Let's have hermitian matrix A. Then these three conditions are equivalent:

1) A is positively definite

[tex]

\forall x \in \mathbb{C}^{n} \ {0} : x^{H}Ax > 0

[/tex]

2) All eigenvalues of A are positive

3) There exists regular matrix U such that

[tex]

A = U^{H}U

[/tex]

Proof:

[tex]

2) \Rightarrow 3)

[/tex]

A is hermitian => [itex]\exists[/itex] unitary matrix R such that

[tex]

A = R^{H}DR

[/tex]

where D is diagonal matrix.

This is what I don't understand - we know that if A is hermitian, there exists R unitary such that

[tex]

R^{-1}AR = D \mbox{ diagonal \\}

[/tex]

[tex]

AR = RD

[/tex]

[tex]

A = RDR^{-1}

[/tex]

And because R is unitary

[tex]

R^{-1} = R^{H}

[/tex]

[tex]

A = RDR^{H}

[/tex]

This isn't imho equivalent to

[tex]

A = R^{H}DR

[/tex]

is it?

Thank you.

1) A is positively definite

[tex]

\forall x \in \mathbb{C}^{n} \ {0} : x^{H}Ax > 0

[/tex]

2) All eigenvalues of A are positive

3) There exists regular matrix U such that

[tex]

A = U^{H}U

[/tex]

Proof:

[tex]

2) \Rightarrow 3)

[/tex]

A is hermitian => [itex]\exists[/itex] unitary matrix R such that

[tex]

A = R^{H}DR

[/tex]

where D is diagonal matrix.

This is what I don't understand - we know that if A is hermitian, there exists R unitary such that

[tex]

R^{-1}AR = D \mbox{ diagonal \\}

[/tex]

[tex]

AR = RD

[/tex]

[tex]

A = RDR^{-1}

[/tex]

And because R is unitary

[tex]

R^{-1} = R^{H}

[/tex]

[tex]

A = RDR^{H}

[/tex]

This isn't imho equivalent to

[tex]

A = R^{H}DR

[/tex]

is it?

Thank you.

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