Is there a real \(2\times2\) matrix \(A\) such that \[ A^{20}= \begin{pmatrix} -1&0\\ 0&-1-\varepsilon \end{pmatrix} \] ? Exhibit such an \(A\) or prove there is none.
The problem does not quantify \(\varepsilon\). For every real \(\varepsilon\neq0\), in particular for every \(\varepsilon>0\), no real \(2\times2\) matrix \(A\) satisfies \(A^{20}=\operatorname{diag}(-1,-1-\varepsilon)\).
Solution
For a real parameter \(\varepsilon\), such a matrix exists if and only if \[ \boxed{\varepsilon=0}. \]
1. Suppose\[ A^{20} = D_\varepsilon \coloneqq \begin{pmatrix} -1&0\\ 0&-1-\varepsilon \end{pmatrix}. \]Then\[ AD_\varepsilon=D_\varepsilon A. \]
Every matrix commutes with each of its powers. Hence \[ AA^{20}=A^{20}A. \] Substitute \(A^{20}=D_\varepsilon\).
2. If \(\varepsilon\neq0\), then every real matrix commuting with\(D_\varepsilon\) is diagonal.
Write \[ A= \begin{pmatrix} a&b\\ c&d \end{pmatrix}. \] The diagonal entries of \(D_\varepsilon\) are distinct when \(\varepsilon\neq0\). Direct multiplication gives \[ AD_\varepsilon-D_\varepsilon A = \begin{pmatrix} 0&-\varepsilon b\\ \varepsilon c&0 \end{pmatrix}. \] If this vanishes and \(\varepsilon\neq0\), then \[ b=c=0. \] Thus \(A\) is diagonal.
3. No solution exists when \(\varepsilon\neq0\).
If a solution existed, steps 1 and 2 would give \[ A= \begin{pmatrix} a&0\\ 0&d \end{pmatrix} \] with \(a,d\in\RR\). Then \[ A^{20} = \begin{pmatrix} a^{20}&0\\ 0&d^{20} \end{pmatrix}, \] whose diagonal entries are nonnegative. This cannot equal \(D_\varepsilon\), whose first diagonal entry is \(-1\).
4. A solution exists when \(\varepsilon=0\).
Let \[ A= \begin{pmatrix} \cos(\pi/20)&-\sin(\pi/20)\\ \sin(\pi/20)&\cos(\pi/20) \end{pmatrix}. \] This is the rotation matrix through angle \(\pi/20\). Powers of rotation matrices add angles, so \[ A^{20} = \begin{pmatrix} \cos\pi&-\sin\pi\\ \sin\pi&\cos\pi \end{pmatrix} = -I = \begin{pmatrix} -1&0\\ 0&-1 \end{pmatrix}. \] When \(\varepsilon=0\), this is exactly \(D_\varepsilon\).
5. Thus existence is equivalent to \(\varepsilon=0\).
Step 3 proves nonexistence for every nonzero real \(\varepsilon\), and step 4 constructs a solution for \(\varepsilon=0\).
6. Q.E.D.
Step 5 answers the question for every real \(\varepsilon\).