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1 Eigenvalue and 2 Eigenvectors

  1. Apr 10, 2010 #1
    Hi all,

    Let's say we have a symmetric matrix A with its corresponding diagonal matrix D. If A has only 1 eigenvalue, how do we show that there exists 2 eigenvectors?

  2. jcsd
  3. Apr 10, 2010 #2


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    What you have written makes no sense. If a matrix has an eigenvalue, then there exist an infinite number of eigenvectors. Do you mean "2 independent eigenvectors"? And are you talking about a 2 by 2 matrix?

    A matrix is "diagonalizable" if and only if it has a "complete set of eigenvectors"- that is, there is a basis for the vector space consisting of eigenvalues of the matrix. If A is an n by n matrix, then it must have n independent eigenvectors. If it has only one eigenvalue, then there must exist n independent eigenvectors corresponding to that one eigenvalue.
  4. Apr 10, 2010 #3
    Hi HallsofIvy,

    I'm sorry, i forgot to mention that it's a 2x2 matrix. Yes, is there a general method to find 2 independent eigenvectors?
  5. Apr 10, 2010 #4


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    In this case, every vector is an eigenvector. Pick one. Then find another vector orthogonal to it.

    If you need to, you can normalize the vectors.
  6. Apr 18, 2010 #5
    A is symmetric

    if A is symmetric then that means A equals its transpose and is of size nxn (THIS IS ALWAYS TRUE)
    if A is symmetric then it has n independent eigenvectors (THIS IS ALWAYS TRUE)

    The matrix factorization of A is SDS^(-1)
    The n columns of S are the n independent eigenvectors and for a symmetric matrix those eigenvectors are orthogonal when A is symmetric and can be made orthonormal (which makes finding the factorization of A a lot easier, via Gram-Schmidt)
    The diagonal entries of the Diagonal matrix D are the eigenvalues associated with the eigenvectors

    So, to find those diagonal entries and those independent eigenvectors the general form is as follows where A is an nxn matrix, x is an n-dimensional vector, and d is a constant, $ is the n-dimensional zero vector and I is the nxn identity matrix.

    det(A-dI)=0 solve for all values of d
    (this is called the characteristic equation which gives the n-degree polynomial which is used to determine the values of d (your eigenvalues that satisfy the equation))

    Edit: Hope this helps, I'm taking intro linear algebra this semester so if anything is wrong please let me know.
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