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

\[A^1 = \begin{bmatrix}a^1 & 0 & 0 \\ 0 & b^1 & 0 \\ 0 & 0 & c^1\end{bmatrix} = \begin{bmatrix}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{bmatrix} = A\]

So, the result is true for n = 1.

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

\[A^m = \begin{bmatrix}a^m & 0 & 0 \\ 0 & b^m & 0 \\ 0 & 0 & c^m\end{bmatrix}\]         ...(1)

Now, we shall check if the result is true for

\[n = m + 1\]

Here,

\[A^{m + 1} = \begin{bmatrix}a^{m + 1} & 0 & 0 \\ 0 & b^{m + 1} & 0 \\ 0 & 0 & c^{m + 1}\end{bmatrix}\]

By definition of integral power of matrix, we have

\[A^{m + 1} = A^m A\]

\[ \Rightarrow A^{m + 1} = \begin{bmatrix}a^m & 0 & 0 \\ 0 & b^m & 0 \\ 0 & 0 & c^m\end{bmatrix}\begin{bmatrix}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{bmatrix} \left[ \text{From eq} . \left( 1 \right) \right]\]

\[ \Rightarrow A^{m + 1} = \begin{bmatrix}a a^m + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + b b^m + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + c c^m\end{bmatrix}\]

\[ \Rightarrow A^{m + 1} = \begin{bmatrix}a^{m + 1} & 0 & 0 \\ 0 & b^{m + 1} & 0 \\ 0 & 0 & c^{m + 1}\end{bmatrix}\]

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.