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Matrix with one eigenvalue

  1. Mar 7, 2010 #1
    Hi, this is my 1st post here and i was wondering if I could get some help

    Suppose wehave a 2x2 matrix A with one eigenvalue [tex]\lambda[/tex], but it is not a scalar matrix. Suppose [tex]\vec{v2}[/tex] is a nonzero vector which is not an eigenvector of A; show that [tex]\vec{v1}[/tex] = (A-[tex]\lambda[/tex])[tex]\vec{v2}[/tex] is an eigenvector of A. Also show that if P is the matrix with columns [tex]\vec{v1}[/tex] and [tex]\vec{v2}[/tex] then P^(-1)AP = [[tex]\lambda[/tex] 1
    0 [tex]\lambda[/tex]]


    3. The attempt at a solution

    I tried calculating (A-[tex]\lambda[/tex])[tex]\vec{v1}[/tex] to try and proove that it is equal to 0, however i end up with it being equal to (A-[tex]\lambda[/tex] I)^2[tex]\vec{v2}[/tex]

    Thank you very much
     
  2. jcsd
  3. Mar 7, 2010 #2

    Dick

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    You know that a matrix solves it's characteristic polynomial, right? What's the characteristic polynomial of a 2x2 matrix with one eigenvalue?
     
  4. Mar 7, 2010 #3
    well if A [ a b the the characteristic polynomial is
    c d]
    [tex]\lambda[/tex]^2 + [tex]\lambda[/tex](-a-d) + (ad -bc) but since we have 1 eigenvalue then the dicriminant must be 0 and hence (a+d)^2-4(ad - bc) = 0

    I already tried doing this and calculating (A-[tex]\lambda[/tex]I)^2 but that leads to nowhere and certainly doesn't equal 0

    (thanks for the quick reply)
     
  5. Mar 7, 2010 #4

    Dick

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    If the discriminant is zero that quadratic must factor into a square. The characteristic polynomial must be (x-lambda)^2.
     
  6. Mar 7, 2010 #5
    So in that case can we say that the minimal polynomial is therefore (x-[tex]\lambda[/tex])^2 and hence we have (A-[tex]\lambda[/tex])^2 is equal to 0,
    therefore we get from the original equation that
    (A-[tex]\lambda[/tex])[tex]\vec{v1}[/tex]=0 and therefore v1 is an eigenvector.
    correct?
     
  7. Mar 7, 2010 #6

    Dick

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    Correct.
     
  8. Mar 7, 2010 #7
    Ok, but what does this tell us about the matrix A, for example in the second part if we take
    P=[v1 v2] = [v11 v21 then we have
    v12 v22]
    P^-1 = v22/det(P) -v21/det(P)
    -v12/det(P) v11/det(P)
    So how does P-1AP = [tex]\lambda[/tex] 1
    0 [tex]\lambda[/tex]
     
  9. Mar 7, 2010 #8

    Dick

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    You are thinking of this on too detailed a level. P^(-1)AP is just the matrix of A in the basis {v1,v2}. Write down Av1 and Av2 and deduce what the matrix must look like.
     
  10. Mar 7, 2010 #9
    ok so we have v1 is eigenvector
    hence Av1 = [tex]\lambda[/tex]v1
    hence (A-[tex]\lambda[/tex])v1 = 0 but we also have v1 = (A-[tex]\lambda[/tex]I)v2

    hence Av1 = A(A-[tex]\lambda[/tex]I)v2 and -[tex]\lambda[/tex]v1 = -[tex]\lambda[/tex](A-[tex]\lambda[/tex]I)v2

    hence by calculating it out i find (A-[tex]\lambda[/tex]I)v1 = (A-[tex]\lambda[/tex])^2 v2
    To get v1 =(A-[tex]\lambda[/tex])v2 which is equivalent to A = v1 + [tex]\lambda[/tex]v2

    but then i m stuck, I dont quite understand how to get A from this because we'll always have it in terms of v1 and v2
     
    Last edited: Mar 7, 2010
  11. Mar 7, 2010 #10

    Dick

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    You WANT to have it in terms of v1 and v2, P^(-1)AP is the matrix of the linear transformation A IN THE BASIS {v1,v2}. You have Av1=lambda*v1 and Av2=v1+lambda*v2, right? Suppose e1=(1,0) and e2=(0,1) were the usual basis and I told you Be1=lambda*e1 and Be2=e1+lambda*e2. Then you could probably tell me the entries of the matrix B, right? What would they be?
     
  12. Mar 7, 2010 #11
    yea in ur case we would have
    B = [tex]\lambda[/tex] 1
    0 [tex]\lambda[/tex]

    but in our case we don't have the components of v1 and v2 we can just right A in terms of v11,v12,v21,v22?
    so would P-1AP cancel out and we're sort of left in the basis e1 e2 to get B?
     
  13. Mar 7, 2010 #12

    Dick

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    Hmm. Not sure I'm getting through. Since you KNOW how A acts in the basis {v1,v2} you KNOW what the matrix of A is in the basis {v1,v2}. It's [[lambda,1],[0,lambda]]. In contrast, you DON'T know what the original entries of A were. The thing you KNOW is called P^(-1)AP.
     
  14. Mar 7, 2010 #13
    yea i think I get it now, after doing the change of basis on A with v1 and v2, then doing P(-1)AP is equivalent to the matrix B in your earlier statement since A and B are the same but in different bases
    correct?
     
  15. Mar 7, 2010 #14

    Dick

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    Yes, A and P^(-1)AP are the same transformation in two different bases.
     
  16. Mar 7, 2010 #15
    great THANK YOU very much, i really appreciated your help :D
     
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