Is Every Scalar Multiple of an Eigenvector Also an Eigenvector?

  • Thread starter Thread starter JayTheGent
  • Start date Start date
  • Tags Tags
    Eigenvector Proof
JayTheGent
Messages
1
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
Back
Top