Question

If A = diag (a, b, c), show that Aⁿ = diag (aⁿ, bⁿ, cⁿ) for all positive integer n.

Answer 1

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

$A^1=\left[\begin{array}{ccc}{a}^1& 0& 0\\ 0& b^1& 0\\ 0& 0& c^1\end{array}\right]=\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]=A$

So, the result is true for n = 1.

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

$A^m=\left[\begin{array}{ccc}{a}^m& 0& 0\\ 0& b^m& 0\\ 0& 0& c^m\end{array}\right]$ ...(1)

Now, we shall check if the result is true for $n=m+1$.
Here,
$A^{m+1}=\left[\begin{array}{ccc}{a}^{m+1}& 0& 0\\ 0& b^{m+1}& 0\\ 0& 0& c^{m+1}\end{array}\right]$

By definition of integral power of matrix, we have

$$\begin{gathered} A^{m+1}=A^{m}A \\ \Rightarrow A^{m+1}=\left[\begin{array}{ccc}{a}^m& 0& 0\\ 0& b^m& 0\\ 0& 0& c^m\end{array}\right]\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow A^{m+1}=\left[\begin{array}{ccc}a{a}^m+0+0& 0+0+0& 0+0+0\\ 0+0+0& 0+b{b}^m+0& 0+0+0\\ 0+0+0& 0+0+0& 0+0+c{c}^m\end{array}\right] \\ \Rightarrow A^{m+1}=\left[\begin{array}{ccc}{a}^{m+1}& 0& 0\\ 0& b^{m+1}& 0\\ 0& 0& c^{m+1}\end{array}\right] \end{gathered}$$

This shows that when the result is true for n = m, it is also true for $n=m+1$.
Hence, by the principle of mathematical induction, the result is valid for any positive integer n.