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  • If AB = BA for any two square matrices, prove by mathematical induction that (AB)^n = A^n B^n.

If AB = BA for any two square matrices, prove by mathematical induction that (AB)^n = A^n B^n.

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By the principle of mathematical induction, we say that if a statement P(n) is true for n = 1, and if we assume P(k) to be true for some random natural number k then if we prove P(k+1) to be true, we can say that P(n) is true for all natural numbers.

We have to prove that (AB)\textsuperscript{n} = A\textsuperscript{n}B\textsuperscript{n}

Let P(n) be the statement: (AB)\textsuperscript{n} = A\textsuperscript{n}B\textsuperscript{n}

So, P(1): (AB)\textsuperscript{1} = A\textsuperscript{1}B\textsuperscript{1}

\\$ \Rightarrow $ P(1) : AB = AB \\ \Rightarrow $ P(1) is true

Let P(k) be true.

\therefore $ (AB)\textsuperscript{k} = A\textsuperscript{k}B\textsuperscript{k} $ \ldots $ (1)

Let’s take P(k+1) now:

\\ \because$ (AB)\textsuperscript{k+1} = (AB)\textsuperscript{k}(AB) \\$ \Rightarrow $ (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k}(AB)

NOTE: As we know, matrix multiplication is not commutative. So we can’t write directly that

A\textsuperscript{k}B\textsuperscript{k}(AB) = A\textsuperscript{k+1}B\textsuperscript{k+1}

But we are given that AB = BA

\\ \therefore $ (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k}(AB) \\$ \Rightarrow $ (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-1}(BAB)

As, AB = BA

\\ \Rightarrow $ (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-1}(ABB) \\$ \Rightarrow $ (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-1}(AB\textsuperscript{2}) \\$ \Rightarrow $ (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-2}(BAB\textsuperscript{2}) \\$ \Rightarrow $ (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-2}(ABB\textsuperscript{2}) \\$ \Rightarrow $ (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-2}(AB\textsuperscript{3})

We observe that one power of B is decreasing while other is increasing. After certain repetitions decreasing power of B will become I

And at last step:

\\ \Rightarrow $ (AB)\textsuperscript{k+1} = A\textsuperscript{k}I(AB\textsuperscript{k+1}) \\$ \Rightarrow $ (AB)\textsuperscript{k+1} = A\textsuperscript{k}AB\textsuperscript{k+1} \\$ \Rightarrow $ (AB)\textsuperscript{k+1} = A\textsuperscript{k+1}B\textsuperscript{k+1}

Thus P(k+1) is true when P(k) is true.

\therefore $ (AB)\textsuperscript{n} = A\textsuperscript{n} B\textsuperscript{n} $ \forall $ n $ \in $ N when AB = BA.

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