Why Do Eigenvectors Stay The Same When A Matrix Is Squared?

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

• Context: Undergrad
• Thread starter Thread starter Aldnoahz
• Start date Start date Sep 11, 2016
• Tags Tags Eigenvectors Matrix

Click For Summary

Discussion Overview

The discussion centers around the relationship between the eigenvectors of a matrix A and its square A² in the context of linear algebra. Participants explore whether A² retains the same eigenvectors as A and seek proofs or clarifications regarding this relationship.

Discussion Character

• Exploratory
• Debate/contested
• Mathematical reasoning

Main Points Raised

• One participant questions the common assumption that A² has the same eigenvectors as A, suggesting that the existing reasoning does not constitute a proof of this claim.
• Another participant argues that it is not possible to prove that A²y = αy and Ax = λx implies x = y, as x and y can be different eigenvectors of A.
• A participant confirms that if Ay = αy, then A²y = α²y, supporting the idea that A² can be expressed in terms of its eigenvalues but does not necessarily share eigenvectors with A.
• One participant acknowledges a misunderstanding in their logic regarding the relationship between A and A².
• Another participant asserts that A² does not necessarily have the same eigenvectors as A, providing an example involving the derivative operator to illustrate that A² can have additional eigenvectors.

Areas of Agreement / Disagreement

Participants express disagreement regarding whether A² has the same eigenvectors as A. Some argue that while eigenvectors of A are also eigenvectors of A², the converse is not necessarily true, indicating multiple competing views on the topic.

Contextual Notes

Participants highlight the limitations of the discussion, including the dependence on specific definitions of eigenvectors and the potential for different eigenvectors arising from the squaring of matrices.

Who May Find This Useful

This discussion may be of interest to students and practitioners of linear algebra, particularly those exploring the properties of eigenvectors and eigenvalues in relation to matrix operations.

Aldnoahz Messages 37 Reaction score 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? Physics news on Phys.org

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fresh_42 Staff Emeritus Science Advisor Homework Helper Insights Author 2025 Award Messages 20,815 Reaction score 28,457 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 Science Advisor Homework Helper Messages 9,419 Reaction score 4,683

Aldnoahz said: therefore A2x = λ2x.

Let B = A2 and α = λ2. Then Bx = αx. That is the definition of x being an eigenvector of B. Aldnoahz Messages 37 Reaction score 1 Yeah I think I had some problems with my logic. Now I understand. Thank you. mathwonk Science Advisor Homework Helper Messages 11,979 Reaction score 2,261 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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