TITLE: How many $3 \times 3$ non-symmetric and non-singular matrices $A$ are there such that $A^{T}=A^2-I$?
QUESTION [5 upvotes]: How many $3 \times 3$ non-symmetric and non-singular matrices $A$ are there such that $A^{T}=A^2-I$?
Note: $I$ denotes the identity matrix of size $3 \times3$ and $A^{T}$ represents the transpose of the matrix $A$.
I took transpose on both sides in given equation to get $A=(A^T)^2-I$ and then I put value of $A^T$ in this equation using $A^{T}=A^2-I$ to get $A^3-2A-I=0$ which ultimately gave $(A^T-A)(A+I)=0$ . How to deal the problem from here? And is there any better approach to tackle this problem? The given answer is $0$.

REPLY [3 votes]: I would like to say why the above dan_fulea's comment is good and Ian's comment (upvoted!) is inappropriate and also why it is assumed that $A$ is invertible. Note also that the hypothesis $n=3$ has nothing to do here. Assume that $A\in M_n(K)$. 
Case 1. $K=\mathbb{R}$. Since $AA^T=A^TA$, $A$ is normal and therefore unitarily diagonalizable; then we may assume that $A=diag(\lambda_i),A^T=diag(\overline{\lambda_i})$ where $\overline{\lambda_i}=\lambda_i^2-1$. The previous equation has only real solutions and, consequently, $A=A^T$. Note that the hypothesis "$A$ invertible" is useless.
Case 2. $K=\mathbb{C}$. Note that $A(A+I)(A^2-A-I)=0$ and $A$ is diagonalizable over $\mathbb{R}$.
If $\det(A)\not= 0$, then $A(A^2-I)$ is invertible and consequently $A(A+I)$ too; we deduce easily that $A^2-A-I=0$ and $A=A^T$.
Assume that $\det(A)=0$ and $A$ non-symmetric; then $0,0^2-1=-1\in spectrum(A)$ and $A^2-A-I\not= 0$.
For $n=2$ a solution is $A_2=1/2\begin{pmatrix}-1&-i\\i&-1\end{pmatrix}$ where $spectrum(A_2)=\{0,-1\}$.
For $n=3$, a solution is  $A_3=diag(A_2,(1+\sqrt{5})/2)$  where $spectrum(A_3)=\{0,-1,(1+\sqrt{5})/2\}$.