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

[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?

 

[fresh_42]

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.

 

[FactChecker]

therefore A²x = λ²x.

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

 

[Aldnoahz]

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

 

[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.