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Question on Eigenspaces

  1. Jan 1, 2008 #1
    Hi there, I'm having a bit of a problem understanding eigenspaces, kernel and span. I've searched the net and wikipedia but there doesn't seem to be any clear examples.

    I have an example in a book that says this:

    Let,
    A =
    [ 2 2 2 ]
    [ 0 2 2 ]
    [ 0 0 2 ]

    I can see the characteristic polynomial = [itex](X - 2)^3[/itex] so 2 is the only eigenvalue.

    It then calculates the generalised eigenspaces: [itex]V_t(2)[/itex]
    [itex]V_1(2) = [/itex]
    ker [ 0 2 2 ]
    .....[ 0 0 2 ]
    .....[ 0 0 0 ]

    The kernel is calculated by row reducing the matrix:

    [itex]V_1(2) = [/itex]
    ker [ 0 1 0 ] = span [1]
    .....[ 0 0 1 ]...........[0]
    .....[ 0 0 0 ]...........[0]

    [itex]V_2(2) = [/itex]
    ker [ 0 0 1 ] = span [1] [0]
    .....[ 0 0 0 ]...........[0] [1]
    .....[ 0 0 0 ]...........[0] [0]


    [itex]V_3(2) = [/itex]
    ker [ 0 0 0 ] = span [1] [0] [0]
    .....[ 0 0 0 ]...........[0] [1] [0]
    .....[ 0 0 0 ]...........[0] [0] [1]

    that is the end of the example.


    so now here are my questions:

    QUESTION 1
    What does it mean by : [itex]V_1(2)[/itex], [itex]V_2(2)[/itex], [itex]V_3(2)[/itex] etc.

    QUESTION 2
    What exactly is the kernal in this example and how is the span calculated from the kernal matrices....??

    QUESTION 3
    which part of this whole question/example is the eigenspace?

    thankyou very much. if this could be explained then it would clear most of the confusion on this topic.

    :)
     
  2. jcsd
  3. Jan 1, 2008 #2

    Hurkyl

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    I'm confused -- is it just that you don't know the definitions, or is there more to your questions?
     
  4. Jan 1, 2008 #3

    HallsofIvy

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    [itex]V_1[/itex] is X-2 (more correctly, X-2I), row reduced, [itex]V_2= (X-2I)^2[/itex], row reduced, and [itex]V_3= (X-2I)^3[/itex]. Of course, [itex]V_3= 0[/itex] because X satisfies its "characteristic equation", (X- 2I)3= 0.

    I don't know what you mean by "the kernel" or "the span". The kernel of any linear matrix, T, is the set of vectors, v, such that Tv= 0. If
    [tex]V_1 x= \left[\begin{array}{ccc}0 & 1 & 0 \\0 & 0 & 1\\0 & 0 & 0\end{array}\right]\left[\begin{array}{c} x \\ y \\ z\end{array}\left]= \left[\begin{array}{c} y \\ z \\ 0\end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \\ 0\end{array}\right][/tex]
    then we must have y= 0 and z= 0. x can be anything so the kernel consists of vectors of the form (x, 0, 0)= x(1, 0, 0). (1, 0, 0) spans that vector space. Similarly for the other two matrices. I would have expected you to have learned "span" and "kernel" long before you start working with eigenvectors.

    Eigen space or "generalized eigenspaces"? The eigen space for A itself is the kernel of [itex]V_1[/itex] which is the set of all vectors (x, 0, 0), spanned by (1, 0, 0). The "generalized eigenspaces" include the kernel of [itex]V_2[/itex], all vectors of the form (x, y, 0) which is spanned by (1, 0, 0) and (0, 1, 0) and the kernel of [itex]V_3[/itex] which is all of R3, spanned, of course, by (1, 0, 0), (0, 1, 0), and (0, 0, 1).
     
  5. Jan 1, 2008 #4

    i dont know the definitions.. i dont know how they got the kernels, the span etc etc
    i also would appreciate what the difference between generalised eigenspace and normal eigenspace is? thanx
     
  6. Jan 1, 2008 #5

    thnx. this was a brilliant explanation :) really helped me :) brilliant
     
  7. Jan 1, 2008 #6
    oh just one question though.

    for V_2(2)
    [tex]V_2x= \left[\begin{array}{ccc}0 & 0 & 1 \\0 & 0 & 0\\0 & 0 & 0\end{array}\right]\left[\begin{array}{c} x \\ y \\ z\end{array}\left]= \left[\begin{array}{c} z \\ 0 \\ 0\end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \\ 0\end{array}\right][/tex]

    here z must be 0... so x and y can be anything... the kernal therefore consists of the vectors of form: {1,0,0,} and {1,1,0}..
    so how does that deduce the span as[tex] \left[\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right]\left[\begin{array}{c} 0 \\ 1 \\ 0\end{array}\right][/tex]
    the vector forms arent the same as the span for me...
    please clear this final confusion :)
     
  8. Jan 1, 2008 #7

    Vid

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    x = t
    y = s
    z = 0

    Therefore, (x,y,z) = (t,s,0) = (t,0,0) + (0,s,0) = t(1,0,0) + s(0,1,0) = span[(1,0,0), (0,1,0)]
     
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