Let $A \in M_2(\mathbb R)$ such that $AA^T=A^TA=I$ and $\det(A)=1$. Which one of the following is necessarily true?

Question

Let $A \in M_2(\mathbb R)$ such that $AA^T=A^TA=I$ and $\det(A)=1$. Which one of the following is necessarily true?

(a)$Av=v$ for some unit vector $v \in \mathbb R^2$.

(b)$Av\neq v$ for some unit vector $v \in \mathbb R^2$.

(c)$||Av|| \neq 1$ for some unit vector $v \in \mathbb R^2$.

(d)None of the above.

I know that if $\lambda_1, \lambda_2 $ are the eigenvalues of $A$ then $\lambda_1\lambda_2=1$ and $1/\lambda_1, 1/\lambda_2$ are the eigenvalues of $A^{-1}$. But here $A^T=A^{-1}$ and since $A,A^T$ have the same eigenvalues $\lambda_1=1/\lambda_1 \implies \lambda_1=\pm 1$ or $\lambda_1=1/\lambda_2$ gives no further information. The given answer is (d) but I wanted to find out what conclusions we can draw from this information.

Answer 1

(a) is false: take $A=\left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$.

(b) is false: take $A=\left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$.

(c) is false: if $\lVert v\rVert=1$, then\begin{align}\lVert Av\rVert^2&=\langle Av,Av\rangle\\&=\langle v,A^TAv\rangle\\&=\langle v,v\rangle\\&=1.\end{align}

So, yes, (d) is true.

Answer 2

Matrices of that satisfy $AA^T = A^TA = I$ are called orthogonal matrices. These matrices are a rotation, reflection, or composition of both. When the determinant is $1$, it is only a rotation, and in $M_2(\mathbb R)$ the matrix therefore has form $$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$ of which eigenvalues can be explicitly computed which exhibit various properties; you will find they have complex modulus of $1$, for example.

For more information about orthogonal matrices and rotation matrices see here: https://en.m.wikipedia.org/wiki/Orthogonal_matrix https://en.m.wikipedia.org/wiki/Rotation_matrix