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Eigenvector for Complex Eigenvalue help

  1. Oct 21, 2011 #1
    In my lecture notes my prof used the eigenvalue c= 1 + i and ended up with the matrix with (5 3+i) as row 1, and the second row is zeroes. After that, he simply wrote that the basis for this eigenvalue c is (3+i,-5) (in column form) without explaining. How did he get that basis? I tried working it out and not sure what he did.
     
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  3. Oct 22, 2011 #2

    HallsofIvy

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    I really don't understand what you are trying to say. You cannot start with an eigenvalue and just make a matrix out of it. Possibly, you are saying that 1+i is an eigenvalue of a given matrix and that the new matrix was constructed of corresponding eigenvectors. Also, it makes no sense to say "a basis for an eigenvalue" since only vector spaces have bases and an eigenvalue is a number. Perhaps you mean "basis for the subspace of all eigenvectors corresponding to the given eigenvalue".


    What was the original matrix for which 1+ i was an eigenvalue? That is essential to understanding this.

    (It is certainly true that the vectors (5, 3+i) and (3+ i, -5) are orthogonal. That may be relevant.)
     
  4. Oct 22, 2011 #3
    Yes, this is what I meant. Sorry for any confusion.

    In the example, A=(4 2) (row 1) and (-5 -2) (row 2) (my apologies, I don't know how to put actually make that into a matrix here). We found the characteristic polynomial, and that the roots were 1+i and 1-i. He took the eigenvalue 1+i, and proceeded to get the eigenvector for that eigenvalue. When he plugged in [tex] ((i+1)I_2 - A)) [/tex] he then row reduced and got the matrix I said in my first post. He then wrote that the basis for this eigenvalue was (3+i, -5). I hope that clarifies. I was just confused on the arithmetic he did to get that basis.
     
  5. Oct 22, 2011 #4

    lurflurf

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    It is because for a real operator eigenvalues and eigenvectors come in complex conjugate pairs.
    so if we have the eigen pair
    1 + i;(3+i,-5)
    we also have the eigen pair
    1 - i;(3-i,-5)
     
  6. Oct 22, 2011 #5

    AlephZero

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    If you row reduce a 2x2 singular matrix, you will always get something of the form
    [tex]\left(\begin{array}{c}a & b \\ 0 & 0\end{array}\right)[/tex]
    And if you want a non-zero solution to
    [tex]\left(\begin{array}{c}a & b \\ 0 & 0\end{array}\right)
    \left(\begin{array}{c}x \\ y \end{array}\right)=
    \left(\begin{array}{c}0 \\ 0 \end{array}\right)[/tex]
    If sould be obvious it will be a multiple of
    [tex]\left(\begin{array}{c}-b \\ a\end{array}\right)[/tex]
     
  7. Oct 22, 2011 #6
    Ok this all makes sense. Thank you
     
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