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Question:

If $A=\operatorname{diag}(a, b, c)$, show that $A^{n}=\operatorname{diag}\left(a^{n}, b^{n}, c^{n}\right)$ for all positive integer $n$.

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

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

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

$A^{m+1}=A^{m} A$

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

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

 

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

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

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