Linear Algebra: 2 eigenfunctions, one with eigenvalue zero

[binbagsss]

Homework Statement

If I have two eigenfunctions of some operator, that are linearly indepdendent e.g $F(x) , G(x)+16F(x)$ and $F(x)$ has eigenvalue $0$, does this mean that $G(x)$ must itself be an eigenfunction?

I thought for sure yes, but the way I particular question I just worked through went seemed to suggest it shouldn't be obvious, so perhaps not always guaranteed too.

Homework Equations

So I have $\hat{P} F(x) = 0 F(x)$
$\hat{P}G(x)=16F(x)+G(x)$

The Attempt at a Solution

[/B]
$=> \hat{P}(16F(x)+G(x))= \hat{P}(16F(x))+\hat{P}(G(x))=0+\hat{P}(16F(x)+G(x))$ therefore $16F(x)+G(x)$ is an eigenfunction with eigenvalue $1$

Intuition says $G(x)$ should be an eigenfunction, I can't think how to show it from the above however.
Thanks

 

[Orodruin]

A linear combination of eigenfunctions is generally not an eigenfunction. It is only an eigenfunction if the eigenvalues are the same.

 

[MathematicalPhysicist]

Perhaps you drank too much vodka... but P(16F(x)+G(x))= itself is not the definition of an eigenvalue equation.

 

[binbagsss]

A linear combination of eigenfunctions is generally not an eigenfunction. It is only an eigenfunction if the eigenvalues are the same.

Totally agree, didn't say that.
I thought if one has eigenvalue zero it may be a special case.

 

[binbagsss]

Perhaps you drank too much vodka... but P(16F(x)+G(x))= itself is not the definition of an eigenvalue equation.

Not this time my friend, unfortunately.

Typo =..0× F(x) + P(G(x))= 16F(x) + G(x)

 

[PeroK]

Homework Statement

If I have two eigenfunctions of some operator, that are linearly indepdendent e.g $F(x) , G(x)+16F(x)$ and $F(x)$ has eigenvalue $0$, does this mean that $G(x)$ must itself be an eigenfunction?

I thought for sure yes, but the way I particular question I just worked through went seemed to suggest it shouldn't be obvious, so perhaps not always guaranteed too.

Homework Equations

So I have $\hat{P} F(x) = 0 F(x)$
$\hat{P}G(x)=16F(x)+G(x)$

The Attempt at a Solution

[/B]
$=> \hat{P}(16F(x)+G(x))= \hat{P}(16F(x))+\hat{P}(G(x))=0+\hat{P}(16F(x)+G(x))$ therefore $16F(x)+G(x)$ is an eigenfunction with eigenvalue $1$

Intuition says $G(x)$ should be an eigenfunction, I can't think how to show it from the above however.
Thanks

You've lost me here. In general, if $F$ and $G$ are eigenfunctions, then $F+G$ is an eigenfunction iff $F$ and $G$ have the same eigenvalue:

$\hat{P}(F + G) = \lambda F + \mu G = \lambda (F+G) + (\mu - \lambda) G$

Which is just what @Orodruin said in post #2.

 

[Orodruin]

Totally agree, didn't say that.
I thought if one has eigenvalue zero it may be a special case.

Why would it be a special case? There is no reason for that. As has already been said, the only relevant information is if the eigenvalues are the same.

 

[binbagsss]

You've lost me here. In general, if $F$ and $G$ are eigenfunctions, then $F+G$ is an eigenfunction iff $F$ and $G$ have the same eigenvalue:

$\hat{P}(F + G) = \lambda F + \mu G = \lambda (F+G) + (\mu - \lambda) G$

Which is just what @Orodruin said in post #2.

I am doing the opposite of this.
I have that f and 16f +g are both eigenfunctions . F has eigenvalue 0.

 

[Orodruin]

I am doing the opposite of this.
I have that f and 16f +g are both eigenfunctions . F has eigenvalue 0.

No, you are really not doing the opposite. Let h = 16f + g and then what you are asking is if g = h - 16 f is an eigenvector if h and f are eigenvectors.

 

[PeroK]

I am doing the opposite of this.
I have that f and 16f +g are both eigenfunctions . F has eigenvalue 0.

You assumed that $F$ and $16F + G$ were eigenfunctions and asked if $G$ was also an eigenfunction. As $F$ is an eigenfunction, so is $-16F$, with the same eigenvalue as $F$. Hence, $G = -16F + (16F + G)$ is an eigenfunction iff $-16F$ and $16F + G$ have the same eigenvalue, i.e. iff $F$ and $16F + G$ have the same eigenvalue.