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

• jack476
In summary, the determinant of the complex conjugate of a matrix A is equal to the complex conjugate of the determinant of A, which can be proven using the fact that the conjugate of a sum is the sum of the conjugates, the conjugate of a product is the product of the conjugates, and one of the definitions of the determinant. This direct approach is faster than using induction.
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}
\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.

jack476 said:

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}
\begin{vmatrix}
a^* & b^*\\
c^* & d^*
\end{vmatrix}
##
I would establish the basis for ##n = 1##, so ##1 \times 1## matrices.
jack476 said:
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.
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:
jack476 said:

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}
\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.

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

What is the definition of det(A*)?

The determinant of a matrix A* is the determinant of the conjugate transpose of A.

How is det(A*) related to det(A)?

Det(A*) is equal to the complex conjugate of det(A), so det(A*) = [det(A)]*.

Can the equality det(A*) = [det(A)]* be proven?

Yes, it can be proven using the properties of determinants and complex numbers.

What are the implications of det(A*) = [det(A)]*?

This equality is important in linear algebra and complex analysis, as it allows for the simplification of calculations involving determinants of complex matrices.

Are there any exceptions to the rule det(A*) = [det(A)]*?

No, this equality holds for all square matrices with complex entries.

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