# Linear Algebra: 2 eigenfunctions, one with eigenvalue zero

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

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

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.

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
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## 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
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Homework Helper
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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.

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
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Homework Helper
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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
PeroK