- #1
Dustinsfl
- 2,281
- 5
Let [tex]\lambda[/tex] be an eigenvalue of [tex]A[/tex] and let [tex]\mathbf{x}[/tex] be an eigenvector belonging to [tex]\lambda[/tex]. Use math induction to show that, for [tex]m\geq1[/tex], [tex](\lambda)^m[/tex] is an eigenvalue of [tex]A^m[/tex] and [tex]\mathbf{x}[/tex] is an eigenvector of [tex]A^m[/tex] belonging to [tex](\lambda)^m[/tex].
[tex]A\mathbf{x}=\lambda\mathbf{x}[/tex]
[tex]p(1): A^1\mathbf{x}=(\lambda)^1\mathbf{x}[/tex]\
[tex]p(k): A^k\mathbf{x}=(\lambda)^k\mathbf{x}[/tex]
[tex]p(k+1): A^{k+1}\mathbf{x}=(\lambda)^{k+1}\mathbf{x}[/tex]
Assume p(k) is true.
Since p(k) is true, [tex]p(k+1): A*(A^k\mathbf{x})[/tex]
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.
[tex]A\mathbf{x}=\lambda\mathbf{x}[/tex]
[tex]p(1): A^1\mathbf{x}=(\lambda)^1\mathbf{x}[/tex]\
[tex]p(k): A^k\mathbf{x}=(\lambda)^k\mathbf{x}[/tex]
[tex]p(k+1): A^{k+1}\mathbf{x}=(\lambda)^{k+1}\mathbf{x}[/tex]
Assume p(k) is true.
Since p(k) is true, [tex]p(k+1): A*(A^k\mathbf{x})[/tex]
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: