Calculezi [tex]A^{2} , A^3, [/tex] etc până găseşti o regulă şi apoi o demonstrezi prin inducţie.
[tex]a) \, A = \left(\begin{array}{ccc}1&i\\0&1\end{array}\right) \\ \\
A^2 = \left(\begin{array}{ccc}1&i\\0&1\end{array}\right)\left(\begin{array}{ccc}1&i\\0&1\end{array}\right) = \left(\begin{array}{ccc}1&2i\\0&1\end{array}\right) \\ \\
A^3 = A^2 \cdot A = \left(\begin{array}{ccc}1&2i\\0&1\end{array}\right)\left(\begin{array}{ccc}1&i\\0&1\end{array}\right)=\left(\begin{array}{ccc}1&3i\\0&1\end{array}\right).[/tex]
Observăm că [tex]A^n[/tex] are forma [tex]\left(\begin{array}{ccc}1&n \cdot i\\0&1\end{array}\right)[/tex].
Presupunem că [tex]A^k=\left(\begin{array}{ccc}1&k \cdot i\\0&1\end{array}\right)[/tex] şi demonstrăm că [tex]A^{k+1} = \left(\begin{array}{ccc}1&(k+1) \cdot i\\0&1\end{array}\right)[/tex]. Avem: [tex]A^{k+1} = A^k \cdot A = \left(\begin{array}{ccc}1&k\cdot i\\0&1\end{array}\right) \cdot \left(\begin{array}{ccc}1&i\\0&1\end{array}\right) = \left(\begin{array}{ccc}1&i+k \cdot i\\0&1\end{array}\right) \\ \\ = \left(\begin{array}{ccc}1&(k+1)i\\0&1\end{array}\right).[/tex].
Deci, conform principiului inducţiei matematice, [tex]A^n = \left(\begin{array}{ccc}1&n \cdot i\\0&1\end{array}\right).[/tex]
La b aplicăm altă regulă (binomul lui Newton):
[tex]A= \left(\begin{array}{ccc}1&1&1\\0&1&1\\0&0&1\end{array}\right) = \underset{I_3}{\underbrace{ \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)}} + \underset{B}{\underbrace{ \left(\begin{array}{ccc}0&1&1\\0&0&1\\0&0&0\end{array}\right)}} = I_3 + B.[/tex]
[tex]B = \left(\begin{array}{ccc}0&1&1\\0&0&1\\0&0&0\end{array}\right) \\ \\
B^2 = \left(\begin{array}{ccc}0&1&1\\0&0&1\\0&0&0\end{array}\right) \left(\begin{array}{ccc}0&1&1\\0&0&1\\0&0&0\end{array}\right)= \left(\begin{array}{ccc}0&0&1\\0&0&0\\0&0&0\end{array}\right) \\ \\
B^3 = \left(\begin{array}{ccc}0&0&1\\0&0&0\\0&0&0\end{array}\right)\left(\begin{array}{ccc}0&1&1\\0&0&1\\0&0&0\end{array}\right) = \left(\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right) = O_3 [/tex]
[tex]\Rightarrow B^p=O_3, \forall p \geq 3.[/tex]
[tex]A^n = (I_3 + B)^n = C_n^0 I_3^n + C_n^1I_3^{n-1}B+C_n^2 I_3^{n-2}B^2 \\ \\
A^n = I_3 + n \cdot I_3 \cdot B + \frac{n(n-1)}{2}B^2 \\ \\
A^n = I_3 + nB + \frac{n(n-1)}{2}B^2 \\ \\
A^n = \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right) +
\left(\begin{array}{ccc}0&n&n\\0&0&n\\0&0&0\end{array}\right) + \left(\begin{array}{ccc}0&0&\frac{n(n-1)}{2}\\0&0&0\\0&0&0\end{array}\right) \\ \\
\Rightarrow A^n = \left(\begin{array}{ccc}1&n&n+\frac{n(n-1)}{2}\\0&1&n\\0&0&1\end{array}\right) [/tex]
Posibil să fi greşit pe undeva..
