Matrices Proof by Induction
Folien in diesem Satz (2)
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- For the matrix A , A^(k+1) = A^k x A
- When showing something is true for n=k+1 , we use the rule above
- Check series proofs by induction for more help
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- Prove by induction that [1 0][5 1]^n = [1 0][5n 1]
- Assume for n=k : [1 0][5 1]^k = [1 0][5k 1]
- Show true for n=k+1, ie: [1 0][5 1]^k+1 = [1 0][5(k+1) 1] , [1 0][5 1]^k+1 = [1 0][5 1]^k x [1 0][5 1] , [1 0][5 1] x [1 0][5k 1] = [1 0][5k+5 1] = [1 0][5(k+1) 1] as required
- Show true for n=1 : LHS ; [1 0][5 1]^1 = [1 0][5 1]^1 RHS ; [1 0][5(1) 1] = [1 0][5 1] and therefore true for n=1
- As the result is true for n=1, and if true for n=k => true for n=k+1 , it is proven by induction for all n∈Z , n≥1