- #1
jack476
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Homework Statement
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.
Homework Equations
None, but the hint provided with the problem suggested using an induction argument.
The Attempt at a Solution
To establish the base case, I show that:
##
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
^*=(ad-cb)^*=a^*d^*-c^*b^* =
\begin{vmatrix}
a^* & b^*\\
c^* & d^*
\end{vmatrix}
##
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.