Prove: (\lambda)^m is an Eigenvalue of A^m

[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.

 

[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)=?

 

[Dustinsfl]

Ok I gotcha.