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 $0$.
Best Answer
We have $$(A^T - A)(A + I)=0$$
Since $A$ is nonsingular, $\det A \neq 0 \implies \det A^T \neq 0 \implies \det (A^2 - I) \neq 0 \implies \det(A+I)\cdot\det(A-I)\neq 0$ $$\implies \det(A+I) \neq 0 \neq \det(A-I)$$
Therefore, $$(A^T - A)(A + I)=0 \implies A^T - A = 0$$ And we're done
Here's something else that I noticed. You arrive at the following equation: $$A^3 - 2A - I=0 \implies A^3 - A = A+I \implies A(A^2 - I) = A+I \implies AA^T = A +I \\ \implies A = AA^T - I$$ Therefore, $A$ is symmetric