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Linear algebra - Diagonalizable matrix

  1. Jan 30, 2012 #1
    This question has a lot data but I don't really know how to connect it all together


    1. The problem statement, all variables and given/known data

    Let A be a 5x5 matrix over R

    3 eigenvectors of A are:
    [itex]u_1=(1,0,0,1,1)[/itex]
    [itex]u_2=(1,1,0,0,1)[/itex]
    [itex]u_3=(-1,0,1,0,0)[/itex]

    also:
    [itex]\rho (2I-A)>\rho (3I-A)[/itex]

    and: [itex]A(1,2,2,1,3)^t=(0,4,6,2,6)^t[/itex]

    prove that A is diagonalizable and find a diagonal matrix similiar to it.
    2. Relevant equations



    3. The attempt at a solution

    What I can make of this is:
    [itex](1,2,2,1,3)^t=u_1 + 2u_2 + 2u_3[/itex]

    (then maybe I can say that P has the eigenvectors as columns and
    [itex]AP(1,2,2,0,0)^t=(0,4,6,2,6)^t[/itex]

    but then what?)

    and because
    [itex]\rho (2I-A)>\rho (3I-A)[/itex]
    [itex]\rho (2I-A)\leq 5[/itex]
    it means that
    [itex]5>\rho (3I-A)[/itex]
    and so
    [itex]det(3I-A)=0[/itex]
    and 3 is an eigenvalue of A.

    I would like some hints/suggestions on what to do.

    Thanks!
     
  2. jcsd
  3. Jan 30, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    A 5 by 5 matrix is diagonalizable if and only if it has 5 independent eigenvectors. You are given three independent eigenvectors. Can you find two more?
     
  4. Jan 31, 2012 #3
    I know how to get the eigenvectors from an existing matrix
    but not from an eigenvalue or anything they gave me... :/
     
  5. Jan 31, 2012 #4
    I think I found one eigenvector,

    if 3 is an eigenvalue of A, then A is similar to a matrix which it's first column is
    (3,0,0,0,0)t

    so an eigenvector of this matrix is (1,0,0,0,0) and thus it's an eigenvector of A as well

    I only need to find one more :D (assuming I'm right).
     
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