3 ms·

Yes. For a linear system (of any kind of equivalence) within an N-dimensional vector space, the Eigenvalues represent the scaling factors across those dimensio

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.g. xyz) coordinate system, the Eigenvalues can be thought of as a kind of multiplier across the unit vectors (Eigenvectors) [[1,0,0][0,1,0][0,0,1]]. This is the "stretching" analog mentioned in the article. FWIW, Eigenvalues are not just a salient property of linear systems (i.e. matrices), but also of higher-order tensors. Finding (or more-often approximating) these "characteristic scaling states" is a critical step in numerical analysis in everything from quantum mechanics, financial hedging strategies, and even consumer product marketing plans. If you've ever represented a system as a series of Markov probability chains, every row of the "convergent/dominant" (if any) state contains an Eigenvalue.

gjm11 17y ago
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 does the coordinate transformation to diagonalize your matrix. 3. I think the paragraph beginning "For the common 3D isometric ..." is rather confused. I certainly am when reading it. Perhaps the problem is in my brain; what exactly do you mean? (Here's the nearest true thing I can think of to what that paragraph says: For many, but not all, linear transformations from a space to itself, there is an orthogonal coordinate system with respect to which the transformation's matrix is diagonal; then the eigenvectors are the axes of that coordinate system, and the eigenvalues are the amounts by which vectors along those axes get stretched.) 4. Tensors are just as linear as matrices. (You can do nonlinear things with a tensor, but then you can with a matrix too.) 5. The probabilities in a Markov chain's stationary state are not eigenvalues. (Well, I'm sure they're eigenvalues of something; any set of numbers can be made the eigenvalues of something; but they aren't, e.g., eigenvalues of the transition matrix.) What you may be thinking of is: a stationary state of a Markov chain is an eigenvector of the transition matrix, with eigenvalue 1; all eigenvalues have absolute value at most 1; if the Markov chain is ergodic (i.e., can get from any state to any other), then there is exactly one stationary state, exactly one eigenvector of eigenvalue 1, and all the other eigenvalues are strictly smaller. This is enough to guarantee convergence to the stationary state. Also: If eigenvalues are as important in dealing with higher-order tensors as they are for the second-order case (i.e., matrices) then that's news to me. Tell me more?
- eru 17y ago
  > 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.) Linear mappings are the homomorphisms between vector spaces. So they are a "kind of equivalence". See for example http://en.wikibooks.org/wiki/Linear_Algebra/Definition_of_Homomorphism http://en.wikibooks.org/wiki/Linear_Algebra/Definition_of_Ho...
  - gjm11 17y ago
    The zero mapping is a homomorphism. I would not call it a "kind of equivalence"; would you?
    - eru 17y ago
      It's a special (and quite trivial) "kind of equivalence". See http://en.wikipedia.org/wiki/Equivalence_class http://en.wikipedia.org/wiki/Equivalence_class
      - gjm11 17y ago
        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?)

gjm11 17y ago

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 does the coordinate transformation to diagonalize your matrix. 3. I think the paragraph beginning "For the common 3D isometric ..." is rather confused. I certainly am when reading it. Perhaps the problem is in my brain; what exactly do you mean? (Here's the nearest true thing I can think of to what that paragraph says: For many, but not all, linear transformations from a space to itself, there is an orthogonal coordinate system with respect to which the transformation's matrix is diagonal; then the eigenvectors are the axes of that coordinate system, and the eigenvalues are the amounts by which vectors along those axes get stretched.) 4. Tensors are just as linear as matrices. (You can do nonlinear things with a tensor, but then you can with a matrix too.) 5. The probabilities in a Markov chain's stationary state are not eigenvalues. (Well, I'm sure they're eigenvalues of something; any set of numbers can be made the eigenvalues of something; but they aren't, e.g., eigenvalues of the transition matrix.) What you may be thinking of is: a stationary state of a Markov chain is an eigenvector of the transition matrix, with eigenvalue 1; all eigenvalues have absolute value at most 1; if the Markov chain is ergodic (i.e., can get from any state to any other), then there is exactly one stationary state, exactly one eigenvector of eigenvalue 1, and all the other eigenvalues are strictly smaller. This is enough to guarantee convergence to the stationary state. Also: If eigenvalues are as important in dealing with higher-order tensors as they are for the second-order case (i.e., matrices) then that's news to me. Tell me more?

eru 17y ago
> 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.) Linear mappings are the homomorphisms between vector spaces. So they are a "kind of equivalence". See for example http://en.wikibooks.org/wiki/Linear_Algebra/Definition_of_Homomorphism http://en.wikibooks.org/wiki/Linear_Algebra/Definition_of_Ho...
- gjm11 17y ago
  The zero mapping is a homomorphism. I would not call it a "kind of equivalence"; would you?
  - eru 17y ago
    It's a special (and quite trivial) "kind of equivalence". See http://en.wikipedia.org/wiki/Equivalence_class http://en.wikipedia.org/wiki/Equivalence_class
    - gjm11 17y ago
      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?)