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What are eigen values?

physlink.com

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Re: What are eigen values?

#41

Post by gjm11 »

eru wrote: post #38
gjm11 wrote: post #28

Earlier quoted context omitted.

The zero mapping is a homomorphism. I would not call it a "kind of equivalence"; would you?

It's a special (and quite trivial) "kind of equivalence". See http://en.wikipedia.org/wiki/Equivalence_class

I know what an equivalence class is, thank you. How does that make the map from (say) R^3 to itself that sends everything to 0 a "kind of equivalence"?

It gives rise to an equivalence relation on R^3 in a fairly natural way, as indeed any function gives rise to an equivalence relation on its domain: x~y iff f(x)=f(y). But that doesn't mean that it is a "kind of equivalence".

Now, obviously, "kind of" is vague enough that saying "an endomorphism of a vector space simply Is Not a 'kind of equivalence'" would be too strong. But I would like to know what, exactly, you mean by calling something a "kind of equivalence", because I'm unable to think of any meaning for that phrase that (1) seems sensible to me and (2) implies that endomorphisms of vector spaces are "kinds of equivalence".

(Looking back at what bonsaitree wrote, I see s/he said "of any kind of equivalence", which I unconsciously typo-corrected to "or any kind of equivalence". But perhaps I misunderstood and bonsaitree meant something else, though I can't think what it might be. bt, if you're reading this: my apologies if I misunderstood, and would you care to clarify if so?)

Re: What are eigen values?

#42

Post by gjm11 »

gjm11 wrote: post #13
bonsaitree wrote: post #11

Earlier quoted context omitted.

Yes. For a linear system (of any kind of equivalence) within an N-dimensional vector space, the Eigenvalues represent the scaling factors across those dimensions when the system's state is represented by an NxN sparse-diagonalized matrix (i.e. all values are 0 except for the main diagonal). Those non-zero values along the main diagonal are its Eigenvalues and its rows are Eigenvectors. For the common 3D isometric (e.…

I'm afraid there are several errors in that. 1. A linear mapping is not a "kind of equivalence" by any reasonable definition. For instance, the function that maps every vector to 0 is a linear mapping, and it has plenty of eigenvectors. (All with eigenvalue 0.) 2. The eigenvectors are not the rows of the diagonalized matrix. They are the rows (or columns, depending on just how you define things) of the matrix that do…

Incidentally: that Markov chain thing? Suppose you make a Markov chain out of all the world's web pages, where each step moves you from a page to a randomly selected page it links to, or (occasionally for a page that has links, and unconditionally for a page with no links) to a completely random page.

Then the entries in that unique eigenvector -- equivalently, the long-term probabilities of landing on each page -- are basically the Google pagerank. (I'm sure Google's pagerank computation has lots of tweaks in it, and they certainly consider things other than pagerank to determine their search results. But pagerank was their original key idea.)

Re: What are eigen values?

#43

Post by CamperBob »

anonymousDan wrote: post #32
CamperBob wrote: post #29

Earlier quoted context omitted.

So there's one part of the explanation on the page that I don't quite get: she uses the example of a coin being turned 360 degrees along some axis as leaving all possible vectors as eigenvectors: "If you rotate a coin by 360 degrees you preserve all directions and so each direction is an eigenvector. Because no stretching has occurred, all of these eigenvectors have eigenvalue 1. Rotating the coin by 60 degrees destr…

Because the original vector is not pointing in the same direction as the final vector (it is at an angle of 60 degrees to it). If you do a 360 degree rotation you get back to where you started. So any number of successive full rotations will have eigenvectors, each with eigenvalue 1. If you do a 180 degree rotation, the resulting vector will be pointing in the opposite direction, and will have eigenvalue -1.

Because the original vector is not pointing in the same direction as the final vector (it is at an angle of 60 degrees to it).

Mmm, no, the axis around which I rotated the coin didn't change at all, by definition.

Re: What are eigen values?

#44

Post by jibiki »

CamperBob wrote: post #43
anonymousDan wrote: post #32

Earlier quoted context omitted.

Because the original vector is not pointing in the same direction as the final vector (it is at an angle of 60 degrees to it). If you do a 360 degree rotation you get back to where you started. So any number of successive full rotations will have eigenvectors, each with eigenvalue 1. If you do a 180 degree rotation, the resulting vector will be pointing in the opposite direction, and will have eigenvalue -1.

Because the original vector is not pointing in the same direction as the final vector (it is at an angle of 60 degrees to it). Mmm, no, the axis around which I rotated the coin didn't change at all, by definition.

You are doing a rotation in 3 dimensions. Surely the original example meant two dimensions. In 3 dimensions, the axis of rotation is indeed an eigenvector.

Re: What are eigen values?

#45

Post by jibiki »

proee wrote: post #35

I still remember the following question on my linear algebra test in college: "Prove that A and A-Inverse have the same Eigen Values and corresponding Eigen Vectors." The solution to this made no sense, but I managed to memorize it so that I could pass the test. Forest through the trees?

It's not quite true, is it...?

  Av = cv        v is an eigenvector, c the corresponding eigenvalue
  (A^-1)Av = (A^-1)cv
  v = c(A^-1)v
  (1/c)v = (A^-1)v
So if c is an eigenvalue of A, then 1/c is an eigenvalue of (A^-1). (c can't be 0 because A is invertible, I think.)

Re: What are eigen values?

#46

Post by eru »

gjm11 wrote: post #41
eru wrote: post #38

Earlier quoted context omitted.

It's a special (and quite trivial) "kind of equivalence". See http://en.wikipedia.org/wiki/Equivalence_class

I know what an equivalence class is, thank you. How does that make the map from (say) R^3 to itself that sends everything to 0 a "kind of equivalence"? It gives rise to an equivalence relation on R^3 in a fairly natural way, as indeed any function gives rise to an equivalence relation on its domain: x~y iff f(x)=f(y). But that doesn't mean that it is a "kind of equivalence". Now, obviously, "kind of" is vague enough…

Yes, any function gives rise to equivalence classes. And linear functions give rise to equivalence classes that preserve structure in vector spaces.

(I guess our discourse has reached its end of usefulness here.)

Re: What are eigen values?

#47

Post by bonsaitree »

gjm11 wrote: post #13
bonsaitree wrote: post #11

Earlier quoted context omitted.

Yes. For a linear system (of any kind of equivalence) within an N-dimensional vector space, the Eigenvalues represent the scaling factors across those dimensions when the system's state is represented by an NxN sparse-diagonalized matrix (i.e. all values are 0 except for the main diagonal). Those non-zero values along the main diagonal are its Eigenvalues and its rows are Eigenvectors. For the common 3D isometric (e.…

I'm afraid there are several errors in that. 1. A linear mapping is not a "kind of equivalence" by any reasonable definition. For instance, the function that maps every vector to 0 is a linear mapping, and it has plenty of eigenvectors. (All with eigenvalue 0.) 2. The eigenvectors are not the rows of the diagonalized matrix. They are the rows (or columns, depending on just how you define things) of the matrix that do…

Days late, but worth the response for accuracy's sake since I just popped in on HN now:

> 1. A linear mapping is not a "kind of equivalence" by any > reasonable definition.

By definition, ANY mapping is an equivalence relation--even if that relation results in a 0 or NAN value.

> 2. The eigenvectors are not the rows of the diagonalized > matrix...

Correct. It depends on one's orientation (columns vs. rows), but the matrix which does the coordinate transforms contains the eigenvectors. Bad wording on my part. The key behind diagonalization is it removes any orientation issues from the relationship by establishing eigenweights across a given vector space.

> 3. I think the paragraph beginning "For the common 3D isometric ..."

Your interpretation is correct and, indeed, my choice of verbiage was poor. I really should have used the words "ortho-normal to a given vector space" which collapses to the common XYZ unit vectors in a linearly partitioned vector space of 6 degrees of freedom (3 translations & 3 rotations)--e.g. classic Cartesian space.

> Tensors are just as linear as matrices.

Most tensor fields are modeled using linear approximations (i.e. matricies of n-dimensions), but the very fact that a tensor itself is being used in the characteristic equations is typically indicative of non-linear behavior in the overall system. For example, in fluid dynamics used to model airflow across a wing or boundary values issues when the Cauchy stress tensor is used for structures undergoing plastic deformation.

I believe that you are conflating Manifolds with Tensors. The latter is a refinement and/or characteristic relation defined upon the former. A non-linear Tensor is defined upon a manifold with one or more non-linear relations. Perhaps you are used to dealing exclusively with metric tensors?

> The probabilities in a Markov chain's stationary state are not eigenvalues.

Here you are spot-on. Indeed it's the eigenvalues of the transition matrix (or convergence for ergodic ones) to which I was referring. Thanks (again) for the correction and further clarification.

Re: What are eigen values?

#48

Post by bonsaitree »

eru wrote: post #46
gjm11 wrote: post #41

Earlier quoted context omitted.

I know what an equivalence class is, thank you. How does that make the map from (say) R^3 to itself that sends everything to 0 a "kind of equivalence"? It gives rise to an equivalence relation on R^3 in a fairly natural way, as indeed any function gives rise to an equivalence relation on its domain: x~y iff f(x)=f(y). But that doesn't mean that it is a "kind of equivalence". Now, obviously, "kind of" is vague enough…

Yes, any function gives rise to equivalence classes. And linear functions give rise to equivalence classes that preserve structure in vector spaces. (I guess our discourse has reached its end of usefulness here.)

Yes. Thanks for the correct re-wording as "or" and "equivalence class".
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