description/proof of that conjugate of unitary matrix by invertible matrix is not necessarily unitary
Topics
About: matrices space
The table of contents of this article
Starting Context
- The reader knows a definition of unitary matrix.
- The reader knows a definition of conjugate of matrix by invertible matrix.
- The reader admits the proposition that each component of any unitary matrix has an absolute value equal to or smaller than \(1\).
Target Context
- The reader will have a description and a proof of the proposition that the conjugate of a unitary matrix by an invertible matrix is not necessarily 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:
\(M\): \(\in \{\text{ the unitary matrices }\}\)
\(N\): \(\in \{\text{ the invertible matrices }\}\)
//
Statements:
not necessarily "\(N M N^{- 1} \in \{\text{ the unitary matrices }\}\)"
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2: Note
Compare with the proposition that the conjugate of any unitary matrix by any unitary matrix is unitary.
A point is that a vectors space endomorphism cannot be characterized by its representative matrix being unitary: it depends on the basis: refer to the proposition that for any module with any \(2\) bases of any same finite cardinality and any module endomorphism, the transition of the endomorphism matrices with respect to the change of the bases is this; of course, if a basis has been chosen, any endomorphism can be characterized by its representative matrix with respect to the basis being unitary.
Especially, when \(V\) is a \(d\)-dimensional \(\mathbb{C}\) vectors space and \(g: GL (V) \to M_d (\mathbb{C})^{\times}\) is the 'groups - homomorphisms' isomorphism defined in the proposition that for any module with any \(d\)-elements basis and the general linear group of the module, there is the canonical 'groups - homomorphisms' isomorphism with respect to the basis from the general linear group onto the group of the invertible \(d \times d\) ring matrices, certainly \(U (d) \subset M_d (\mathbb{C})^{\times}\) is a subgroup and \(g^{- 1} (U (d)) \subseteq GL (V)\) is a subgroup, but \(g^{- 1} (U (d))\) depends on the choice of basis, because \(g^{- 1} (M)\) is represented by \(N M N^{- 1}\) with respect to another basis while \(N M N^{- 1}\) is not necessarily unitary. So, it is odd to say "\(U (d)\) is a subgroup of \(GL (V)\)." even if that is interpreted to mean that \(g^{- 1} (U (d))\) is a subgroup of \(GL (V)\), unless the basis is specified.
While there are the proposition that for any linear map from any complex-Euclidean-normed complex Euclidean vectors space into itself, the map is orthogonal if and only if the canonical representative matrix is unitary and the proposition that for any linear map from any Euclidean-normed Euclidean vectors space into itself, the map is orthogonal if and only if the canonical representative matrix is orthogonal, they are because the canonical bases are chosen.
3: Proof
Whole Strategy: Step 1: see a counterexample.
Step 1:
Let \(M = \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix}\) and \(N = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\).
\(M\) is indeed unitary, because \(M^* = \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix}\) and \(M^* M = M M^* = \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix} = I\).
\(N\) is indeed invertible, because \(N^{- 1} = \begin{pmatrix} 1 & - 1 \\ 0 & 1 \end{pmatrix}\).
\(N M N^{- 1} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix} \begin{pmatrix} 1 & - 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & - 1 \\ 0 & - 1 \end{pmatrix} \begin{pmatrix} 1 & - 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & - 2 \\ 0 & - 1 \end{pmatrix}\), which is not unitary, by the proposition that each component of any unitary matrix has an absolute value equal to or smaller than \(1\).
