TITLE: Does this $3\times 3$ matrix exist?
QUESTION [12 upvotes]: Does a real $3\times 3$ matrix $A$ that satisfies conditions $\operatorname{tr}(A)=0$ and $A^2+A^T=I$ ($I$ is an identity matrix) exist?
Thank you for your help.

REPLY [5 votes]: I consider the case where $A$ is restricted to real matrices so $A^\top=A^*$.
Let  $\lambda$ be any eigenvalue of $A$, and let $x$, $\|x\| > 0$ be a corresponding eigenvector.
Multiply on the left by $x^*$ and on the right by $x$ to get
$x^* (A^2 x) + (Ax)^*x = \|x\|^2$, i.e., $\lambda^2\|x\|^2 + \overline{\lambda}\|x\|^2 = \|x\|^2$. 
As the right hand side is non-zero we must have $\lambda \neq 0$ and $ \lambda^2 + \overline{\lambda} = 1.$ 
Let $\lambda = a + ib$ with $a,b$ real. We get $$a^2 -b^2 + a = 1$$ and $$2ab - b = 0.$$ Note, $a \neq 1/2$ as then $a^2 + a - 1 = b^2$, a consequence of the first equation becomes impossible. SO the second equation above implies $b=0$ and all eigenvalues are real and are roots of $x^2 + x - 1 = 0$. The rest follows as above solution by Michael Grant, i.e., the trace being the sum of three numbers in $\{ \frac{-1 \pm \sqrt{5}}{2} \}$ cannot be $0$.