# If $A$ is a $2 \times 2$ real matrix such that $\det (A) = 1$ and $A^n = I$ show that $A ^tA = I$

If $$A$$ is a $$2 \times 2$$ real matrix such that $$\det (A) = 1$$ and $$A^n = I$$ show that $$A ^tA = I$$

IDEA: Since $$\text{det}(A) = 1$$ according to the Cayley-Hamilton theorem, it is true that $$A^2-\text{tr}(A)A+\text{det}(A)=0$$ then $$A^{-1}=\text{tr}(A)I-A$$, just show that $$A^ {-1} = A^{t}$$, another way is to show that the columns of $$A$$ form an orthonormal system of $$\mathbb{R} ^ 2$$ but I don't see a way to test Can anyone give a suggestion ? Thank you.

• Another idea: show that $A$ is diagonalizable and reduce to the case where $A$ is diagonal. Commented Aug 7, 2020 at 17:58
• $A^n = I$ for what values of $n$? Commented Aug 7, 2020 at 20:58

This isn't true. E.g. when $$A=\frac{1}{\sqrt{2}}\pmatrix{2&-1\\ 2&0}$$, we have $$\det(A)=1$$ and $$A^8=I$$, but $$A$$ is not an orthogonal matrix: $$A^TA=\pmatrix{4&-1\\ -1&\frac12}\ne I$$.
• $$A$$ is similar to a real orthogonal matrix;
• if $$AA^T=A^TA$$ (i.e. if $$A$$ is also normal), it is an orthogonal matrix.