Is it possible to find two real \(2\times2\) matrices \(A,B\) such that \(A^2=B^2=\operatorname{Id}\) (the identity matrix), but \(AB\) has eigenvalues \(2\) and \(1/2\)?
Solution
Set \[ A\coloneqq \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}, \qquad B\coloneqq \begin{pmatrix} \frac54&\frac34\\ -\frac34&-\frac54 \end{pmatrix}. \]
1. One has \(A^2=I_2\) and \(B^2=I_2\).
The identity \(A^2=I_2\) is immediate. Also, \[ B^2 = \begin{pmatrix} \frac{25}{16}-\frac9{16} & \frac{15}{16}-\frac{15}{16}\\ -\frac{15}{16}+\frac{15}{16} & -\frac9{16}+\frac{25}{16} \end{pmatrix} = \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} =I_2. \]
2. The product is\[ AB= \begin{pmatrix} \frac54&\frac34\\ \frac34&\frac54 \end{pmatrix}. \]
This follows by direct matrix multiplication.
3. The matrix \(AB\) has eigenvalues \(2\) and \(1/2\).
By step 2, \[ AB \begin{pmatrix} 1\\ 1 \end{pmatrix} = 2 \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad AB \begin{pmatrix} 1\\ -1 \end{pmatrix} = \frac12 \begin{pmatrix} 1\\ -1 \end{pmatrix}. \] Thus \(2\) and \(1/2\) are eigenvalues of \(AB\).
4. Therefore such matrices exist.
Step 1 shows that the displayed \(A\) and \(B\) are involutions, and step 3 shows that their product has the required eigenvalues.
5. Q.E.D.
Step 4 answers the existence question affirmatively.