Power of matrix and power of eigenvalue

[bmanbs2]

Assuming that [tex]k\geq0[/tex],

How does one prove that when [tex]A[/tex] has an eigenvaule [tex]\lambda[/tex] that [tex]A^{k}[/tex] has an eigenvalue [tex]\lambda^{k}[/tex]?

 

[disregardthat]

This is pretty straight-forward. [tex]Av=\lambda v[/tex] for some vector v, so try to calculate [tex]A^kv[/tex].

 

[bmanbs2]

Matrix raised to certain power has exponent raised to same power proof.

Homework Statement

Prove that when [tex]A[/tex] has an eigenvaule [tex]\lambda[/tex] that [tex]A^{k}[/tex] has an eigenvalue [tex]\lambda^{k}[/tex]?

Homework Equations

None

The Attempt at a Solution

Tried to show that [tex]A^{k}X^{k} = \lambda^{k}X^{k}[/tex], but [tex]X^{k}[/tex] isn't possible as [tex]X[/tex] is a vector.

 

[Dick]

Realizing A^k*X^k makes no sense is a good start. But A^2(X)=A(A(X))=A(lambda*X)=lambda*(A(X))=lambda*(lambda*X)=lambda^2*X makes sense, doesn't it?

 

[bmanbs2]

Yes...yes it does indeed. Thank you