Linear Algebra: 2 eigenfunctions, one with eigenvalue zero

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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
 

Answers and Replies

  • #2
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.
 
  • #3
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.
 
  • #4
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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.
Totally agree, didn't say that.
I thought if one has eigenvalue zero it may be a special case.
 
  • #5
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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.
Not this time my friend, unfortunately.

Typo =..0× F(x) + P(G(x))= 16F(x) + G(x)
 
  • #6
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.
 
  • #7
Orodruin
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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.
 
  • #8
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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.
 
  • #9
Orodruin
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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.
 
  • #10
PeroK
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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.

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.
 

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