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Math Help - Find Eigenvalues

  1. #1
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    Find Eigenvalues

    Given a matrix with components

    <br />
(A^{\pm})_{ij} = \delta_{ij} \pm a_i a_j<br />

    where a is a constant real vector, find the eigenvalues of the matrix.

    I can't really see where to start with this - any hints to see the obvious that I may have missed?
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  2. #2
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    Quote Originally Posted by thelostchild View Post
    Given a matrix with components

    <br />
(A^{\pm})_{ij} = \delta_{ij} \pm a_i a_j<br />

    where a is a constant real vector, find the eigenvalues of the matrix.

    I can't really see where to start with this - any hints to see the obvious that I may have missed?

    What is A^{\pm}\,,\,a_i\,,\,a_j? I suppose \delta_{ij} is Kronecker's delta, but for that...

    Tonio
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  3. #3
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    Quote Originally Posted by thelostchild View Post
    Given a matrix with components

    <br />
(A^{\pm})_{ij} = \delta_{ij} \pm a_i a_j<br />

    where a is a constant real vector, find the eigenvalues of the matrix.

    I can't really see where to start with this - any hints to see the obvious that I may have missed?
    \delta_{ij} = 0 for all i \neq j

    So this is a diagonal matrix, that is if I have understood your notation correctly.
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  4. #4
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    Sorry for the misunderstanding with notation
     A^{\pm} are two nxn matrices

    and  a_i are the components of an n component real vector.

    Just looking at it again I can see it can be written as

     A^{\pm} = I \pm a a^T

    If that helps, although I still cant see a way of doing this
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  5. #5
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    The formula uses the Kronecker delta:

    \delta_{ij} =\left \{\begin{matrix} 1, & \mbox{if } i=j \\ 0, & \mbox{if } i \ne j \end{matrix}\right.

    That is a short way of writing

    (A^{\pm})_{ij}= \begin{bmatrix}a_{11}=1 & a_{12}=0 & ...& a_{1n}=0\\a_{21}=0 & a_{22}=1 & ... &a_{2n}=0 \\ ...&...&...&...\\a_{n1}=0 & a_{n2}=0 & ... & a_{nn}=1\end{bmatrix} \pm a_i a_j

    But it doesn't make a whole lot of sense.

    Quote Originally Posted by thelostchild View Post
    Sorry for the misunderstanding with notation
     A^{\pm} are two nxn matrices

    and  a_i are the components of an n component real vector.

    Just looking at it again I can see it can be written as

     A^{\pm} = I \pm a a^T

    If that helps, although I still cant see a way of doing this
    Yes, if "a" is just an nxn matrix then you have to find the two resulting matrices here:  A^{\pm} = I \pm a a^T = B_1, B_2 and then you solve the characteristic equation for this matrix ; i.e., det (\lambda I - B_i) to find its eigenvalues.
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  6. #6
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    I would recommend looking at "easy" cases to start with (always a good strategy).

    For n= 1, this operator is just "1+ a" and its operation is just multiplying the number x: (1+a)x= \lambda x for \lambda= 1+ a, of course.

    For n= 2, this operator can be represented by the matrix \begin{bmatrix}1+ a_1^2 & a_1a_2 \\ a_1a_2 & 1+ a_2^2\end{bmatrix}. What are its eigenvalues?

    For n= 3, this operator can be represented by the matrix \begin{bmatrix}1+ a_1^2 & a_1a_2 & a_1a_3 \\ a_1a_2 & 1+ a_2^2 & a_2a_3 \\ a_1a_3 & a_2a_3 & 1+ a_3^2\end{bmatrix}.
    What are its eigenvalues?
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  7. #7
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    Ok so looking at the results for those two cases I assume the general result is a repeated eigenvalue of n=1 for (n-1) times and then

    <br />
\lambda=1 + \sum_{i=1}^{n} a_{i}^{2}<br />

    Thanks for the help
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  8. #8
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    Quote Originally Posted by thelostchild View Post
    Ok so looking at the results for those two cases I assume the general result is a repeated eigenvalue of n=1 for (n-1) times and then

    <br />
\lambda=1 + \sum_{i=1}^{n} a_{i}^{2}<br />

    Thanks for the help
    Well, I'm impressed! I suggested looking at simple cases as a general method of getting ideas. I hadn't actually solved the problem myself,
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    Quote Originally Posted by thelostchild View Post
    Ok so looking at the results for those two cases I assume the general result is a repeated eigenvalue of n=1 for (n-1) times and then

    <br />
\lambda=1 + \sum_{i=1}^{n} a_{i}^{2}<br />

    Thanks for the help

    This doesn't seem to make sense: what are these a_i's ?? The entries of a general n x n matrix are usually double indexed, so what are those elements whose squares' sum are you taking?

    Tonio
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  10. #10
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    Quote Originally Posted by tonio View Post
    This doesn't seem to make sense: what are these a_i's ?? The entries of a general n x n matrix are usually double indexed, so what are those elements whose squares' sum are you taking?

    Tonio
    Sorry if I wasn't clear above, a is an n component vector therefore it only has one index.
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