Prove that if $\det(A) \in \{1,-1\}$ then $AA^t = A^tA = I_n$.
I have no clue, to be fair. I am trying to prove orthogonal polynomials have a determinant $1$. Any help?
2026-03-30 10:11:45.1774865505
How to prove that $\det(A) \in \{1,-1\} \implies AA^t = A^tA = I_n$?
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As stated, this is false: Let $A$ be defined by
$$A = \left(\begin{array}{cc} 2 & 0 \\ 0 & \frac 1 2\end{array}\right)$$
Then $\det(A) = 2 \cdot \frac 1 2 - 0 = 1$, but $AA^T = A^2 \ne I_2$.