- #1
twoflower
- 368
- 0
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.
Last edited: