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Finding Eigenvectors and Eigenvalues

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data
    The Matrix A is as follows
    A= [4 -4 0
    2 -2 0
    -2 5 3]

    and has 3 distinct eigenvalues λ1<λ2<λ3
    Let Vi be the unique eigenvector associated with λi with a 1 as its first nonzero component.
    Let
    D = [ λ1 0 0
    0 λ2 0
    0 0 λ3]

    and P= [v1|v2|v3]

    Find D and P.



    2. Relevant equations
    not going to right it out but I know you find eigenvectors by (λI-A)X = 0
    and eigenvalues by det(λI-A).



    3. The attempt at a solution

    So I already found D with the eigenvalues as {0,2,3}

    I tried to solve for P and got

    [-1 -2 0
    1 1 0
    1 1 1]
    but this is wrong.

    I am kinds confused b/c the textbook says (λI-A)X=0 but my professor uses (A-λI)X=0 in class!
     
  2. jcsd
  3. Apr 29, 2014 #2

    jbunniii

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    The columns of ##P = [v_1 | v_2 | v_3]## should be eigenvectors of ##A##. But calculating ##Av_1## and ##Av_2## shows that the first two columns are not eigenvectors.

    Both are correct. One is simply the negative of the other. Since a vector is zero if and only if its negative is zero, the equations are equivalent.
     
  4. Apr 29, 2014 #3
    What do you mean? That the eigenvectors that I calculated v1,v2,v3 are not correct? Or that they are not actual eigenvectors?

    What should I do then?
     
  5. Apr 29, 2014 #4

    jbunniii

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    They are not eigenvectors:
    $$Av_1 = \begin{bmatrix}4 & -4 & 0 \\
    2 & -2 & 0 \\
    -2 & 5 & 3 \\
    \end{bmatrix}
    \begin{bmatrix}
    -1 \\ 1 \\ 1 \end{bmatrix}
    =
    \begin{bmatrix}
    -8 \\ -4 \\ 10 \end{bmatrix}
    $$
    which is not a multiple of
    $$\begin{bmatrix}
    -1 \\ 1 \\ 1 \end{bmatrix}
    $$
    Similarly, ##A v_2## is not a multiple of ##v_2##.
     
  6. Apr 29, 2014 #5
    So then what is/how do i find P?
     
  7. Apr 29, 2014 #6

    jbunniii

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    The columns of ##P## should be eigenvectors of ##A##. Your eigenvalues are correct. How did you calculate the corresponding eigenvectors? Maybe you just made an arithmetic mistake.
     
  8. Apr 29, 2014 #7
    So I took each value 0,2,3 and did (A-λI)X = 0

    For zero:
    (4-0) -4 0
    2 (-2-0) 0
    -2 5 (3-0)

    then I reduced and got
    v1 = -1,-1,1
    and did same process for the other values and got

    v2=-2,-1,1
    v3=0,0,1

    Everytime I solve this problem I get the same values, but the signs change but the numbers are still the same. Have you tried solving it? Is this what you got? What could I be doing wrong?
     
  9. Apr 29, 2014 #8

    jbunniii

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    OK, that's fine. Using this ##v_1## we calculate ##Av_1 = 0##, so ##v_1## is an eigenvalue associated with the eigenvalue ##0##. In your original post you had v1 = [-1, 1, 1], so maybe you made a sign error last time you calculated it.
    Nope, using this ##v_2## I calculate ##A v_2 = [0, 0, -3]##, which is not a multiple of ##[-2, -1, 1]##.
    This is fine, assuming it goes with the eigenvalue 3.
    I think you're making sign errors in some of your calculations. Can you show how you computed ##v_2##?
     
  10. Apr 29, 2014 #9
    Okay here is calculation again for v2:
    λ=2

    (4-2) -4 0
    2 (-2-2) 0
    -2 5 1


    simplified:
    2 -4 0
    2 -4 0
    -2 5 1

    row 1 minus row 2
    2 -4 0
    0 0 0
    -2 5 1

    row 1 plus row 3

    2 -4 0
    0 0 0
    0 1 1

    divide row one by 2

    1 -2 0
    0 0 0
    0 1 1

    x= 2t
    y= t
    z= -t

    thus v2 = 2,1,-1
     
  11. Apr 29, 2014 #10

    jbunniii

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    OK, that calculation is correct. You originally had ##v_2 = [-2, 1, 1]##, so you probably made a sign error in your earlier attempt.

    So let's summarize: you have eigenvalues ##0,2,3## with corresponding eigenvectors ##v_1 = [-1, -1, 1]##, ##v_2 = [2, 1, -1]##, and ##v_3 = [0, 0, 1]##. So your ##P## matrix could be
    $$P = [v_1 | v_2 | v_3] = \begin{bmatrix}
    -1 & 2 & 0 \\
    -1 & 1 & 0 \\
    1 & -1 & 1 \end{bmatrix}$$
    However, there is one more thing the original problem asked for:
    Your ##v_1## has ##-1## as its first component, and your ##v_2## has ##-2##. So you have to normalize your vectors to get ##1## as the first nonzero component. (##v_3## is OK as is.)
     
  12. Apr 29, 2014 #11
    Wait but doesn't normalizing a vector mean taking v1/||v1|| ?

    How would that give me a 1 as the first nonzero component?
    Also if I use v2 = 2,1,-1 do I also need to normalize it if the 2 isn't negative?
     
  13. Apr 29, 2014 #12

    jbunniii

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    Sorry, I meant you need to scale the vectors so that the first nonzero component is 1. So for example, if you had an eigenvector ##v = [-3, 1, -5]## you would multiply it by -1/3 to get ##[1, -1/3, 5/3]##.

    Note that any nonzero multiple of an eigenvector is still an eigenvector associated with the same eigenvalue, because if ##Av = \lambda v## then multiplying ##v## by a constant scalar ##c##, we get ##A(cv) = c(Av) = c(\lambda v) = \lambda(cv)##.

    Generally there's no need to scale the eigenvectors in any particular way unless the problem requests it. Probably your instructor wants them scaled with a 1 in the first nonzero component so your answer is guaranteed to match his/hers if it is correct.
     
  14. Apr 29, 2014 #13
    Okay so scaling would give me:
    1 2 0
    1 1 0
    -1 -1 1
    Is this correct?
     
  15. Apr 29, 2014 #14

    jbunniii

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    The first and third columns are OK. The second column's first component is 2, not 1, so it is not scaled as requested.
     
  16. Apr 29, 2014 #15
    whoops forget to scale that one! Thanks for all your help I really appreciate it! For some reason the scaling thing didn't register with me on what the professor wanted and I was confused with the sign changing.
     
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