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  1. #1
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    Prove unitary matrices require orthonormal basis

    Problem: Let A be an nXn matrix with complex entries. Prove that AA*=I if and only if the rows of A form an orthonormal basis for C^n

    Thoughts: If AA*=I then A*=A^-1, but I don't understand how (or if) that implies that the rows of A are an orthonormal basis, or how the reverse is true, or, for that matter, much of anything involving linear algebra, gah!

    Thanks for any help.
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  2. #2
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    Quote Originally Posted by kaelbu View Post
    Problem: Let A be an nXn matrix with complex entries. Prove that AA*=I if and only if the rows of A form an orthonormal basis for C^n

    Thoughts: If AA*=I then A*=A^-1, but I don't understand how (or if) that implies that the rows of A are an orthonormal basis, or how the reverse is true, or, for that matter, much of anything involving linear algebra, gah!
    If you have a matrix product AB, then the definition of matrix multiplication says that the (i,j)-element of AB is the inner product of row i of A with the complex conjugate of column j of B. (The reason for the complex conjugation is that the formula for matrix multiplication gives (AB)_{ij} = \textstyle\sum a_{ik}b_{kj}, but the formula for an inner product uses the complex conjugate of the second vector: \langle x,y\rangle = \textstyle\sum x_k\overline{y_k}.)

    In the adjoint matrix A*, column j is the complex conjugate of row j of A.

    Putting those two facts together, you see that the (i,j)-element of AA* is the inner product of row i of A with row j of A. If that is equal to the (i,j)-element of the identity matrix then ... .
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Putting those two facts together, you see that the (i,j)-element of AA* is the inner product of row i of A with row j of A. If that is equal to the (i,j)-element of the identity matrix then ... .
    (AA^*)_{ij}= <x_i,x_j>
    if  i=j <x_i, x_i> must = 1 therefore ||x_i||=1 -> normal
    if i \not=j <x_i, x_j> must = 0 therefore x_j is perpendicular to  x_i -> orthogonal
    => orthonormal

    Going the opposite direction:
    if the rows of A are orthonormal
    <x_i,x_j> = 0
    except when i=j when <x_i, x_j>=1
    therefore  (AA^*)_{ij} = a_i \overline{a_j} = 1 if  i=j
    or = 0 if  i \not=j which is the definition of the identity.

    One more question: is this enough exposition? That is, did I prove it well enough?
    Thank you!
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    Quote Originally Posted by kaelbu View Post
    (AA^*)_{ij}= <x_i,x_j>
    if  i=j <x_i, x_i> must = 1 therefore ||x_i||=1 -> normal
    if i \not=j <x_i, x_j> must = 0 therefore x_j is perpendicular to  x_i -> orthogonal
    => orthonormal

    Going the opposite direction:
    if the rows of A are orthonormal
    <x_i,x_j> = 0
    except when i=j when <x_i, x_j>=1
    therefore  (AA^*)_{ij} = a_i \overline{a_j} = 1 if  i=j
    or = 0 if  i \not=j which is the definition of the identity.

    One more question: is this enough exposition? That is, did I prove it well enough?
    Yes, that looks good to me.
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