If A is a matrix such that $A^{2}+A+2I=O$ ,then $A$ can't be skew symmetric. 
If $A$ is a matrix such that $A^{2}+A+2I=O$, then $A$ can't be skew symmetric. (True/false)

When $A$ is odd order matrix then the statement is true, since $A$ is non singular. $( |A||A+I| = (-2)^{n})$ and skew symmetric matrix of odd order is singular.
How to check for even case$?$ 
 A: Let $A$ be a matrix such that 
$$A^2+A+2I=0\tag{1}.$$ Then also
$$0=0^{\top}=(A^2+A+2I)^{\top}=(A^{\top})^2+A^{\top}+2I.$$
Suppose $A$ is skew-symmetric, so that $A^{\top}=-A$. Then it follows that
$$0=(-A)^2+(-A)+2I=A^2-A+2I,\tag{2}$$
and subtracting this from $(1)$ then yields $2A=0$ and hence $A=0$. But clearly this does not satisfy $(1)$, a contradiction. So $A$ can't be skew symmetric.
A: $A^{T}=-A$. Taking transpose in the given equation we get $(-A)^{2}-A+2I=0$. subtracting this from the given equation we get $A=0$ but then the given equation fails. 
A: $$A^2+A+I=0\implies A^3=I\\\implies (A^T)^3=I\implies -A^3=I$$both together imply the impossible equality $I=O$ which is a contradiction.
A: More generally. 
Let $K$ be an algebraic closed field with $charac(K)\not= 2$ and $A\in M_n(K)$ be skew symmetric. 
$\textbf{Proposition}$. Let $P\in K[x]$ s.t. $P(A)=0$. Then, necessarily, $P$ contains the factor $x$ or a factor in the form $x^2-a^2$ where $a\in K^*$. 
$\textbf{Proof}$. The spectrum of $A$ is in the form $\{a_1,-a_1,\cdots,a_k,-a_k,0_{n-2k}\}$ where the $(a_i)$ are in $K^*$ and are not necessarily distinct.
