description/proof of that transpose of unitary matrix is unitary
Topics
About: matrices space
The table of contents of this article
Starting Context
- The reader knows a definition of unitary matrix.
- The reader admits the proposition that for any complex matrix, the Hermitian conjugate of the transpose of the matrix is the transpose of the Hermitian conjugate of the matrix.
- The reader admits the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order.
- The reader admits the proposition that for any ring of \(n \times n\) field matrices, if any element multiplied by any another element from left or right is \(1\), the another element is the inverse of the element.
Target Context
- The reader will have a description and a proof of the proposition that the transpose of any unitary matrix is unitary.
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:
\(M\): \(\in \{\text{ the unitary matrices } \}\)
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Statements:
\(M^t \in \{\text{ the unitary matrices } \}\)
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2: Proof
Whole Strategy: Step 1: see that \({M^t}^* M^t = I\).
Step 1:
\({M^t}^* M^t = {M^*}^t M^t\), by the proposition that for any complex matrix, the Hermitian conjugate of the transpose of the matrix is the transpose of the Hermitian conjugate of the matrix, \(= (M M^*)^t\), by the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order, \(= I^t\), because \(M\) is unitary, \(= I\).
\({M^t}^* = {M^t}^{- 1}\), by the proposition that for any ring of \(n \times n\) field matrices, if any element multiplied by any another element from left or right is \(1\), the another element is the inverse of the element.
So, \(M^t\) is unitary.
