If $A = [\begin{matrix} cosθ & i sinθ \\ i sinθ & cosθ \end{matrix}]$ , then prove by principle of mathematical induction that

$A^{n} = [\begin{matrix} \cos n θ & i sinθ \\ i \sin n θ & \cos n θ \end{matrix}]$ for all n ∈ N.

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Solution

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have

$A^{1} = [\begin{matrix} \cos 1 θ & i \sin 1 θ \\ i \sin 1 θ & \cos 1 θ \end{matrix}] = [\begin{matrix} \cos θ & i \sin θ \\ i \sin θ & \cos θ \end{matrix}] = A$

Thus, the result is true for n=1.

Step 2: Let the result be true for n = m. Then,

$A^{m} = [\begin{matrix} \cos m θ & i \sin m θ \\ i \sin m θ & \cos m θ \end{matrix}]$

Now we shall show that the result is true for $n = m + 1$ .
Here,
$A^{m + 1} = [\begin{matrix} \cos (m + 1) θ & i \sin (m + 1) θ \\ i \sin (m + 1) θ & \cos (m + 1) θ \end{matrix}]$ ...(1)

By definition of integral power of matrix, we have
$A^{m + 1} = A^{m} A \Rightarrow A^{m + 1} = [\begin{matrix} \cos m θ & i \sin m θ \\ i \sin m θ & \cos m θ \end{matrix}] [\begin{matrix} \cos θ & i \sin θ \\ i \sin θ & \cos θ \end{matrix}] [From eq . (1)] \Rightarrow A^{m + 1} = [\begin{matrix} \cos m θ . \cos θ + i \sin m θ . i \sin θ & \cos m θ . i \sin θ + i \sin m θ . \cos θ \\ i \sin m θ . \cos θ + \cos m θ . i \sin θ & i \sin m θ . i \sin θ + \cos m θ . \cos θ \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos m θ . \cos θ - \sin m θ . \sin θ & i (\cos m θ . \sin θ + \sin m θ . \cos θ) \\ i (\sin m θ . \cos θ + \cos m θ \sin θ) & - \sin m θ . \sin θ + \cos m θ . \cos θ \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos m θ . \cos θ - \sin m θ . \sin θ & i (\cos m θ . \sin θ + \sin m θ . \cos θ) \\ i (\sin m θ . \cos θ + \cos m θ \sin θ) & \cos m θ . \cos θ - \sin m θ . \sin θ \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos (m θ + θ) & i \sin (m θ + θ) \\ i \sin (m θ + θ) & \cos (m θ + θ) \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos (m + 1) θ & i \sin (m + 1) θ \\ i \sin (m + 1) θ & \cos (m + 1) θ \end{matrix}]$

This shows that when the result is true for n = m, it is true for $n = m + 1$ .
Hence, by the principle of mathematical induction, the result is valid for all n $\in N$ .

Disclaimer: n is missing before $θ$ in a₁₂ in Aⁿ.

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