# I Why do eigenvectors stay the same when a matrix is squared?

1. Sep 11, 2016

### Aldnoahz

I am new to linear algebra but I have been trying to figure out this question. Everybody seems to take for granted that for matrix A which has eigenvectors x, A2 also has the same eigenvectors?

I know that people are just operating on the equation Ax=λx, saying that A2x=A(Ax)=A(λx) and therefore A2x = λ2x. However, in my opinion, this is not a proof proving why A2 and A have the same eigenvectors but rather why λ is squared on the basis that the matrices share the same eigenvectors.

If someone can prove that A2 and A have the same eigenvectors by using equations A2y=αy and Ax=λx, and proceeding to prove y=x, I will be very much convinced that these two matrices have the same eigenvectors.

Or are there any other convincing proofs to show this result?

2. Sep 11, 2016

### Staff: Mentor

You already have said everything that can be said here. I don't know why you don't regard it as a proof.
There is no way of proving $A^2y=\alpha y \; \wedge \; Ax= \lambda x \; \Rightarrow \; x=y$ because it is not true.
E.g. $x$ and $y$ can simply be two different (linear independent) eigenvectors of $A$.

What you can prove is $Ay=\alpha y \; \Rightarrow \; A^2y=\alpha^2 y$ which is done by the equation you posted.

3. Sep 12, 2016

### FactChecker

Let B = A2 and α = λ2. Then Bx = αx. That is the definition of x being an eigenvector of B.

4. Sep 12, 2016

### Aldnoahz

Yeah I think I had some problems with my logic. Now I understand. Thank you.

5. Sep 17, 2016

### mathwonk

i am not sure what you have concluded but it is not true that A^2 has the same eigenvectors as A, since it can have more. E.g. take D the derivative acting on polynomials of degree ≤ one. Then D^2 = 0 and thus has x as an eigenvector, since D^2x = 0, but D does not since Dx = 1. Of course an eigenvector of A is also an eigenvector of A^2, "trivially", as proved above, but the converse is false.

Last edited: Sep 17, 2016