1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integrating of an exponential of a matrix product

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I try to solve this integral with with parameter x as a member of this scale:(-∞ , +∞)
    I=∫∏dx exp(-0.5XAX + XB)=∫∏dx exp( Ʃ-0.5xa[j]x[j] +Ʃ xb )
    In which a[j] and b are components of telated matrix and vector and the first sum is on i and j ranges from 1 to N .Also X and B are two vector with N component and A is a N*N matrix,so the integral is over all x (which denote components of X).

    2. Relevant equations
    Gaussian integrals in the same scale obey this equation:
    ∫dx exp( -ax^2 ) =√(π/a)

    3. The attempt at a solution

    Using a change in variables like X=Y + CB with CA=Ac=1 should be appropriate:

    -0.5(XAX) + XB=(-0.5)(Y+CB)A(Y+CB)+(Y+CB)B=(-0.5)(YAY+YACB+CBAY+(CB)^2)+YB+CB^2
    = (-0.5)(YAY+YACB+CBAY+(CB)^2)+YACB+CB^2
    =(-0.5)(YAY+CBAY+(CB)^2-YACB + 2CBB)

    The problem is there is no YY=Ʃyy to help me change this problem to a Gaussian problem.Another problem is with YACB and CBAY that arenot compatible with this equation:
    (A+B)^2=A^2+B^2+AB+BA .

    Thank you for noticing.
  2. jcsd
  3. Mar 29, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your exponent is -(1/2)Q, where Q = <x,Ax> - 2<b,x> (writing the inner product sum u[j]*v[j] as <u,v>). We need to assume A is *symmetric*; if not, we can replace it by a new A that IS symmetric, and work with that. Writing x = y + c (c a constant vector) we have Q = <y+c,A(y+c)> -2<b,y+c> = <y,Ay> + <y,Ac> + <c,Ay> + <c,Ac> - 2<b,y> = <y,Ay> + <c,Ac> +<y,2Ac-2b>. If we take Ac=b we eliminate linear terms and just have Q = <y,Ay> + <c,Ac>.

    Now, we must assume A is positive-definite; otherwise, the integral will not be convergent. It follows that A is invertible, so c = Inverse(A)b can be found. Furthermore, we can perform a Cholesky factorization of A, to write A = U^T U, where U^T = transpose of U and U is upper-triangular with all u[i,i] > 0. Thus, if U[1] = u[1,1]y[1] + u[1,2]y[2] + ... + u[1,n]y[n], U[2] = u[2,2]y[2] + u[2,3]y[3] + ... + u[2,n]y[n],..., U[n] = u[n,n]y[n] we have <y,Ay> = U[1]^2 + U[2]^2 + ... + U[n]^2. We can change the variables of integration from y[1], y[2], ..., y[n] to U[1], U[2],...,U[n], to get:
    [tex] \int_{R^n} \exp{\left[-\frac{1}{2}<x,Ax> + <b,x>\right]} \, dx_1 \cdots dx_n
    = C\, \exp{\left(-\frac{1}{2}<c,Ac>\right)} \int_{R^n} \exp{\left[-\frac{1}{2} (U_1^2 + \cdots + U_n^2)\right]} \, dU_1 \cdots dU_n, [/tex]
    where C is the Jacobian you get by changing variables from y to U.

    Last edited: Mar 29, 2012
  4. Apr 1, 2012 #3
    Please explain some points:

    1)Using X=Y+C,Q will Change this way:
    I noticed that with the last sentence the final form for exponent -0.5Qwill produce
    <C,AC>/2 instead of -<C,AC>/2.

    2)It seems to me that :
    Q= <y,Ay> + <y,Ac> + <c,Ay> + <c,Ac> - 2<b,y> -2<b,c>
    = <y,Ay> + <c,Ac> +<c,Ay> + <y,Ac> - 2<b,y> -2<b,c>
    = <y,Ay> + <c,Ac> +<y,2Ac-2b> -2<b,c> - <y,Ac>
    By taking Ac=b we obtain Q= <y,Ay> + <c,Ac> - <y,Ac> -2<b,c>

    Please explain how you wrote Q = <y,Ay> + <c,Ac>

  5. Apr 1, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You are supposed to know that for real symmetric A we have [itex]<c,Ay> = <Ac,y> = <y,Ac> = <Ay,c>. [/itex] However, you are right in noting that I incorrectly dropped the term 2<b,c>, so we should have [itex] Q = <y,Ay> + <c,Ac> +2<y,Ac-b> - 2<b,c> = <y,Ay> + <c,Ac> - 2<b,c>[/itex] if we take [itex] Ac=b[/itex]. In fact, this gives [itex] <c,Ac> - 2<b,c> = -<b,c>, [/itex] so
    [tex] \int_{R^n} \exp(-Q/2) \, dx_1 \cdots dx_n = C\, \exp(-<b,c>) \int_{R^n} \exp(-(U_1^2 + \cdots + U_n^2)/2) \, dU_1 \cdots dU_n. [/tex]

    Last edited: Apr 2, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook