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

[jack476]

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_nn, that det(A_nn*) = det(A_nn)*

Then for the inductive step, I use the fact that for any matrix M_nn, det(M_n+1,n+1) = αdet(M_nn) + 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 M_nn is a factor in the equation for the determinant of M_{n+1, n+1}), along with the fact that complex conjugation is distributive under complex addition and multiplication, so that:

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.

 

[S.G. Janssens]

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:
$\begin{vmatrix} a & b\\ c & d \end{vmatrix} ^*=(ad-cb)^*=a^*d^*-c^*b^* = \begin{vmatrix} a^* & b^*\\ c^* & d^* \end{vmatrix}$

I would establish the basis for $n = 1$, so $1 \times 1$ matrices.

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

Then for the inductive step, I use the fact that for any matrix M_nn, det(M_n+1,n+1) = αdet(M_nn) + 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 M_nn is a factor in the equation for the determinant of M_{n+1, n+1}), along with the fact that complex conjugation is distributive under complex addition and multiplication, so that:

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.

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.

 

[Ray Vickson]

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_nn, that det(A_nn*) = det(A_nn)*

Then for the inductive step, I use the fact that for any matrix M_nn, det(M_n+1,n+1) = αdet(M_nn) + 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 M_nn is a factor in the equation for the determinant of M_{n+1, n+1}), along with the fact that complex conjugation is distributive under complex addition and multiplication, so that:

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.

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:
$$\det(A) = \sum_{i_1, i_2,\ldots, i_n} \epsilon(i_1,i_2, \ldots, i_n) a_{1 i_1} a_{2 i_2} \cdots a_{n i_n},$$
where
$$\epsilon(i_1,i_2, \ldots, i_n) = \begin{cases} +1, & i_1, i_2, \ldots i_n \; \text{is an even permutation of } \: 1,2, \ldots, n \\ -1, & i_1, i_2, \ldots i_n \; \text{is an odd permutation of } \: 1,2, \ldots, n \\ 0, & \text{otherwise} \end{cases}$$