TITLE: Existence of some type matrix
QUESTION [9 upvotes]: Is there square matrix $A$ of size $3$ with real entries such that 
$$
\operatorname{tr}(A)=0\text{ and }A^2+A^T=I.
$$ 
I have proved that there is not with size $2$ using definition of  "trace", but for size $3$ it becomes complicated.
Here is the sketch of the proof for $2$.
$$
a_{11}+a_{22}=0,\\
a_{11}^2+a_{12}a_{21}+a_{11}=1,\\
a_{11}a_{12}+a_{12}a_{22}+a_{21}=0,\\
a_{21}a_{11}+a_{21}a_{22}+a_{12}=0,\\
a_{12}a_{21}+a_{22}^2+a_{22}=1.
$$
Putting $a_{11}=-a_{22}$ it is easy to see that above inequalities can't be true together. 
Thanks!

REPLY [2 votes]: The matrix $A$ is normal (and hence diagonalizable), since
$$AA^T=A(I-A^2)=(I-A^2)A=A^TA.$$
Further it is easy to show (see vadim123's answer) that $A$ satisfies the equation
$$A^4-2A^2+A=0,$$
hence its eigenvalues $\lambda_{1,2,3}$ satisfy it as well. 
The possibilities to get zero trace are then $\lambda_1=\lambda_2=\lambda_3=0$ (immediately excluded as $A$ cannot be zero) and $\lambda_1=1$, $\lambda_2=\frac{-1+\sqrt5}{2}$, $\lambda_3=\frac{-1-\sqrt5}{2}$. In the latter case $A^T=I-A^2$ has zero eigenvalue which leads to a contradiction.