Symmetric and Skew Symmetric Matrices $V_0=$ the set of $2\times2$ skew symmetric matrices
I know that any element of $V_0$ has a characteristic polynomial that will not factor over the real numbers, and therefore has no eigenvectors.  Why is this so?  How would I go about showing this?
 A: Let us examine the chacteristic polynomial $p_S(\lambda)$ of $S$, a skew-symmetric $2 \times 2$ matrix.  As is well-known, and in fact easily derived, we have
$p_S(\lambda) = \lambda^2 - \text{Tr}(S)\lambda + \det (S). \tag{1}$
Writing
$S = [s_{ij}], \tag{2}$
we see that the skew-symmetry of $S$ implies the diagonal entries $s_{ii} = 0$,
since $S^T = -S$ implies $s_{ii} =-s_{ii}$ for $i = 1,2$.  Thus the trace of $S$ vanishes, and furthermore we have
$\det(S) = -s_{12}s_{21} = s_{12}^2 >0 \tag{3}$
if $S \ne 0$.  Thus
$p_S(\lambda) = \lambda^2 + s_{12}^2; \tag{4}$
but the roots of (4) are $\pm i\vert s_{12} \vert$, both purely imaginary, so $p_A(\lambda)$ is irreducuble over $\Bbb R$.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: Here's one way of getting there:
Let $A$ be a real, skew-symmetric matrix. Let $\lambda$ be a real eigenvalue of $A$, and $v$ an associated eigenvector.  We then note that
$$
\lambda(v \cdot v) = (\lambda v) \cdot v = (Av) \cdot v 
\\= v^T (Av) = v^T A v = -v^T A^T v = -(Av)^T v \\
= v\cdot (-Av) = v\cdot(-\lambda v) = -\lambda (v \cdot v)
$$
So, if $\lambda$ is a real eigenvalue, we have $\lambda \|v\|^2 = -\lambda \|v\|^2$, which (since $v$ is by definition non-zero) implies that $\lambda = 0$.
So, if $A$ is skew-symmetric, the only eigenvalue it can have is $0$.
A: Suppose $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ is skew-symmetric. Then, $A+A^T = \begin{bmatrix}2a&b+c\\b+c&2d\end{bmatrix} = 0$.
What does this tell you about $a,b,c,d$? This lets you write $A$ in terms of just $b$. 
From there, it should be easy to show that $A$ has no real eigenvalues. 
A: The set of real skew-symmetric $2\times 2$ matrices is a $1$-dimensional real vector space. Given that if a matrix is diagonalisable, so are all its scalar multiples (with the same eigenvectors, and eigenvalues multiplied by the scalar in question), it suffices to test the condition for a single nonzero real skew-symmetric $2\times 2$ matrix. Choosing $(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix})$ for simplicity one sees its characteristic polynomial is $X^2+1$ and has no real roots: also the matrix represents a rotation by a quarter turn which clearly has no real eigenvalues (it fixes no lines).
Of course, as was commented, this only shows that nonzero real skew-symmetric $2\times 2$ matrices are not diagonalisable, and the zero matrix is of course diagonalisable.
A: If $A$ is a $2 \times2$ skew symmetric metric, then $A$ has the form $$A = \begin{bmatrix}0&a\\-a&0\end{bmatrix}$$ 
Supose $\lambda_1,\lambda_2$ are  the eigen values of $A.$
Then $tr(A)=\lambda_1+\lambda_2=0,$ and $det(A)=\lambda_1\lambda_2=a^2.$
Clearly the eigen values are $\lambda_1=ia$ and $\lambda_2=-ia.$
Therefor no non zero real vectors $v$ such that $$Av= \lambda v$$ for real $a \not =0. $
