5 ms·

Can you give an example of two eigenvectors that do not lie on the same line? (I am curious, I am not a mathematician)

stablemap 9y ago
What he gives is the identity matrix—multiply on the right by any 2x1 vector and you will get that vector back. So any such vector, provided it is not the zero vector, is an eigenvector with eigenvalue 1.
btilly 9y ago
[1 0] and [0 1] are both eigenvectors of the identity matrix, yet not on the same line. I highly recommend http://www.axler.net/DwD.html http://www.axler.net/DwD.html for developing a good intuition about what eigenvalues and eigenvectors actually are.
- dragandj 9y ago
  I obviously formulated this wrong. Will rephrase, but doesn't the intuiton generally hold that when you find an eigenvector for one eigenvalue, then you can construct infinite number of eigenvectors by scaling that one?
  - btilly 9y ago
    Yes, that is trivially true. In fact any linear combination of eigenvectors for the same eigenvalue will be an eigenvector for that eigenvalue. Another important point to appreciate is that eigenvalues and eigenvectors frequently require complex numbers. For example the eigenvalues of [[0 1] [-1 0]] are i and -i. This geometrically correlates with some kind of "rotation".
  - kgwgk 9y ago
    The problem arises when there is degeneration (some eigenvalues are repeated). See for example https://www.ch.ntu.edu.tw/~byjin/course/ChemMath/handouts/degeneracy.pdf https://www.ch.ntu.edu.tw/~byjin/course/ChemMath/handouts/de... [edit: the rest of my original comment was not correct] If an eigenvalue is unique, the corresponding eigenvectors are on a line. If it appears twice, the corresponding eigenvectors are [edit: could be] on a plane. In general, they span [edit: could span] a subspace of dimension equal to the multiplicity.
    - btilly 9y ago
      Not true. The matrix [[1 1] [0 1]] has the eigenvalue 1 repeated twice but the eigenvectors are on a line. The generalized eigenvectors are on a plane. The case where the eigenvectors do not span the space corresponds to matrices whose Jordan normal form has upper triangular bits.
    - neutronicus 9y ago
      The dimension of the space spanned by the eigenvectors associated with a degenerate eigenvalue is bounded above by the algebraic multiplicity of the eigenvalue and bounded below by one.
  - gizmo686 9y ago
    You seem to have reworded to "All eigenvectors that correspond to one a unique eigenvalue lie on the same line, but have different magnitudes." This is technically true if you interpret "unique" to mean "multiplicity of one", but still seems confusing, especially as you have not yet discussed the multiplicity of eigenvalues. I would consider rephrasing the article as you do in the above comment, "when you find an eigenvector for one eigenvalue, then you can construct infinite number of eigenvectors by scaling that one".
- agandy 9y ago
  The book based on this paper is one of the best text books I've read. I highly recommend reading it if you're looking for a solid understanding of linear algebra.
  - tptacek 9y ago
    As someone who worked their way through Strang (we did a little study group and did the homework assigned to some class at Rutgers), I'd love it if someone could take a moment to explain what they like so much about Axler. I've tried a couple times to slog my way through Axler and had not much success.
    - tnecniv 9y ago
      I have both Axler and Strang. I learned from Axler but occasionally use Strang as a reference. I find that Axler emphasizes the algebra part of linear algebra while Strang emphasizes the linear part. One isn't better than the other but one might make more sense to different people.
    - tanderson92 9y ago
      Axler avoids determinants with a passion. For good reason, too, because they are not helpful in understanding how spaces interact. I forget how Strang proves the spectral theorem but compared to Lax the pedagogy could not be more different. Determinants are really nice tools for short proofs, but not helpful in understanding. But Axler is the way you want to think about linear algebra as a practitioner.
    - pvg 9y ago
      Axler is easier to digest if you have some familiarity with the tools and the jargon but hated/forgot everything in your college linear algebra course. A (very imperfect) analogy might be something like the GoF book vs Peter Norvig's essay on design patterns.
    - jacobolus 9y ago
      I’m not in the target audience and have neither learned nor taught from Axler’s book, so I can’t tell you precisely what people like about it, but it is aimed at pure math undergraduates who are interested in using linear algebra in much more abstract contexts, and focuses more on proofs than many introductory linear algebra textbooks. Not sure if they’re useful (I haven’t watched them), but Axler made a series of videos about the core content of his book https://www.youtube.com/watch?v=lkx2BJcnyxk&list=PLGAnmvB9m7zOBVCZBUUmSinFV0wEir2Vw&index=1 https://www.youtube.com/watch?v=lkx2BJcnyxk&list=PLGAnmvB9m7...
- deleted 9y ago
  [deleted]

What he gives is the identity matrix—multiply on the right by any 2x1 vector and you will get that vector back. So any such vector, provided it is not the zero vector, is an eigenvector with eigenvalue 1.

btilly 9y ago

[1 0] and [0 1] are both eigenvectors of the identity matrix, yet not on the same line. I highly recommend http://www.axler.net/DwD.html http://www.axler.net/DwD.html for developing a good intuition about what eigenvalues and eigenvectors actually are.

dragandj 9y ago
I obviously formulated this wrong. Will rephrase, but doesn't the intuiton generally hold that when you find an eigenvector for one eigenvalue, then you can construct infinite number of eigenvectors by scaling that one?
- btilly 9y ago
  Yes, that is trivially true. In fact any linear combination of eigenvectors for the same eigenvalue will be an eigenvector for that eigenvalue. Another important point to appreciate is that eigenvalues and eigenvectors frequently require complex numbers. For example the eigenvalues of [[0 1] [-1 0]] are i and -i. This geometrically correlates with some kind of "rotation".
- kgwgk 9y ago
  The problem arises when there is degeneration (some eigenvalues are repeated). See for example https://www.ch.ntu.edu.tw/~byjin/course/ChemMath/handouts/degeneracy.pdf https://www.ch.ntu.edu.tw/~byjin/course/ChemMath/handouts/de... [edit: the rest of my original comment was not correct] If an eigenvalue is unique, the corresponding eigenvectors are on a line. If it appears twice, the corresponding eigenvectors are [edit: could be] on a plane. In general, they span [edit: could span] a subspace of dimension equal to the multiplicity.
  - btilly 9y ago
    Not true. The matrix [[1 1] [0 1]] has the eigenvalue 1 repeated twice but the eigenvectors are on a line. The generalized eigenvectors are on a plane. The case where the eigenvectors do not span the space corresponds to matrices whose Jordan normal form has upper triangular bits.
  - neutronicus 9y ago
    The dimension of the space spanned by the eigenvectors associated with a degenerate eigenvalue is bounded above by the algebraic multiplicity of the eigenvalue and bounded below by one.
- gizmo686 9y ago
  You seem to have reworded to "All eigenvectors that correspond to one a unique eigenvalue lie on the same line, but have different magnitudes." This is technically true if you interpret "unique" to mean "multiplicity of one", but still seems confusing, especially as you have not yet discussed the multiplicity of eigenvalues. I would consider rephrasing the article as you do in the above comment, "when you find an eigenvector for one eigenvalue, then you can construct infinite number of eigenvectors by scaling that one".
agandy 9y ago
The book based on this paper is one of the best text books I've read. I highly recommend reading it if you're looking for a solid understanding of linear algebra.
- tptacek 9y ago
  As someone who worked their way through Strang (we did a little study group and did the homework assigned to some class at Rutgers), I'd love it if someone could take a moment to explain what they like so much about Axler. I've tried a couple times to slog my way through Axler and had not much success.
  - tnecniv 9y ago
    I have both Axler and Strang. I learned from Axler but occasionally use Strang as a reference. I find that Axler emphasizes the algebra part of linear algebra while Strang emphasizes the linear part. One isn't better than the other but one might make more sense to different people.
  - tanderson92 9y ago
    Axler avoids determinants with a passion. For good reason, too, because they are not helpful in understanding how spaces interact. I forget how Strang proves the spectral theorem but compared to Lax the pedagogy could not be more different. Determinants are really nice tools for short proofs, but not helpful in understanding. But Axler is the way you want to think about linear algebra as a practitioner.
  - pvg 9y ago
    Axler is easier to digest if you have some familiarity with the tools and the jargon but hated/forgot everything in your college linear algebra course. A (very imperfect) analogy might be something like the GoF book vs Peter Norvig's essay on design patterns.
  - jacobolus 9y ago
    I’m not in the target audience and have neither learned nor taught from Axler’s book, so I can’t tell you precisely what people like about it, but it is aimed at pure math undergraduates who are interested in using linear algebra in much more abstract contexts, and focuses more on proofs than many introductory linear algebra textbooks. Not sure if they’re useful (I haven’t watched them), but Axler made a series of videos about the core content of his book https://www.youtube.com/watch?v=lkx2BJcnyxk&list=PLGAnmvB9m7zOBVCZBUUmSinFV0wEir2Vw&index=1 https://www.youtube.com/watch?v=lkx2BJcnyxk&list=PLGAnmvB9m7...
deleted 9y ago
[deleted]