description/proof of that for square complex matrix, determinant of Hermitian conjugate of matrix is complex conjugate of determinant of matrix
Topics
About: field
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any square complex matrix, the determinant of the Hermitian conjugate of the matrix is the complex conjugate of the determinant of the matrix.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(M\): \(\in \{\text{ the } n \times n \mathbb{C} \text{ matrices }\}\)
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Statements:
\(det M^* = \overline{det M}\)
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2: Note
As an immediate corollary, for any unitary \(M\), \(\vert det M \vert = 1\), because as \(M^* M = I\), \(det (M^* M) = det I = 1\), but the left hand side is \(det M^* det M = \overline{det M} det M = \vert det M \vert^2\).
3: Proof
Whole Strategy: Step 1: see that \(det M^* = \overline{\sum_{\sigma \in S_n} sgn (\sigma) M^{\sigma_1}_1 ... M^{\sigma_n}_n} = \overline{det M^t} = \overline{det M}\).
Step 1:
\(det M^* = \sum_{\sigma \in S_n} sgn (\sigma) {M^*}^1_{\sigma_1} ... {M^*}^n_{\sigma_n}\), by the definition of determinant, \(= \sum_{\sigma \in S_n} sgn (\sigma) \overline{M^{\sigma_1}_1} ... \overline{M^{\sigma_n}_n} = \overline{\sum_{\sigma \in S_n} sgn (\sigma) M^{\sigma_1}_1 ... M^{\sigma_n}_n} = \overline{\sum_{\sigma \in S_n} sgn (\sigma) {M^t}^1_{\sigma_1} ... {M^t}^n_{\sigma_n}} = \overline{det M^t}\).
\(= \overline{det M}\), by the proposition that for any square commutative ring matrix, the determinant of the transpose of the matrix is the determinant of the matrix.
