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=$$\lambda$$v

The Attempt at a Solution

Av=$$\lambda$$v
(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 = $$\lambda$$v.
Then A²v = A(Av) = A($$\lambda$$v) = $$\lambda (\lambda$$v). = $$\lambda^2$$v.

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

 

[soc4ward14]

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

 

[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($$\lambda$$v) can = $$\lambda$$($$\lambda$$v?

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