A=(a_{ij})_{3 \times 3}$ satisfies $A^T=-A$.  This means that $a_{ii}=-a_{ii}$ or equivalently $a_{ii}=0$ for all $i \in \{1,2,3\}$.
We know the trace of $A$ is the sum of its eigenvalues, which is $0$ since the main diagonal of $A$ comprises of zeroes.
Since a $3 \times 3$ matrix has $3$ (not necessarily distinct) eigenvalues (given by the roots of the characteristic polynomial [a degree $3$ polynomial with leading coefficient either $1$ or $(-1)^3$, depending on how it's defined]), there is precisely one other eigenvalue $\lambda$ and we must have $2i-2i+\lambda=0$, which implies $\lambda=0$.
A: Here's another way of looking at it, which I thought of after posting my previous answer.
The approach here seemed sufficiently different that I felt a separate answer was merited.  Hope I'm not overdoing it.  Anyway . . . 
First, note that for any skew-symmetric, real matrix, the only possible real eigenvalue is zero.  For, if $K$ is real skew-symmetric, so that $K^T = -K$, and if
$v \ne 0$ is an eigenvector with real eigenvalue $\lambda$, so that
$Kv = \lambda v, \tag{1}$
we have
$\langle Kv, v \rangle = \lambda \langle v, v \rangle, \tag{2}$
but
$\langle Kv, v \rangle = \langle v, K^Tv \rangle = \langle v, -Kv \rangle = -\langle Kv, v \rangle, \tag{3}$
or
$\lambda \langle v, v \rangle = -\lambda \langle v, v \rangle, \tag{4}$
and since $\langle v, v \rangle \ne 0$ this shows that 
$\lambda = - \lambda, \tag{5}$
whence
$\lambda = 0. \tag{6}$
Here I take $\langle \cdot, \cdot \rangle$ to be the standard inner product on the real vector space on which $K$ operates.
Now argue as follows:  since $2i$ is an eigenvalue of $K$, and the characteristic polynomial of $K$ is real, $-2i$ is an eigenvalue as well.  This means
$\lambda^2 + 4 \mid p_K(\lambda), \tag{7}$
where
$p_K(\lambda) = det(\lambda - K) \tag{8}$
is the characteristic polynomial of $K$.  But $\deg p_K(\lambda) = 3$ in the present case, so the quotient polynomial $q(\lambda)$ is of degree $1$, i.e. we must have
$p_K(\lambda) = (\lambda^2 + 4)q(\lambda) \tag{9}$
with
$q(\lambda) = \lambda - a, \tag{10}$
$a$ real.  But this of course implies $a$ is a real eigenvalue of $K$, whence 
we must have
$a = 0. \tag {11}$
Well, I hope this answer is more than mere mathematical logorrhea, and it sheds a little more light on the subject at hand.
Cheers!
