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

  1. Sep 11, 2016 #1
    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. jcsd
  3. Sep 11, 2016 #2


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    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.
  4. Sep 12, 2016 #3


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    Let B = A2 and α = λ2. Then Bx = αx. That is the definition of x being an eigenvector of B.
  5. Sep 12, 2016 #4
    Yeah I think I had some problems with my logic. Now I understand. Thank you.
  6. Sep 17, 2016 #5


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    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

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