# Principle of math induction Show with math induction `A^nB - BA^n = nA^n` in natural `n`   `A = [[0,1],[0,0]]` `B = [[1,0],[0,2]]`

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In the following text `0` denotes zero matrix `[[0,0],[0,0]].`

Before we begin math induction let's show that

`A^n=0`, `forall ngeq2`                                                                (1)

`A^2=AcdotA=0`

Now, no mater with what we multiply zero matrix we will get zero matrix.

`A^n=A^2A^(n-2)=0cdotA^(n-2)=0`

Math induction

1. We need to show that the expression is true for `n=1` (it's obviously true for `n=0.` ` `)

`AB-BA=[[0,1],[0,0]][[1,0],[0,2]]-[[1,0],[0,2]][[0,1],[0,0]]``=[[0,2],[0,0]]-[[0,1],[0,0]]` `=1cdot[[0,1],[0,0]]=1A`

2. Let' assume that `A^kB-BA^k=kA^k` holds for all `k leq n`.

3. Since `n geq 2` (we have already proven case `n=1` ) we can use (1)

`A^(n+1)B-BA^(n+1)= 0B-B0=0=(n+1)0=(n+1)A^(n+1)`

which completes induction and proves that equation holds for all natural numbers.

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