Orthogonal matrix sufficient proof

Question

To show that a square matrix $F$ is orthogonal, is it sufficient to show that $F^TF = I$ or is it also necessary to show $\det(F)= \pm1$?

Answer 1

All you need to show a square matrix $A$ is orthogonal is $A^T A = I$ (or equivalently, $A A^T = I$ or also equivalently, $A^{-1} = A^T$).

The fact that $det(A) = \pm 1$ follows from this, since $det(A^T A) = det(A^T) det(A) = det(A) det(A) = det(A)^2$ and $det(I) = 1$. So, $det(A)^2 = 1$ or $det(A) = \pm 1$.

Answer 2

It is not necessary to show that $det(F) = \pm1$

Square matrix $F$ is orthogonal if
$$F^{-1} = F^T$$
But we know $$ F^T \cdot F = I $$

And
$$ F^{-1} \cdot F = I $$ So
$$ F^T \cdot F = I $$ $$ (F^T \cdot F)\cdot F^{-1} = (I)\cdot F^{-1} $$ $$ F^T\cdot(F\cdot F^{-1}) = F^{-1} $$ $$ F^T\cdot(I) = F^{-1} $$ $$ F^T = F^{-1} $$