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Prove that det(A*) = [det(A)]*

  1. Feb 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove that the determinant of A*, the matrix of the complex conjugates of the elements of A, is equal to the complex conjugate of the determinant.

    2. Relevant equations
    None, but the hint provided with the problem suggested using an induction argument.

    3. The attempt at a solution
    To establish the base case, I show that:
    a & b\\
    c & d
    ^*=(ad-cb)^*=a^*d^*-c^*b^* =
    a^* & b^*\\
    c^* & d^*

    For my inductive hypothesis, I claim that for the square matrix Ann, that det(Ann*) = det(Ann)*

    Then for the inductive step, I use the fact that for any matrix Mnn, det(Mn+1,n+1) = αdet(Mnn) + β, where α is some constant and β is the sum of all of the other terms in that determinant (that is, I use the fact that the determinant of Mnn is a factor in the equation for the determinant of Mn+1, n+1), along with the fact that complex conjugation is distributive under complex addition and multiplication, so that:

    det(An+1, n+1)* = [αdet(Ann) + β]* = α*det(Ann)* + β* = det(An+1, n+1*)

    So that the inductive step is completed, and therefore for all nxn matrices of complex elements, the determinant of the complex conjugate matrix is the complex conjugate of that matrix's determinant.

    Any feedback would be greatly appreciated.
  2. jcsd
  3. Feb 10, 2016 #2


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    I would establish the basis for ##n = 1##, so ##1 \times 1## matrices.
    I understand the idea, and it is close. However, I prefer to be a bit more explicit. Why don't you write ##\text{det}\,{M_{n+1,n+1}}## as a cofactor expansion along, say, the first row, so
    \text{det}\,{M_{n+1,n+1}} = \sum_{k=1}^{n + 1}{(-1)^{1 + k}P_{1,k}}
    where ##P_{1,k}## is the ##(1,k)##-minor (an ##n \times n## determinant) and hence ##(-1)^{1 + k}P_{1,k}## is the corresponding cofactor.
    Last edited: Feb 10, 2016
  4. Feb 10, 2016 #3

    Ray Vickson

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    As Krylov has pointed out, you are very nearly done. However, wouldn't a direct, non-indiction approach be faster? All we need to know are three things: (1) the conjugate of a sum is the sum of the conjugates; (2) the conjugate of a product is the product of the conjugates; and (3) one of the definitions of ##\det(A)## is:
    [tex] \det(A) = \sum_{i_1, i_2,\ldots, i_n} \epsilon(i_1,i_2, \ldots, i_n) a_{1 i_1} a_{2 i_2} \cdots a_{n i_n}, [/tex]
    [tex] \epsilon(i_1,i_2, \ldots, i_n) = \begin{cases} +1, & i_1, i_2, \ldots i_n \; \text{is an even permutation of } \: 1,2, \ldots, n \\
    -1, & i_1, i_2, \ldots i_n \; \text{is an odd permutation of } \: 1,2, \ldots, n \\
    0, & \text{otherwise}
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