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A matrix with Repeated eigenvalues and its corresponding eigenvectors.

  1. May 17, 2014 #1
    1. The problem statement, all variables and given/known data

    I am asked to find the diagonal matrix of eigenvalues, D, and the matrix of corresponding eigenvectors, P, of the following matrix:
    [tex]
    \begin{pmatrix}
    1 & 0 & 0\\
    0 & 1 & -2\\
    0 & 0 & -1
    \end{pmatrix}
    [/tex]

    2. Relevant equations



    3. The attempt at a solution
    We just started this topic, and so far i haven't had much trouble answering questions like these, but this question is different.

    I found the eigenvalues to be λ = 1 (with algebraic multiplicity 2) and λ = -1

    for λ = 1 the matrix is [tex]
    \begin{pmatrix}
    0 & 0 & 0\\
    0 & 0 & 0\\
    0 & 0 & -2
    \end{pmatrix}
    [/tex] after putting in row echelon form. This is where my confusion begins, solving gives z=0, and since x nor y are described, they can take any value I please (to my understanding). so for the first eigenvector i let x=1 and y=0, then i let x=0,y=1 to give [tex]
    \begin{bmatrix}
    1\\
    0\\
    0
    \end{bmatrix}
    [/tex] and [tex]
    \begin{bmatrix}
    0\\
    1\\
    0
    \end{bmatrix}
    [/tex] respectively.
    For λ = -1 the matrix is [tex]
    \begin{pmatrix}
    2 & 0 & 0\\
    0 & 2 & -2\\
    0 & 0 & 1
    \end{pmatrix}
    [/tex] which gives x=y=z=0 which confuses me even more as this is [tex]
    \begin{bmatrix}
    0\\
    0\\
    0
    \end{bmatrix}
    [/tex]

    So after all that I said D = [tex]
    \begin{pmatrix}
    1 & 0 & 0\\
    0 & 1 & 0\\
    0 & 0 & -1
    \end{pmatrix}
    [/tex] which im fairly sure is correct
    and P = [tex]
    \begin{pmatrix}
    1 & 0 & 0\\
    0 & 1 & 0\\
    0 & 0 & 0
    \end{pmatrix}
    [/tex] which im fairly sure isnt (Mathematica told me at least one of the parts was wrong and i think its this one).

    Could someone please help me understand this example, and forgive me for the bad formatting, im not sure how to keep everything in the same line.
     
  2. jcsd
  3. May 17, 2014 #2
    Dont worry i think i found my mistake. I got the matrix for λ=-1 wrong and hence confused the shiz out of myself.
    Just checking the correct matrix for λ= -1 would be [tex]
    \begin{bmatrix}
    0\\
    1\\
    1
    \end{bmatrix}
    [/tex] right?
     
  4. May 17, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, 1 is a "double" eigenvalue and -1 is an eigenvalue. I find it simplest to use the definition of "eigenvalue" to find the eigenvectors: [itex]\lambda[/itex] is an eigenvalue of linear transformation A if and only if there exist a non-zero vector, v, such that [itex]Av= \lambda v[/itex]. (Of course, then, [itex](A-\lambda)v= 0[/itex] which is what you used.)

    For eigenvalue "1":
    [tex]\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}x \\ y- 2z \\ -z\end{pmatrix}= \begin{pmatrix}x \\ y \\ z \end{pmatrix}[/tex]

    so we have x= x, y- 2z= y, and -z= z. The last equation gives z= 0 and the first two equations, x= x and y= y, are always true. Any vector of the form (x, y, 0)= x(1, 0, 0)+ y(0, 1, 0) is an eigenvalue corresponding to eigenvalue 1.


    For eigenvalue "-1":
    [tex]\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}x \\ y- 2z \\ -z\end{pmatrix}= \begin{pmatrix}-x \\ -y \\ -z \end{pmatrix}[/tex]

    so that we must have x= -x, y- 2z= -z, and -z= -z. The first equation gives x= 0, the third is true for any z and the second gives y= z. Any eigenvector corresponding to eigenvalue -1 is of the form (0, z, z)= z(0, 1, 1).

    It is easy to see that the matrix [tex]P= \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}[/tex]
    has inverse [tex]P^{-1}= \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix}[/tex]


    and that
    [tex]P^{-1}AP= \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}[/tex]
     
    Last edited: May 17, 2014
  5. May 17, 2014 #4

    pasmith

    User Avatar
    Homework Helper

    You may want to reconsider the claim [itex]P^{-1} = I[/itex].
     
  6. May 17, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Thanks, I copied the wrong matrix from my work! I went back and edited my post to correct that.
     
  7. May 19, 2014 #6
    Thank you for the alternate way of approaching these types of questions halls. When, for example, you say x=-x which leads to x=0 I assume this is because the only possible way for this to be true is if x=0 ?
     
  8. May 19, 2014 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Isn't that what "x satisfies the equation" means?

    More specifically, I observed that adding x to both sides of "x= -x" gives "2x= 0" and then, dividing both sides by 2, "x= 0".
     
  9. May 19, 2014 #8
    Oh wow, please excuse that last comment. I have no Idea what I was thinking.... lol.
     
  10. May 19, 2014 #9
    One more thing, I noticed that mathematica allowed any matrix, P, of the form
    ##
    \begin {pmatrix}
    a & c & 0 \\
    b & d & 1 \\
    0 & 0 & 1
    \end {pmatrix}
    ##
    As long as (a, b,0) =/= h (c, d,0) where h is a scalar multiple. Is this because the eigenvectors for the eigenvalue of 1 could be of the form (a, b,0) and (c, d,0) as long as they are not scalar multiples of eachother?
     
  11. May 21, 2014 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I don't know what you mean by "Mathematica allowed". But, in general, 1 is NOT necessarily an eigenvalue for such a matrix.
     
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