Is Every Scalar Multiple of an Eigenvector Also an Eigenvector?

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    Eigenvector Proof
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Homework Help Overview

The discussion revolves around the properties of eigenvectors in linear algebra, specifically whether every scalar multiple of an eigenvector is also an eigenvector. Participants are examining the implications of the eigenvalue equation A*v=λ*v and exploring the conditions under which scalar multiples maintain the eigenvector property.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the validity of the original poster's reasoning regarding scalar multiplication of eigenvectors. Some are suggesting that the proof needs to be approached from both directions, while others are emphasizing the importance of the properties of linear operators.

Discussion Status

The discussion is active, with participants providing guidance on the need for a more thorough proof. There is recognition of the original poster's attempt, but also a clear indication that further exploration of the topic is necessary. Multiple interpretations of the eigenvector property are being considered.

Contextual Notes

Some participants note the importance of not dividing by the scalar c without considering its implications, as well as the need to clarify the definitions and properties of linear operators in this context.

JayTheGent
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Hello PF, brand new member here.

A question about a proof:

If A*v=λ*v, then w = c*v is also an eigenvector of A.

This seems really simple to me, but perhaps I am doing it incorrectly:

A*c*v=λ*c*v, divide both sides by c and you are left with your original eigenvector of A. Am I missing something here?
 
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JayTheGent said:
Hello PF, brand new member here.

A question about a proof:

If A*v=λ*v, then w = c*v is also an eigenvector of A.

This seems really simple to me, but perhaps I am doing it incorrectly:

A*c*v=λ*c*v, divide both sides by c and you are left with your original eigenvector of A. Am I missing something here?

What you've shown is that if cv is an eigenvector, then v is an eigenvector. You need to show the other way. In addition, you can't always divide by c.

Also, this should be in homework help.
 
notice that subspaces are closed under scaling. Or just factor out the c as the matrix cId.
 
JayTheGent said:
Hello PF, brand new member here.

A question about a proof:

If A*v=λ*v, then w = c*v is also an eigenvector of A.

This seems really simple to me, but perhaps I am doing it incorrectly:

A*c*v=λ*c*v, divide both sides by c and you are left with your original eigenvector of A. Am I missing something here?

Welcome to PF!:smile:
Do not use * when applying an operator on a vector. The operator A assigns a vector u to vector v. Write u=A(v). In case v is an eigenvector of operator A, A(v)= λ*v. The right side is a product - a vector multiplied by a scalar, but the left-hand side is not.

You can use the property of linear operators that A(cv)=c A(v) (c is a scalar).

ehild
 
Last edited:
JayTheGent said:
Hello PF, brand new member here.

A question about a proof:

If A*v=λ*v, then w = c*v is also an eigenvector of A.

This seems really simple to me, but perhaps I am doing it incorrectly:

A*c*v=λ*c*v, divide both sides by c and you are left with your original eigenvector of A. Am I missing something here?
As you have been told, what you shown is that "If cv is an eigenvector of A with eigenvalue λ then so is v". To prove the other way, reverse your proof: from Av= λv, multiply both sides by c.
 

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