description/proof of that positive-definite Hermitian matrix can be transformed to identity by unitary matrix multiplied by positive diagonal matrix from right
Topics
About: matrix
The table of contents of this article
Starting Context
- The reader knows a definition of positive-definite Hermitian matrix.
- The reader knows a definition of unitary matrix.
- The reader knows a definition of orthogonal matrix.
- The reader admits the proposition that any Hermitian matrix can be diagonalized by a unitary matrix to be inevitably real.
Target Context
- The reader will have a description and a proof of the proposition that any positive-definite Hermitian matrix can be transformed to the identity by a unitary matrix multiplied by a positive diagonal matrix from right.
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:
\(M\): \(\in \{\text{ the positive-definite Hermitian matrices }\}\)
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Statements:
\(\exists U \in \{\text{ the unitary matrices }\}, \exists D \in \{\text{ the positive diagonal matrices }\} ((U D)^* M (U D) = I)\)
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\((U D)^* M (U D) = D^* U^* M U D = D^t U^{-1} M U D\), because \(U\) is unitary and \(D\) is real. \(= D U^{-1} M U D\), because \(D\) is symmetric.
When \(M\) is real symmetric, which is Hermitian, \(U\) can be taken to be orthogonal, which is unitary, and \((U D)^* M (U D) = (U D)^t M (U D) = I\).
2: Proof
Whole Strategy: Step 1: see that there is a \(U\) such that \(M' = U^* M U = \begin{pmatrix} \lambda_1 & 0 & ... & 0 \\ 0 & \lambda_2 & ... & 0 \\ ... \\ 0 & ... & 0 & \lambda_n \end{pmatrix}\) is a positive diagonal matrix; Step 2: take \(D = \begin{pmatrix} 1 / \sqrt{\lambda_1} & 0 & ... & 0 \\ 0 & 1 / \sqrt{\lambda_2} & ... & 0 \\ ... \\ 0 & ... & 0 & 1 / \sqrt{\lambda_n} \end{pmatrix}\), and see that \((U D)^* M (U D) = I\).
Step 1:
There is a \(U\) such that \(M' = U^* M U = \begin{pmatrix} \lambda_1 & 0 & ... & 0 \\ 0 & \lambda_2 & ... & 0 \\ ... \\ 0 & ... & 0 & \lambda_n \end{pmatrix}\) is a real diagonal matrix, by the proposition that any Hermitian matrix can be diagonalized by a unitary matrix to be inevitably real.
As \(M\) is positive-definite, each \(\lambda_j\) is positive: otherwise, \(v^* (U^* M U) v\) would be non-positive for a nonzero \(v\) and \(= (U v)^* M (U v)\) with \(U v\) nonzero with non-positive result, a contradiction: \(U v \neq 0\) because \((U v)^* (U v) = v^* U^* U v = v^* I v = v^* v \neq 0\).
Step 2:
Let us define \(D := \begin{pmatrix} 1 / \sqrt{\lambda_1} & 0 & ... & 0 \\ 0 & 1 / \sqrt{\lambda_2} & ... & 0 \\ ... \\ 0 & ... & 0 & 1 / \sqrt{\lambda_n} \end{pmatrix}\).
\((U D)^* M (U D) = D^* U^* M U D = D^* (U^* M U) D = \begin{pmatrix} 1 / \sqrt{\lambda_1} & 0 & ... & 0 \\ 0 & 1 / \sqrt{\lambda_2} & ... & 0 \\ ... \\ 0 & ... & 0 & 1 / \sqrt{\lambda_n} \end{pmatrix} \begin{pmatrix} \lambda_1 & 0 & ... & 0 \\ 0 & \lambda_2 & ... & 0 \\ ... \\ 0 & ... & 0 & \lambda_n \end{pmatrix} \begin{pmatrix} 1 / \sqrt{\lambda_1} & 0 & ... & 0 \\ 0 & 1 / \sqrt{\lambda_2} & ... & 0 \\ ... \\ 0 & ... & 0 & 1 / \sqrt{\lambda_n} \end{pmatrix} = I\).
