- #1

jack476

- 328

- 125

## 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

*A*, that

_{nn}*det(A*

_{nn}*) = det(A_{nn})*Then for the inductive step, I use the fact that for any matrix

*M*, 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

_{nn}, det(M_{n+1,n+1}) = αdet(M_{nn}) + β*M*is a factor in the equation for the determinant of

_{nn}*M*), along with the fact that complex conjugation is distributive under complex addition and multiplication, so that:

_{n+1, n+1}*det(A*

_{n+1, n+1})* = [αdet(A_{nn}) + β]* =*α*det(A*_{nn})* + β* =*det(A*

_{n+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.