Must every linear operator have eigenvalues? If so, why?

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Discussion Overview

The discussion centers on whether every linear operator must have eigenvalues, exploring this question within the context of finite-dimensional vector spaces over the complex and real numbers, as well as infinite-dimensional spaces. Participants delve into theoretical implications, definitions, and related mathematical principles.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants argue that every linear operator on a finite-dimensional vector space over the complex numbers must have at least one eigenvalue, while acknowledging that eigenvalues could be zero.
  • Others point out that in finite-dimensional spaces over the real numbers, certain operators, such as rotation matrices, may not have eigenvalues because their eigenvalues are complex.
  • A participant mentions that in infinite-dimensional spaces, exceptions may arise, citing the momentum operator in quantum mechanics, where eigenstates are not normalizable.
  • There is a discussion about using the pigeonhole principle to argue for the existence of eigenvalues, with some participants seeking clarification on its application in this context.
  • Some participants propose that the characteristic polynomial of a linear operator cannot be constant, suggesting that this leads to the existence of eigenvalues.
  • Concerns are raised regarding the necessity of non-trivial eigenvectors associated with eigenvalues, with some asserting that the definition implies a non-trivial solution must exist.
  • One participant suggests that proving the existence of eigenvalues is equivalent to proving that every polynomial has a root in its field of coefficients, linking it to the Fundamental Theorem of Algebra.

Areas of Agreement / Disagreement

Participants express differing views on the existence of eigenvalues for linear operators, particularly between finite-dimensional spaces over the complex numbers and real numbers. The discussion remains unresolved regarding the implications for infinite-dimensional spaces and the necessity of non-trivial eigenvectors.

Contextual Notes

Participants note that the discussion hinges on the algebraic properties of the underlying field, with implications for the existence of eigenvalues depending on whether the field is algebraically closed.

ygolo
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It seems to me that http://en.wikipedia.org/wiki/Schur_decomposition" relies on the fact that every linear operator must have at least one eigenvalue...but how do we know this is true?

I have yet to find a linear operator without eigenvalues, so I believe every linear operator does have at least one eigenvalue.

Still how does one prove it?

Since we are looking for solutions to (A-Ia)|V>=|0>, wouldn't it be possible the A-Ia is always nonsingular and that the equation has only trivial solutions?
 
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On a finite-dimensional vector space V over the complex numbers, it should be obvious that any linear operator must have eigenvalues, although some or all of those eigenvalues might be zero. Since the operator is a map from V to itself (or a subset), one can use the pigeonhole principle to show that at least one vector in V must be parallel to its image.

Over the real numbers, some operators do not have eigenvalues (e.g. rotation matrices in R^2), because the eigenvalues happen to be complex. But I don't know if this really counts.

On an infinite-dimensional vector space, they may be other subtleties that allow an exception. For example, in quantum mechanics, the momentum operator is linear, but its eigenstates are not normalizable, and hence not technically part of the Hilbert space. That is, the eigenvectors are not actually members of the vector space, and so it might be reasonable to say that the corresponding eigenvalues do not actually exist.
 
Ben Niehoff said:
On a finite-dimensional vector space V over the complex numbers, it should be obvious that any linear operator must have eigenvalues, although some or all of those eigenvalues might be zero. Since the operator is a map from V to itself (or a subset), one can use the pigeonhole principle to show that at least one vector in V must be parallel to its image.

Point for Clarification: How does on use the pigeonhole principle in this case? It seems like you could be mapping an uncountably infinite number of vectors (though from a space of finite dimension).

What are the pigeonholes? (the vector directions? I don't really trust intuitions based on the arrow representations of vectors)

...and how do you assign the images of the operation to them?

---------

I just thought of something. If we can prove that a complex polynomial must have complex roots, then we can apply it to the characteristic polynomial.

...and this is, I believe (for non-constant polynomials), The Fundamental Theorem of Algebra.

http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

Because of the diagonal entries of |A-aI|, I believe, the determinant will have to be a non-costant polynomial
 
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For every Eigenvalue must there be a non-trivial eigenvector? If so, why?

So then the next question becomes:

For every Eigenvalue must there be a non-trivial eigenvector? If so, why?
 


ygolo said:
So then the next question becomes:

For every Eigenvalue must there be a non-trivial eigenvector? If so, why?


I think you cannot have one without the other. It is in the definition.
 


I wonder why the science advisers and pf mentors have so far not replied this thread. Too basic?

I am actually not sure I answered my first question fully, since I did not rigorously prove that the characteristic polynomial is non-constant.

Reb said:
I think you cannot have one without the other. It is in the definition.

Is it? I think its slightly more subtle. Ax=ax, x could be trivial, that is |0>, even if there are non-trivial a, such that |A-aI|=0.

The solution to (A-aI)x=|0>, could have both parts be 0, no?
 
No, part of the definition is that Ax = ax holds for some nontrivial x.

The characteristic polynomial cannot be constant. This is easy to prove by expanding the determinant |A-aI| by minors. Specifically, if A is an NxN matrix, then what is the coefficient of a^N?

I wonder why the science advisers and pf mentors have so far not replied this thread. Too basic?

What are we, chopped liver? And as far as I know, there is nobody being paid to sit at their desk and read everything that happens on PF so that they can help everybody who asks.
 
ygolo said:
Point for Clarification: How does on use the pigeonhole principle in this case? It seems like you could be mapping an uncountably infinite number of vectors (though from a space of finite dimension).

What are the pigeonholes? (the vector directions? I don't really trust intuitions based on the arrow representations of vectors)

http://golem.ph.utexas.edu/category/2007/05/linear_algebra_done_right.html

The pigeon holes are the dimensions of the space. We've only got n dimensions. If we apply a transformation T over and over again to a point, we'll produce a linearly dependent set of vectors. I only have a fluff idea of how the rest of the argument works, but I got that much of it. The articles explains the rest.
 
Ben Niehoff said:
No, part of the definition is that Ax = ax holds for some nontrivial x.

Well then the proof by the fundamental theorem of algebra falls short of proving the exisence of an eigenvalue then doesn't it?

Ben Niehoff said:
The characteristic polynomial cannot be constant. This is easy to prove by expanding the determinant |A-aI| by minors. Specifically, if A is an NxN matrix, then what is the coefficient of a^N?

Good point!

Ben Niehoff said:
What are we, chopped liver? And as far as I know, there is nobody being paid to sit at their desk and read everything that happens on PF so that they can help everybody who asks.

I was just curious, but I appreciate yours and the others' help.

Tac-Tics said:
http://golem.ph.utexas.edu/category/2007/05/linear_algebra_done_right.html

The pigeon holes are the dimensions of the space. We've only got n dimensions. If we apply a transformation T over and over again to a point, we'll produce a linearly dependent set of vectors. I only have a fluff idea of how the rest of the argument works, but I got that much of it. The articles explains the rest.

I'll try and decipher this after work.
 
  • #10
ygolo said:
Well then the proof by the fundamental theorem of algebra falls short of proving the exisence of an eigenvalue then doesn't it?

Are we talking in general or just numeral matrices under the reals?
 
  • #11
John Creighto said:
Are we talking in general or just numeral matrices under the reals?

General complex linear operators.

---
Silly me. If |A-aI|=0 then A-aI is sigular, and therefore not of full rank, and therefore has a (non trivial) null-space, which means A has an eigenvector.

Forgive me, it has been 13 years since I took linear algebra, and 12 years since Complex Analysis.
 
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  • #12
By the way, the question of whether every linear operator (on a finite-dimensional F-vector space) has an eigenvalue is actually equivalent to the question of whether every polynomial with coefficients in F has a root in F.

I.e. if you can prove what the OP is asking for (over \mathbb{C}) without using the fundamental theorem of algebra, then you can actually use your argument to prove the FTA!
 
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  • #13
ygolo said:
It seems to me that http://en.wikipedia.org/wiki/Schur_decomposition" relies on the fact that every linear operator must have at least one eigenvalue...but how do we know this is true?

I have yet to find a linear operator without eigenvalues, so I believe every linear operator does have at least one eigenvalue.

Still how does one prove it?

Since we are looking for solutions to (A-Ia)|V>=|0>, wouldn't it be possible the A-Ia is always nonsingular and that the equation has only trivial solutions?

For finite dimensional vector space there is a classic theorem. It shows that your question boils down to whether a certain polynomial has roots in the underlying scalar field of the vector space. If the field is algebraically close-e.g. the complex numbers-then there is always a root.

If the field is not algebraically closed then there may not be eigenvalues. For instance rotation of the plane by 90 degrees has no real eigenvalues.
 
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  • #14
I started to think about this problem with the following quasi-informal statement

"Every linear operator can be represented as a matrix..."

which is for my taste, a very very very beautiful theorem. By the way I originally misstated it. The fields and the spaces are left out for a good reason. to just focus on the necessary part
 
  • #15
Every linear operator, over finite dimensional vector space V, can be represented by a matrix, given a specific basis on that vector space. (The matrix depends upon the choice of basis.) Given that, then you can write the "characteristic equation" for the matrix as a polynomial equation and the use the fundamental theorem of algebra: that every such equation over the complex numbers has a solution.
 
  • #16
ygolo said:
It seems to me that http://en.wikipedia.org/wiki/Schur_decomposition" relies on the fact that every linear operator must have at least one eigenvalue...but how do we know this is true?

I have yet to find a linear operator without eigenvalues, so I believe every linear operator does have at least one eigenvalue.

Still how does one prove it?

Since we are looking for solutions to (A-Ia)|V>=|0>, wouldn't it be possible the A-Ia is always nonsingular and that the equation has only trivial solutions?

Rotation by 90 degrees in the plane has no real eigenvalues - only complex.
Why do you think that every linear operator has at least one eigen value? Can you show me your argument?
 
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  • #17
In my opinion, the most revealing way to see this theorem for finite dimensional vector spaces is the following classical argument.

If L:V -> V is a linear map of the n dimensional vector space,V, then the vectors L^n(v),L^n-1(v),...,L(v),V are linearlly dependent so some linear combination of them is zero. This linear combination says that some polynomial in L sends V to zero. Consider the ideal generated by all such polynomials.

Since the ring of polynomials over a field is a principal ideal domain, there is a polynomial that divides all of the others in this ideal. This polynomial must be zero on the entire vector space since the above construction using the vector,V, can be performed on any vector.

If this polynomial it has a root over F, then it factors as (L - a)P. The linear factor (L - a) must be zero on some non-zero vector so a is an eigen-value and the vector is an eigen-vecor.
 

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