# Math induction

1. Apr 17, 2010

### Dustinsfl

Let $$\lambda$$ be an eigenvalue of $$A$$ and let $$\mathbf{x}$$ be an eigenvector belonging to $$\lambda$$. Use math induction to show that, for $$m\geq1$$, $$(\lambda)^m$$ is an eigenvalue of $$A^m$$ and $$\mathbf{x}$$ is an eigenvector of $$A^m$$ belonging to $$(\lambda)^m$$.

$$A\mathbf{x}=\lambda\mathbf{x}$$

$$p(1): A^1\mathbf{x}=(\lambda)^1\mathbf{x}$$\

$$p(k): A^k\mathbf{x}=(\lambda)^k\mathbf{x}$$

$$p(k+1): A^{k+1}\mathbf{x}=(\lambda)^{k+1}\mathbf{x}$$

Assume p(k) is true.

Since p(k) is true, $$p(k+1): A*(A^k\mathbf{x})$$

Not sure if this is correct path to take to the end result. Let alone, if it is if I am going about it right.

Last edited: Apr 17, 2010
2. Apr 17, 2010

### Dick

You were doing ok, except you don't want A^(k+1)(lambda). You want p(k+1): A^(k+1)(x)=A*(A^k(x))=A*(lambda^k*x)=lambda^k*A(x)=???

3. Apr 17, 2010

Ok I gotcha.