1152: Positive-Definite Hermitian Matrix Can Be Transformed to Identity by Unitary Matrix Multiplied by Positive Diagonal Matrix from Right

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description/proof of that positive-definite Hermitian matrix can be transformed to identity by unitary matrix multiplied by positive diagonal matrix from right

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Topics

About: matrix

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the positive-definite Hermitian matrices }\}$
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Statements:
$\mathrm{\exists }U\in \{\text{ the unitary matrices }\},\mathrm{\exists }D\in \{\text{ the positive diagonal matrices }\}((UD{)}^{\ast }M(UD)=I)$
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$(UD{)}^{\ast }M(UD)=D^{\ast }{U}^{\ast }MUD=D^t{U}^{-1}MUD$, because $U$ is unitary and $D$ is real. $=D{U}^{-1}MUD$, because $D$ is symmetric.

When $M$ is real symmetric, which is Hermitian, $U$ can be taken to be orthogonal, which is unitary, and $(UD{)}^{\ast }M(UD)=(UD{)}^{t}M(UD)=I$.

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2: Proof

Whole Strategy: Step 1: see that there is a $U$ such that $M^{\prime }=U^{\ast }MU=\left(\begin{array}{cccc}{\lambda }_1& 0& ...& 0\\ 0& {\lambda }_2& ...& 0\\ ...\\ 0& ...& 0& {\lambda }_n\end{array}\right)$ is a positive diagonal matrix; Step 2: take $D=\left(\begin{array}{cccc}1/\sqrt{{\lambda }_1}& 0& ...& 0\\ 0& 1/\sqrt{{\lambda }_2}& ...& 0\\ ...\\ 0& ...& 0& 1/\sqrt{{\lambda }_n}\end{array}\right)$, and see that $(UD{)}^{\ast }M(UD)=I$.

Step 1:

There is a $U$ such that $M^{\prime }=U^{\ast }MU=\left(\begin{array}{cccc}{\lambda }_1& 0& ...& 0\\ 0& {\lambda }_2& ...& 0\\ ...\\ 0& ...& 0& {\lambda }_n\end{array}\right)$ 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, $v^{\ast }(U^{\ast }MU)v$ would be non-positive for a nonzero $v$ and $=(Uv{)}^{\ast }M(Uv)$ with $Uv$ nonzero with non-positive result, a contradiction: $Uv\ne 0$ because $(Uv{)}^{\ast }(Uv)=v^{\ast }{U}^{\ast }Uv=v^{\ast }Iv=v^{\ast }v\ne 0$.

Step 2:

Let us define $D:=\left(\begin{array}{cccc}1/\sqrt{{\lambda }_1}& 0& ...& 0\\ 0& 1/\sqrt{{\lambda }_2}& ...& 0\\ ...\\ 0& ...& 0& 1/\sqrt{{\lambda }_n}\end{array}\right)$.

$(UD{)}^{\ast }M(UD)=D^{\ast }{U}^{\ast }MUD=D^{\ast }(U^{\ast }MU)D=\left(\begin{array}{cccc}1/\sqrt{{\lambda }_1}& 0& ...& 0\\ 0& 1/\sqrt{{\lambda }_2}& ...& 0\\ ...\\ 0& ...& 0& 1/\sqrt{{\lambda }_n}\end{array}\right)\left(\begin{array}{cccc}{\lambda }_1& 0& ...& 0\\ 0& {\lambda }_2& ...& 0\\ ...\\ 0& ...& 0& {\lambda }_n\end{array}\right)\left(\begin{array}{cccc}1/\sqrt{{\lambda }_1}& 0& ...& 0\\ 0& 1/\sqrt{{\lambda }_2}& ...& 0\\ ...\\ 0& ...& 0& 1/\sqrt{{\lambda }_n}\end{array}\right)=I$.

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References

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