definition of \(n \times n\) unitary matrices group
Topics
About: group
About: matrices space
The table of contents of this article
Starting Context
- The reader knows a definition of %ring name% matrices space.
- The reader knows a definition of unitary matrix.
- The reader knows a definition of group.
Target Context
- The reader will have a definition of \(n \times n\) unitary matrices group.
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:
\( \mathbb{C}\): \(= \text{ the complex numbers field }\)
\( \{M\}\): \(= \text{ the } \mathbb{C} \text{ matrices space }\)
\( n\): \(\in \mathbb{N} \setminus \{0\}\)
\(*U (n)\): \(= \{M \in \{M\} \vert M \in \{\text{ the } n \times n \text{ unitary matrices }\}\}\)
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Conditions:
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2: Note
Let us see that \(U (n)\) is indeed a group.
Let \(M_1, M_2, M_3 \in U (n)\) be any.
\(M_1 M_2 \in U (n)\), because \((M_1 M_2)^* = {M_2}^* {M_1}^*\), by the proposition that the Hermitian conjugate of the product of any complex matrices is the product of the Hermitian conjugates of the constituents in the reverse order, \(= {M_2}^{- 1} {M_1}^{- 1}\), so, \((M_1 M_2)^* M_1 M_2 = {M_2}^{- 1} {M_1}^{- 1} M_1 M_2 = {M_2}^{- 1} I M_2 = {M_2}^{- 1} M_2 = I\) and \(M_1 M_2 (M_1 M_2)^* = M_1 M_2 {M_2}^{- 1} {M_1}^{- 1} = M_1 I {M_1}^{- 1} = M_1 {M_1}^{- 1} = I\), so, \((M_1 M_2)^* = (M_1 M_2)^{- 1}\).
1) \((M_1 \bullet M_2) \bullet M_3 = M_1 \bullet (M_2 \bullet M_3)\): by the proposition that for any ring, the multiplications of any matrices over the ring are associative.
2) \(i \in U (n)\) (called 'identity element') such that \(i \bullet M_1 = M_1 \bullet i = M_1\): the identity matrix, \(I\), is in \(U (n)\), because \(I^* = I = I^{-1}\), and \(I M_1 = M_1 I = M_1\).
3) \({M_1}^{- 1} \in U (n)\) (called 'inverse element of \(M_1\)') such that \({M_1}^{- 1} \bullet M_1 = M_1 \bullet {M_1}^{- 1} = I\): \({M_1}^* = {M_1}^{- 1} \in U (n)\), because \({{M_1}^*}^* = M_1\) and \({M_1}^* M_1 = I = M_1 {M_1}^*\), so, \({{M_1}^*}^{- 1} = M_1 = {{M_1}^*}^*\).
So, \(U (n)\) is a group.
