Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Total potential energy in matrix

  1. Dec 5, 2004 #1
    Consider the matrix A=[3 -1; -1 3]. (a) Can the quadratic form x'*A*x evaluate to zero for some non-zero vector x? (x is a 2x1 column vector, and x' means x transposed: usual Matlab notation). (b) Does the equation A*x=b, b arbitrary, have a unique solution? If it does, prove it. (c) Show that the total potential energy f(x)=1/2*x'*A*x-b'*x has a minimum for x that solves the linear equations A*x=b.

    This is the question which i'm approached with
    my answer is as following
    a) No, because it's a symetric matrix only x'Ax = 0 only if vector x = 0.
    b) Yes , it is not underdetermined and det does not equal 0
    c) iono

    i was wondering what you guys thought
     
    Last edited: Dec 6, 2004
  2. jcsd
  3. Dec 5, 2004 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    a) Compute [itex]x^tAx[/itex] and you will obtain an expression in terms of the entries of x, and this expression, as is given, will be a quadratic form. With this expression, you should easily be able to see that unless x is the zero vector, the expression will not be equal to zero (if x is to have real entries).

    b) Compute det(A). Note that it is non-zero. Therefore, A is invertible, so there exists an inverse of A, [itex]A^{-1}[/itex]. If [itex]Ax = b[/itex], then [itex]A^{-1}Ax = A^{-1}b[/itex], so [itex]x = A^{-1}b[/itex], so clearly x exists and is unique.

    c) The best way I can think to do this is to compute f(x) to obtain an expression in terms of the entries of x. Essentially, f is a function from [itex]\mathbb{R}^2[/itex] to [itex]\mathbb{R}[/itex]. Compute the Jacobian of f at x, and find the x such that the Jacobian is zero. For all those points, compute the Hessian, and show that there is only one that has a positive definite Hessian. Calculate the value of f at this point. Also, calculate the value of [itex]f(A^{-1}b)[/itex], and show that the two values are the same, thereby showing that the vector where f has it's minimum is indeed the unique solution to [itex]Ax = b[/itex].

    If you are unfamiliar with any of the terminology above, you can easily look it up at wikipedia.com or mathworld.com, and then, if you're still stuck, ask for further clarification.
     
  4. Dec 6, 2004 #3
    wow.. thanks.. that was very detailed
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Total potential energy in matrix
  1. Total number of zeros. (Replies: 9)

Loading...