Question
If \[ A = \text{diag}(a, b, c) = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \] show that \[ A^n = \text{diag}(a^n, b^n, c^n) \] for all positive integers \(n\).
Solution (Mathematical Induction)
Step 1: Base Case (n = 1)
\[ A^1 = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} = \begin{bmatrix} a^1 & 0 & 0 \\ 0 & b^1 & 0 \\ 0 & 0 & c^1 \end{bmatrix} \] ✔ True for \(n=1\)Step 2: Assume for \(n = k\)
\[ A^k = \begin{bmatrix} a^k & 0 & 0 \\ 0 & b^k & 0 \\ 0 & 0 & c^k \end{bmatrix} \]Step 3: Prove for \(n = k+1\)
\[ A^{k+1} = A^k \cdot A \] \[ = \begin{bmatrix} a^k & 0 & 0 \\ 0 & b^k & 0 \\ 0 & 0 & c^k \end{bmatrix} \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \]Step 4: Multiply
\[ = \begin{bmatrix} a^{k+1} & 0 & 0 \\ 0 & b^{k+1} & 0 \\ 0 & 0 & c^{k+1} \end{bmatrix} \]Step 5: Conclusion
✔ True for \(k+1\) \[ \Rightarrow A^n = \begin{bmatrix} a^n & 0 & 0 \\ 0 & b^n & 0 \\ 0 & 0 & c^n \end{bmatrix} \]Final Result
\[
A^n = \text{diag}(a^n, b^n, c^n)
\]
Hence proved.