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Question about eigenvectors

  1. Nov 28, 2005 #1
    I am studying for a test that I have today and just need a quick anwser. the problem is find the eigenvector of [8 -10;2 -1] where ; means skip to a new line. So I get lamda=4,3 then try to find eigenvector for lamda=4. I get the vector [5/2;1] which works but the anwser in the book is [2;5]. Both anwsers seem to be right but is there some rule to say that the book anwser is better.
     
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  3. Nov 28, 2005 #2

    TD

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    Eigenvectors are always determined up to an arbitrary (non-zero) multiplication factor. The book apparently preferred not to have fractions but the smallest possible integers, which is usually done. Both are fine though :smile:
     
  4. Nov 28, 2005 #3

    matt grime

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    its answer doesn't have any fractions in it, that is considered better, just as you should replace 1/sqrt(2) with sqrt(2)/2. Its answer is also valid even if we only consider vectors with entries in the integers (which we may want to) so it is more general. Your answer is fine if we remember that we CAN divide by two, but it holds in greater generality than that.

    Just remember that if you can you should clear denominators in an answer, and since e-vectors are only defined up to scalar multiples why not? PERSONAL OPINION and NOT in any way authoritative is that I wouldn't mark you down for your answer. But that is purely MY opinion and is not necessarily that of your teacher/marker.

    actually, have just noticed your answer and the books are not compatible, bnut that is probably a typo

    (5/2,1) is not a scalar multiple of (2,5)

    (5/2,1) and (5,2) yes or
    (2/5,1) and (2,5) also ok
     
  5. Nov 28, 2005 #4
    Thanks a lot! Ill stick to using the smallest possible integers since thats what the book does and I wouldnt want to loose points on something so easy. The anwser in the book was [5,2] I just typed it wrong.
     
  6. Nov 28, 2005 #5

    HallsofIvy

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    As pointed out before, if v is an eigenvector corresponding to eigenvalue [itex]\lambda[/itex] the so is any multiple of v. Yes, [5/2, 1] is an eigenvector corresponding to eigenvalue 4. Multiplying that vector by 2, so is [5, 2]. Some people prefer to have integer components (which is not always possible), some people prefer to write eigenvectors so that have unit length. Since [5,2] has length [itex]\sqrt{29}[/itex] an unit eigenvector would be [tex]\left[\frac{5}{\sqrt{29}},\frac{2}{\sqrt{29}}\right][/tex]
     
  7. Nov 29, 2005 #6
    Why would you want to use the unit vector. We never really went over that in class but there is a question on a lab that uses it. Where we had to show that the vectors are mutually orthogonal. I dont have the exact question with me but Im still working on the anwser.
     
  8. Nov 30, 2005 #7

    HallsofIvy

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    If you have a basis for your vector space where the vectors in the basis are "orthonormal"- orthogonal and have length 1- then calculations are much simplified. If, in addition, your basis vectors are all eigenvectors of some linear transformation, then working with that linear transformation is very simple.
     
  9. Nov 30, 2005 #8
    Im home now so I can give you the exact question.

    Given matrix a=[3,3,2; 3,6,5; 2,5,11]

    a) determine eigenvalues.
    This part I can do just fine

    b) Determine unit eigenvectors with magnitude one.
    So I just take each eigenvector and divide by the lenght?

    c) Show that the unit eigenvectors are mutually orthogonal.
    This I have no idea what to do. Ive been reading through the book and it looks like I have to show that two unit vectors mutiplied together are zero.
     
  10. Nov 30, 2005 #9

    TD

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    Indeed, just as HallsofIvy illustrated.

    Yes, if you mean the scalair product (inner product).
     
  11. Nov 30, 2005 #10

    HallsofIvy

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    Good! That's the hard part!

    Yes, that's exactly right.

    Reading the book is always a good idea! Two vectors are orthogonal if and only if their dot product is equal to 0. Becareful to say "dot" product- there are several different kinds of multiplication defined for vectors. Take the dot product of each pair of vectors and see what you get.
     
  12. Nov 30, 2005 #11
    Thanks a lot! I tried it out and the dot products do equal zero.
     
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