Linear Algebra Proof- Pleaseeeee help

[soc4ward14]

Homework Statement

Let A be an n x n matrix with eigenvalue \lambda. Prove that \lambda^2 is an eigenvalue of A^2 and that if v is an eigenvector of A, then v is also an eigenvector for A^2.

Homework Equations

Av=\lambdav

The Attempt at a Solution

Av=\lambdav
(A*A)V=(\lambda * \lambda)v
so then v will be an eigenvector to A^2 when \lambda^2

 

[hgfalling]

Seems you have the right track, although you could be a little more precise:

Let v be an eigenvector of the nxn matrix A.

We have Av = \lambdav.
Then A²v = A(Av) = A(\lambdav) = \lambda (\lambdav). = \lambda^2v.

Hence \lambda^2 is an eigenvalue of A² and v is an eigenvector corresponding to \lambda^2.

 

[soc4ward14]

thank you but how come A(\lambdav) can = \lambda(\lambdav?

 

[hgfalling]

Any nonzero scalar multiple of an eigenvector is also an eigenvector, and is associated with the same eigenvalue.

 

[radou]

thank you but how come A(\lambdav) can = \lambda(\lambdav?

A(λv) = (by the linearity of A!) = λA(v) = λ λv = λ^2 v.