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Eigenvectors of a 3x3 matrix (specifically diagonalizing), but need a clarifying hint

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data
    find the diagonalizing matrix P of the matrix A
    A=:
    1 3 0
    0 2 0
    0 0 2



    2. Relevant equations



    3. The attempt at a solution
    so i do the whole A-[tex]\lambda[/tex]I3 thing and i find my eigenvalues to be [tex]\lambda = 1, 2[/tex]
    when i do [tex]\lambda[/tex] = 1, i get the matrix
    0 3 0
    0 1 0
    0 0 1

    which turns into
    0 1 0
    0 0 1
    0 0 0

    i guess here's my main question. can i have any eigenvectors from here that satisfy this equation? i'm getting a different solution than the book, but it's been a while since i've taken linear algebra (this is a differential equations class and this parts review), and i seem to recall that this doesn't need to be like the book (i.e. the book presents one possible solution; as long as my solution is linearly independent, it should be valid).
    ...
    okay, so i just plugged it in and found out i'm incorrect.

    so how do i find the solution to
    0 1 0 | 0
    0 0 1 | 0
    0 0 0 | 0?
     
  2. jcsd
  3. Feb 27, 2010 #2

    vela

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    Re: eigenvectors of a 3x3 matrix (specifically diagonalizing), but need a clarifying

    What equations correspond to that augmented matrix?
     
  4. Feb 27, 2010 #3

    Dick

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    Re: eigenvectors of a 3x3 matrix (specifically diagonalizing), but need a clarifying

    And the first row of your second matrix should be [-1,0,0].
     
  5. Mar 1, 2010 #4
    Re: eigenvectors of a 3x3 matrix (specifically diagonalizing), but need a clarifying

    why? i thought the whole point of doing the eigenvalue was to get that to be zero? (at least in this part, when the particular eigenvalue cancels that out)
     
  6. Mar 1, 2010 #5

    Dick

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    Re: eigenvectors of a 3x3 matrix (specifically diagonalizing), but need a clarifying

    And I thought you were subtracting the eigenvalue from the matrix diagonal.
     
  7. Mar 1, 2010 #6
    Re: eigenvectors of a 3x3 matrix (specifically diagonalizing), but need a clarifying

    the original matrix:
    1 3 0
    0 2 0
    0 0 2

    eigenvalues 1, 2, 2

    subtracting 1*identity

    0 3 0
    0 1 0
    0 0 1

    isn't that right?
     
  8. Mar 1, 2010 #7

    Dick

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    Re: eigenvectors of a 3x3 matrix (specifically diagonalizing), but need a clarifying

    Yes, it is. I was thinking one of your matrices was the one where you subtracted 2*identity. Now you need to find the null space of that matrix.
     
  9. Mar 1, 2010 #8
    Re: eigenvectors of a 3x3 matrix (specifically diagonalizing), but need a clarifying

    I believe that was his original problem - finding the null space.
     
  10. Mar 2, 2010 #9

    HallsofIvy

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    Re: eigenvectors of a 3x3 matrix (specifically diagonalizing), but need a clarifying

    I have never particularly liked that method of finding eigenvectors- I prefer working directly from the definition.

    It is obvious that the eigenvalues of A are 1 and 2 with 2 being a double root of the eigenvalue equation.

    Any eigevector corresponding to eigenvalue 1 must satify
    [tex]\begin{bmatrix}1 & 3 & 0 \\ 0 & 2 & 0 \\ 0 & 2 & 0\end{bmatrix}\begin{bmatrix} x \\ y \\ z\end{bmatrix}= \begin{bmatrix} x \\ y \\ z\end{bmatrix}[/tex]

    Multiplying that out we get x+ 3y= x, 2y= y, and 2z= z. The only solutions to the last two equations are y= z= 0 and any value of x then satifies the first. An eigenvector corresponding to eigenvalue 1 is <x, 0, 0>= x<1, 0, 0>.

    Any eigenvector corresponding to eigenvalue 2 must satisfy
    [tex]\begin{bmatrix}1 & 3 & 0 \\ 0 & 2 & 0 \\ 0 & 2 & 0\end{bmatrix}\begin{bmatrix} x \\ y \\ z\end{bmatrix}= \begin{bmatrix} 2x \\ 2y \\ 2z\end{bmatrix}[/tex]

    Multiplying that out, we get the 3 equations x+ 3y= 2x, 2y= 2y, and 2z= 2z. The first equation gives us x= 3y and any values of y and z satisfy the last two. An eigenvector corresponding to eigenvalue 2 must be of the form <3y, y, z> = y<3, 1, 0>+ z<0, 0, 1>. That is, {<1, 0, 0>, <3, 1, 0>, <0, 0, 1>} is a basis for the entire space consisting of eigenvectors of A. If we take P to be the matrix having those vectors as columns:
    [tex]P= \begin{bmatrix}1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]
    then P-1AP is a diagonal matrix.



    2. Relevant equations



    3. The attempt at a solution
    so i do the whole A-[tex]\lambda[/tex]I3 thing and i find my eigenvalues to be [tex]\lambda = 1, 2[/tex]
    when i do [tex]\lambda[/tex] = 1, i get the matrix
    0 3 0
    0 1 0
    0 0 1

    which turns into
    0 1 0
    0 0 1
    0 0 0

    i guess here's my main question. can i have any eigenvectors from here that satisfy this equation? i'm getting a different solution than the book, but it's been a while since i've taken linear algebra (this is a differential equations class and this parts review), and i seem to recall that this doesn't need to be like the book (i.e. the book presents one possible solution; as long as my solution is linearly independent, it should be valid).
    ...
    okay, so i just plugged it in and found out i'm incorrect.

    so how do i find the solution to
    0 1 0 | 0
    0 0 1 | 0
    0 0 0 | 0?[/QUOTE]
     
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