Results 1 to 6 of 6

Math Help - Hermitian scalar product relationship

  1. #1
    Newbie
    Joined
    Dec 2006
    Posts
    9

    Hermitian scalar product relationship

    Hi all i could use a little help. Thanks in advance.

    Prove that if A is Hermitian than (Ax, y) = (x, Ay), where x and y are any vectors.

    what ive tried:

    (Ax, y) = summation(A.x_i . conj(y_i); i = 1...n)

    (x, Ay) = summation(x_i . conj(A . y_i); i =1...n) but A = A* so
    (x, Ay) = summation(x_i . conj(A*, y_i); i =1...n) and A* = conj(trans(A))
    note*[trans = transpose]
    (x, Ay) = summation(x_i . conj(conj(trans(A)), y_i); i =1...n) which can be simplified to
    (x, Ay) = summation(x_i . trans(A) . conj(y_i); i =1....n) using distributive property of matrix multiplication
    (x, Ay) = summation(trans(A) . x_i . conj(y_i); i =1....n)

    I must be missing something because i cannot guarantee A = trans(A) or A = conj(A) i think.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,053
    Thanks
    368
    Awards
    1
    Quote Originally Posted by OkashiiKen View Post
    Hi all i could use a little help. Thanks in advance.

    Prove that if A is Hermitian than (Ax, y) = (x, Ay), where x and y are any vectors.

    what ive tried:

    (Ax, y) = summation(A.x_i . conj(y_i); i = 1...n)

    (x, Ay) = summation(x_i . conj(A . y_i); i =1...n) but A = A* so
    (x, Ay) = summation(x_i . conj(A*, y_i); i =1...n) and A* = conj(trans(A))
    note*[trans = transpose]
    (x, Ay) = summation(x_i . conj(conj(trans(A)), y_i); i =1...n) which can be simplified to
    (x, Ay) = summation(x_i . trans(A) . conj(y_i); i =1....n) using distributive property of matrix multiplication
    (x, Ay) = summation(trans(A) . x_i . conj(y_i); i =1....n)

    I must be missing something because i cannot guarantee A = trans(A) or A = conj(A) i think.

    Thanks.
    A matrix (operator) A is Hermitian iff A^{\dagger} = A.

    So given two column vectors x, y:
    (Ax, y) \equiv x^{\dagger} \cdot A^{\dagger}y = \sum_j x_j^* (A^{\dagger}y)_j

    But A^{\dagger} = A so:
    (Ax, y) \equiv x^{\dagger} \cdot A^{\dagger}y = \sum_j x_j^* (A^{\dagger}y)_j  = \sum_j x_j^* (Ay)_j = \sum_{ij} x_j^* A_{ji}y_i

    Similarly:
    (x, Ay) = x^{\dagger} \cdot Ay = \sum_{j}x_j^*(Ay)_j = \sum_{ij} x_j^* A_{ji}y_i

    Since these two expressions are equal, (Ax, y) = (x, Ay).

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by OkashiiKen View Post
    Hi all i could use a little help. Thanks in advance.

    Prove that if A is Hermitian than (Ax, y) = (x, Ay), where x and y are any vectors.

    what ive tried:

    (Ax, y) = summation(A.x_i . conj(y_i); i = 1...n)

    (x, Ay) = summation(x_i . conj(A . y_i); i =1...n) but A = A* so
    (x, Ay) = summation(x_i . conj(A*, y_i); i =1...n) and A* = conj(trans(A))
    note*[trans = transpose]
    (x, Ay) = summation(x_i . conj(conj(trans(A)), y_i); i =1...n) which can be simplified to
    (x, Ay) = summation(x_i . trans(A) . conj(y_i); i =1....n) using distributive property of matrix multiplication
    (x, Ay) = summation(trans(A) . x_i . conj(y_i); i =1....n)

    I must be missing something because i cannot guarantee A = trans(A) or A = conj(A) i think.

    Thanks.
    (Ax,b)=\sum_i \left(\sum_j (A_{i,j}x_j) \right) \overline{y_i}
    = \sum_i \left(\sum_j (A_{i,j}x_j) \overline{y_i}\right)
    = \sum_i \left(\sum_j x_j \overline{\overline{A_{i,j}}y_i}\right)

    Now change the order of summation:

    (Ax,b)= \sum_j \left(\sum_i x_j \overline{\overline{A_{i,j}}y_i}\right)= (x,\overline{A}^{\,t} y) =(x,A^H y)

    and since A is Hermitian A=A^H so:

    (Ax,b) =(x,A y).

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2006
    Posts
    9
    Quote Originally Posted by topsquark View Post
    A matrix (operator) A is Hermitian iff A^{\dagger} = A.

    So given two column vectors x, y:
    (Ax, y) \equiv x^{\dagger} \cdot A^{\dagger}y = \sum_j x_j^* (A^{\dagger}y)_j

    But A^{\dagger} = A so:
    (Ax, y) \equiv x^{\dagger} \cdot A^{\dagger}y = \sum_j x_j^* (A^{\dagger}y)_j  = \sum_j x_j^* (Ay)_j = \sum_{ij} x_j^* A_{ji}y_i

    Similarly:
    (x, Ay) = x^{\dagger} \cdot Ay = \sum_{j}x_j^*(Ay)_j = \sum_{ij} x_j^* A_{ji}y_i

    Since these two expressions are equal, (Ax, y) = (x, Ay).

    -Dan
    Thanks, but i am having a little difficulty following.... :/

    i dont understand this operator:
    A^{\dagger} = A
    I had understood a hermitian matrix to be one such that
    A* = A and a_{ij} = \overline{a_{ji}}

    Your first step made some leaps i cant quite seem to follow especially this one:
    (x, Ay) = x^{\dagger} \cdot Ay = \sum_{j}x_j^*(Ay)_j = \sum_{ij} x_j^* A_{ji}y_i
    I can mostly follow after this step, but I dont understand this equation. Could you elaborate this step for me? Thank you.

    Quote Originally Posted by CaptainBlack View Post
    (Ax,b)=\sum_i \left(\sum_j (A_{i,j}x_j) \right) \overline{y_i}
    = \sum_i \left(\sum_j (A_{i,j}x_j) \overline{y_i}\right)
    = \sum_i \left(\sum_j x_j \overline{\overline{A_{i,j}}y_i}\right)

    Now change the order of summation:

    (Ax,b)= \sum_j \left(\sum_i x_j \overline{\overline{A_{i,j}}y_i}\right)= (x,\overline{A}^{\,t} y) =(x,A^H y)

    and since A is Hermitian A=A^H so:

    (Ax,b) =(x,A y).

    RonL
    This solution is much more clear to me. Although i am still confused at one step.
    (Ax,b)= \sum_j \left(\sum_i x_j \overline{\overline{A_{i,j}}y_i}\right)= (x,\overline{A}^{\,t} y) =(x,A^H y)
    I cant follow this:
      \sum_j \left(\sum_i x_j \overline{\overline{A_{i,j}}y_i}\right)= (x,\overline{A}^{\,t} y)
    When i work it out:
     \sum_j \left(\sum_i x_j \overline{\overline{A_{i,j}}y_i}\right)= \sum_j \left(\sum_i x_j A_{i,j}\overline y_i\right)=(x,\overline A y)
    Am I missing the transpose because we are summing across the i direction instead of the j direction? I have a problem when working from (x, Ay)

    I keep showing that
     (Ax, y) = (x, A^T y)

    Thanks again.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,053
    Thanks
    368
    Awards
    1
    Quote Originally Posted by OkashiiKen View Post
    Thanks, but i am having a little difficulty following.... :/

    i dont understand this operator:
    A^{\dagger} = A
    I had understood a hermitian matrix to be one such that
    A* = A and a_{ij} = \overline{a_{ji}}

    Your first step made some leaps i cant quite seem to follow especially this one:
    (x, Ay) = x^{\dagger} \cdot Ay = \sum_{j}x_j^*(Ay)_j = \sum_{ij} x_j^* A_{ji}y_i
    I can mostly follow after this step, but I dont understand this equation. Could you elaborate this step for me? Thank you.
    I could elaborate, but I suspect you will be more comfortable following CaptainBlack's solution. We are essentially saying the same thing, just in a different notation. I am not as familiar with the notation you and CaptainBlack are using, so I'll leave it to him.

    For the sake of clarity I will mention that the \dagger operator is defined (virtually always, at least in Physics) as the Hermitian conjugate, or transpose conjugate, operator: Acting on a square matrix A, A_{ij}^{\dagger} = A_{ji}^*. (Where, to be clear, * is the complex conjugate operator.) So the Hermitian scalar product acting on two vectors x (a row vector) and y (a column vector) is (x, y) = x^{\dagger} \cdot y, where \cdot is the usual (Euclidean) scalar product.

    -Dan
    Last edited by topsquark; January 6th 2007 at 02:27 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Dec 2006
    Posts
    9
    Thanks Topsquark.

    I feeling pretty slow today....forgot that

     x^T y = \sum_i x_i y_i

    Thank you CaptainBlack i can follow your solution now! Both of you were a big help. Now i can move on to my next problem...proving all eigenvalues of a hermitian matrix A are real. Thanks for all your help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: September 7th 2010, 09:03 PM
  2. Relationship between scalar curvature and the volume of balls
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 9th 2010, 05:57 PM
  3. multivariable differential for inner product(scalar product)?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 23rd 2009, 05:40 PM
  4. Hermitian, complex inner product space
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 16th 2008, 08:44 AM
  5. Dot and Scalar Product
    Posted in the Advanced Applied Math Forum
    Replies: 8
    Last Post: September 9th 2008, 08:03 PM

Search Tags


/mathhelpforum @mathhelpforum