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

Question

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

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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.

Answer 1

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: $A^TA=\pmatrix{4&-1\\ -1&\frac12}\ne I$.

However, it can be shown that with the given assumptions,

• $A$ is similar to a real orthogonal matrix;
• if $AA^T=A^TA$ (i.e. if $A$ is also normal), it is an orthogonal matrix.