# Linear Algebra Proof- Pleaseeeee help!

1. Apr 29, 2010

### soc4ward14

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Av=$$\lambda$$v

3. 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

2. Apr 29, 2010

### 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 $$\lambda^2$$.

3. Apr 29, 2010

### soc4ward14

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

4. Apr 29, 2010

### hgfalling

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

5. Apr 30, 2010