 Solution

Use mathematical induction on $n$. For $n=0$, the identity matrix gives $(1,2{)}^T=(L_1,L_0{)}^T$. If $A^n(1,2{)}^T=(L_{n+1},L_n{)}^T$, applying $A$ and using part (a) gives $A^{n+1}(1,2{)}^T=(L_{n+2},L_{n+1}{)}^T$. ThereforeThe base case is essential: the starting vector represents consecutive Lucas numbers, not two arbitrary initial values.