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What is the associated eigenvalue?

  1. Apr 14, 2008 #1
    1. The problem statement, all variables and given/known data

    V is an eigenvector of the nxn matrix A, with a eigenvalue of 4. explain why V is a eigenvector of A^2+2A+3I. what is the associated eigenvalue?

    2. Relevant equations



    3. The attempt at a solution

    is the eigenvalue of A^2+2A+3I=21?
     
    Last edited by a moderator: Mar 7, 2013
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  3. Apr 14, 2008 #2

    quasar987

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    What does it mean that V is an eigenvector of A with eigenvalue 4?

    Answer that, and then ask yourself what happens when you multiply the matrix (A^2+2A+3I) by V from the right.

    The answer for the eigenvalue is not 21.
     
  4. Apr 14, 2008 #3
    I don't understand how A^2+2A+3I is populated in what I assume is a 2x2 matrix? If you tell me that it will help me a lot. I don't understand how a polynomial populates a matrix
     
  5. Apr 14, 2008 #4
    to answer you question. A is a matrix (nxn) v is a nonzero vector in R^n. Av is a scalar multiple of lamda, Av=landav. lamda is the eigenvalue. the unknown here is A either its a matrix that i do know the vectors of or it's a variable that populates the matrix that is used to calculate the eigenvalue...right? so as you say A^2+2A+3I is the matrix how does it populated matrix...what is its dimensions?
     
  6. Apr 14, 2008 #5

    quasar987

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    I don't understand your use of the work "populated".

    Here, A is an nxn matrix. And as you probably know, multiplying two nxn matrices or adding two nxn matrices, or multiplying an nxn matrix by a scalar gives out another nxn matrix. So (A^2+2A+3I) is an nxn matrix.

    Now, what is (A^2+2A+3I)v, knowing that Av=4v?
     
  7. Apr 15, 2008 #6
    eigenvalue

    looks like it equals 4...this has been helpful. I've had a misconception that coefficients in A^2+2A+3I would be entered into the nxn matrix in some manner..e.g. [[A^2, 2], [3, 0]] or if i knew the vectors of A i could try to calculated the eigenvalue. so is 4 the correct solution?
     
  8. Apr 15, 2008 #7

    quasar987

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    No, it isn't.

    Use the fact that multiplication between matrix and vector is distributive. Meaning that for A, B two nxn matrices and u in R^n, (A+B)u=Au+Bu.
     
  9. Apr 15, 2008 #8

    HallsofIvy

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    Saying that v is an eigenvector of A with eigenvalue 4 means that Av= 4v.

    What is (A2+ 2A+ 3I)v= A(Av)+ 2Av+ v?
     
  10. Apr 15, 2008 #9
    eigenvalue

    ok i get what you've been saying about using the distributive rule (thanks) A^2v+2Av+3v (is 3v correct 3I*v=3v). so are we at this point A^2v+2Av+3v=4v? if yes then what? thanks for your help.
     
  11. Apr 15, 2008 #10

    quasar987

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    Why "A^2v+2Av+3v=4v"? What you know is that Av=4v.

    So use that to write A^2v+2Av+3v = A(4v)+2(4v)+3v=... (you finish)
     
  12. Apr 15, 2008 #11

    HallsofIvy

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    No, I didn't say that at all. Av= 4v. What is A(Av)= A(4v)= 4Av? What is 2Av?
     
  13. Apr 15, 2008 #12
    A(4v)+2(4v)+3v=4(Av)+8v+3v=4(Av)+8v+3v=4(4v)+8v+3v=16v+8v+3v=27v does that make sense?
     
  14. Apr 15, 2008 #13

    quasar987

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    Plenty.
     
  15. Apr 15, 2008 #14
    thank you. i'll do better with the next question!
     
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