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Problem 1. Let \( k \) be a field of characteristic \( \neq 2 \), and \( A \) and \( B \) be two 2 by 2 matrices on \( k \). Then prove that \(A^{2} - 2AB + B^{2} = O\) implies \((A-B)^{2} = O\).

Solution. Let \( C = A - B \). Then we have to show that \( C^{2} = O \). By Cayley-Hamilton Theorem, \begin{equation*} C^{2} = (\mathrm{tr} \, C)C - (\det C) I \end{equation*} and it suffices to show that both \( \mathrm{tr} \, C = 0 \) and \( \det C = 0 \). Observe first that \begin{equation*} C^{2} = A^{2} - AB - BA + B^{2} = AB - BA = CA - AC. \end{equation*} We first show that \( \det C = 0 \). Suppose not. Then \( C \) is invertible and \begin{equation*} I = C^{-1}C^{2}C^{-1} = C^{-1}(CA - AC)C^{-1} = AC^{-1} - C^{-1}A. \end{equation*} But taking trace to both sides, we have \begin{equation*} 2 = \mathrm{tr} \, I = \mathrm{tr} \, \left( AC^{-1} - C^{-1}A \right) = 0, \end{equation*} which is a contradiction unless \( \mathrm{char} \, k = 2 \). So we must have \( \det C = 0 \). Then \begin{equation*} (\mathrm{tr} \, C)C = C^{2} = CA - AC. \end{equation*} Taking trace again, we have \begin{equation*} (\mathrm{tr} \, C)^{2} = \mathrm{tr} \, \left( (\mathrm{tr} \, C)C \right) = \mathrm{tr} \, ( CA - AC ) = 0. \end{equation*} Thus \( \mathrm{tr} \, C = 0 \), completing the proof.

Problem 2. Let \( SO(2) \) be the special orthogonal group of degree 2 over \( \mathbb{R} \), that is, the family of matrices of the form \begin{equation*} R(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin\theta & \cos\theta \end{pmatrix} \end{equation*} for \( \theta \in \mathbb{R} \). Suppose \( A \) is a 2 by 2 real matrix satisfying \( A^{2} - 2a A + I = O \) for some \( a \) and \( A^{n} \in SO(2) \) for some \( n \in \mathbb{N} \). Then find an appropriate condition for \( a \) such that \( A \) also satisfies \( A \in SO(2) \).

Solution. Note that \( a \) is real, since so is \( A \). Then we divide into several cases.

Case 1. Assume \( a = \pm 1 \). Then we have \begin{equation*} x^{m} = (x - a)^{2}q_{m} (x) + m a^{m-1} x - (m-1) a^{m}. \end{equation*} Plugging \( A \) and \( m = nk \) for a positive integer \( k \), we obtain \begin{equation*} R(k\varphi) = nk a^{nk-1} A - (nk-1) a^{nk} I. \end{equation*} Now divide both sides by \( nk a^{nk-1} \) and take \( k \to \infty \). Then we obtain \( A = aI \in SO(2) \) as desired.

Case 2. Now assume \( |a|<1 \). We can write \( a = \cos \theta \) for some real \( \theta \). Clearly we have \( \sin \theta \neq 0 \). We also observe that \( A \) cannot be a constant multiple of \( I \). If not, we have \( \lambda^{2} - 2a\lambda + 1 = 0 \) for \( A = \lambda I \), which is impossible since \( \lambda \) is real. Finally, together with Cayley-Hamilton Theorem, this implies that \( \mathrm{tr} \, A = 2a \) and \( \det A = 1 \). Now by \begin{equation*} x^{2} - 2a x + 1 = (x - e^{i\theta})(x - e^{-i\theta}), \end{equation*} we have \begin{equation*} x^{m} = (x^{2} - 2ax + 1)q_{m} (x) + \frac{\sin m\theta}{\sin\theta} x - \frac{\sin (m-1)\theta}{\sin \theta}. \end{equation*} Plugging \( x = A \) and \( m = n \) gives \begin{equation*} R(\varphi) = A^{n} = \frac{\sin n\theta}{\sin\theta} A - \frac{\sin (n-1)\theta}{\sin \theta} I, \end{equation*} or equivalently \begin{equation*} A \sin n\theta = R(\varphi) \sin\theta + I \sin (n-1)\theta. \end{equation*} Taking trace to both sides, we obtain \begin{equation*} \sin n\theta \cos \theta = \cos\varphi \sin\theta + \sin (n-1)\theta. \end{equation*} This implies \begin{equation*} A \sin n\theta = \begin{pmatrix} \sin n\theta \cos\theta & - \sin \varphi \sin \theta \\ \sin \varphi \sin \theta & \sin n\theta \cos \theta \end{pmatrix}. \end{equation*} So taking \( \det \), we have \begin{equation*} \sin^{2} n\theta = \sin^{2} n\theta \cos^{2}\theta + \sin^{2}\varphi \sin^{2} \theta, \end{equation*} which implies \( \sin \varphi = \pm \sin n\theta \). But recall that \( \theta \) is chosen only to satisfy \( a = \cos \theta \), so we may adjust the sign so that \( \sin \varphi = \sin n \theta \). For this choice of \( \theta \), we have \begin{equation*} A \sin n\theta = R(\theta) \sin n\theta. \end{equation*} Thus \( A \in SO(2) \) whenever \( \sin n \theta \neq 0 \).

Case 3. Finally we assume \( |a| > 1 \). By replacing \( A \) by \( -A \) if needed, we may assume \( a > 1 \) without altering the fact that \( A^{n} \in SO(2) \). Write \( a = \cosh \theta \) for some real \( \theta > 0 \). Then by \begin{equation*} x^{2} - 2a x + 1 = (x - e^{\theta})(x - e^{-\theta}), \end{equation*} we have \begin{equation*} x^{m} = (x^{2} - 2ax + 1)q_{m} (x) + \frac{\sinh m\theta}{\sinh\theta} x - \frac{\sinh (m-1)\theta}{\sinh \theta}. \end{equation*} Plugging \( x = A \) and \( m = nk \) gives \begin{equation*} R(k \varphi) \sinh \theta = A \sinh nk \theta - I \sinh (nk-1)\theta, \end{equation*} Dividing both sides by \( \sinh nk \theta \) and taking \( k \to \infty \), we have \( A = O \), a contradiction.

Therefore,

• \( |a| \leq 1 \) and
• \( A \in SO(2) \) if either \( a = \pm 1 \) or \( A^{n} \neq \pm I \).

We make a final remark: In the Case 2, the condition that \( \sin n\theta \neq 0 \), or in other words, \( A^{n} \neq \pm I \) is indispensible, by observing the following counterexample: \begin{equation*} A = \begin{pmatrix} \cos \theta & -p \sin \theta \\ p^{-1}\sin\theta & \cos\theta \end{pmatrix} \end{equation*} for some \( p \neq 0, 1 \) and \( n\theta \in 2\pi \mathbb{Z} \).