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Bilinear forms & Symmetric bilinear forms

  1. Nov 13, 2007 #1
    1) Let f: V x V -> F be a symmetric bilinear form on V, where F is a field.
    Suppose B={v1,...,vn} is an orthogonal basis for V
    This implies f(vi,vj)=0 for all i not=j
    =>A=diag{a1,...,an} and we say that f is diagonalized.

    Now I don't understand the red part, i.e. how does orthongality imply f(vi,vj)=0 for all i not=j?


    2) For the following symmetric matrix A,
    A=[1 -1 -1
    -1 1 -1
    -1 -1 1]
    a) find an orthogonal matrix P such that (P^T)AP is diagonal.
    b) Let f be the bilinear form on R^3 that has matrix A relative to the basis B={(1,1,1),(1,0,1),(0,1,-1)}. Use the matrix P from part a to find a basis of R^3 relative to which f is represented by a diagonal matrix. Also write out the corresponding diagonalized expression for f.

    I got part a),
    P= [1/sqrt3 1/sqrt2 1/sqrt6
    1/sqrt3 -1/sqrt2 1/sqrt6
    1/sqrt3 0 -2/sqrt6]
    is orthogonal such that (P^T)AP=diag{-1,2,2} is diagonal.
    Now part b is the headache, I don't even know how to start, can someone please help me? It seems really challenging, but I am sure someone here knows how to solve it.

    Thanks a million!
  2. jcsd
  3. Nov 14, 2007 #2
    Can someone please help me?
  4. Nov 14, 2007 #3


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    Staff Emeritus
    Science Advisor

    It doesn't- there certainly exist symmetric bilinear forms that are not "diagonalized" just because we pick an orthonormal basis. Perhaps you have misunderstood or miscopied something. A symmetric bilinear operator can always be written as a symmetric matrix in any basis and it is always possible to choose an orthonormal basis so that matrix is diagonal. But it must be chosen carefully, it is not true for every orthonormal basis.

    In fact, this problem wouldn't make sense if the forgoing were true! If a symmetric bilinear form is diagonal in any orthonormal basis, they wouldn't be asking you to find such a basis!

    Okay, I presume that you found that A has a double eigenvalue of 2 and a single eigenvalue of 1 and have found 3 independent vectors corresponding to those eigenvalues and used those vectors as columns to construct the matrix P. Well, that's the whole point isn't it! You've already done all the work. If you use THOSE eigenvectors as basis vectors, then A is diagonal with the eigenvalues on the diagonal. The columns of the matrix P are the basis vectors sought in b and D is just the diagonal matrix with the eigenvalues of A, 2, 2, -1, on the diagonal.
  5. Nov 19, 2007 #4

    2b) So is the answer to this part going to be the columns of P?
    But isn't there a difference between diagonalizing A and diagonalizing the bilinear form f? I don't get why the answer is true...
    Also, what would the diagonalized expression for f look like?

    By the way, the question says "Let f be the bilinear form on R^3...", so f is not necessarily a symmetric bilinear form, right?
    Theorem: A bilinear form f is symmetric if and only if EVERY matrix that represents f is symmetric.
    So using this theorem, ONE matrix (i.e. the matrix A) that represents f is symmetric doesn't imply that f is symmetric.

    I am so confused, please help me...
    Last edited: Nov 19, 2007
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