Let \(T\) be any \(2 \times 2\) matrix with eigenvalues \(-1, 3\). For each positive integer \(n\), the numbers \(a_n\) and \(b_n\) are such that \(T^{\,n+1} = a_n T + b_n I\), where \(I\) is the \(2 \times 2\) identity matrix. Then \(a_n\) and \(b_n\) satisfy the recurrence relations:
\(a_{n+1} = 2 a_n + b_n;\ b_{n+1} = 3 a_n\)
The eigenvalues \(-1\) and \(3\) give the characteristic polynomial \((\lambda+1)(\lambda-3) = \lambda^{2} - 2\lambda - 3 = 0\). By the Cayley-Hamilton theorem, \(T^{2} = 2T + 3I\).
Starting from \(T^{\,n+1} = a_n T + b_n I\), multiply both sides by \(T\): \(T^{\,n+2} = a_n T^{2} + b_n T = a_n(2T + 3I) + b_n T = (2 a_n + b_n) T + 3 a_n I\).
Comparing with \(T^{\,n+2} = a_{n+1} T + b_{n+1} I\) gives \(a_{n+1} = 2 a_n + b_n\) and \(b_{n+1} = 3 a_n\).
Hence, the recurrence relations are \(a_{n+1} = 2 a_n + b_n;\ b_{n+1} = 3 a_n\).
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