Equivalent conditions for hermitian matrix

[twoflower]

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

1) A is positively definite

<br /> \forall x \in \mathbb{C}^{n} \ {0} : x^{H}Ax &gt; 0<br />

2) All eigenvalues of A are positive
3) There exists regular matrix U such that

<br /> A = U^{H}U<br />

Proof:

<br /> 2) \Rightarrow 3)<br />

A is hermitian => \exists unitary matrix R such that

<br /> A = R^{H}DR<br />
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

<br /> R^{-1}AR = D \mbox{ diagonal \\}<br />

<br /> AR = RD<br />

<br /> A = RDR^{-1}<br />

And because R is unitary

<br /> R^{-1} = R^{H}<br />

<br /> A = RDR^{H}<br />

This isn't imho equivalent to

<br /> A = R^{H}DR<br />

is it?

Thank you.

 

[matt grime]

I can't quite figure out the best answer, but it as along the general lines of "so what?" Are you claiming R is unique and that it must be written as AR=RD? what is wrong with saying there is a unitary S such that SA=DS? they are equivalent.

but i can't tell what your problem is; indeed RDR^H and R^HDR are almost always different, but so what?

 

[twoflower]

I can't quite figure out the best answer, but it as along the general lines of "so what?" Are you claiming R is unique and that it must be written as AR=RD?

I'm not sure whether R is unique. Anyway, I have theorem (the one I wrote into another post):

A is hermitian => there exists R unitary such that

<br /> R^{-1}AR\mbox{ is diagonal matrix}<br />

From this I can get:

<br /> R^{-1}AR = D<br />

<br /> AR = RD<br />

<br /> A = RDR^{-1} = RDR^{H}<br />

But, the proof of the theorem about equivalent conditions for hermitian matrix says, that

A is hermitian => there exists R unitary such that

<br /> A = R^{H}DR<br />


How can I achieve this? I can't see that..

what is wrong with saying there is a unitary S such that SA=DS? they are equivalent.

Well, that's the point. Why are they equivalent? I just want to use the theorem

A is hermitian => there exists R unitary such that

<br /> R^{-1}AR\mbox{ is diagonal matrix}<br />

which I know is true.

 

[matt grime]

You are assuming that the R's in both those theorems are the same; they are not and there is no reason to be. rename the R as S in the first part and we know

A is hermitian if there is a unitary S such that

S*AS=E (using * for the H operation)

where E is diagonal with real entries

Ie A=SES*

right?

now the therem you wnat to prove states as a acondition that there is a unitary R and a diagoanl real D such that A=R*DR

well, they are equivalent if I let D=E and R=S*

there is nothing that states the diagonaliztaion of a hermitian matrix is unique (it isn't; reorder the basis).

you at least see they are equivalent? why must those two R's be the same? Or for that matter, the D?

 

[twoflower]

You are assuming that the R's in both those theorems are the same; they are not and there is no reason to be. rename the R as S in the first part and we know

A is hermitian if there is a unitary S such that

S*AS=E (using * for the H operation)

where E is diagonal with real entries

Ie A=SES*

right?

now the therem you wnat to prove states as a acondition that there is a unitary R and a diagoanl real D such that A=R*DR

well, they are equivalent if I let D=E and R=S*

there is nothing that states the diagonaliztaion of a hermitian matrix is unique (it isn't; reorder the basis).

you at least see they are equivalent? why must those two R's be the same? Or for that matter, the D?

Thank you very much Matt, now I'm completely clear about that matter. My fault that I was interpreting the theorem too literally and the substitution R = S* didn't come to my mind. Your answers teach me more that studying from lecture notes

 

[matt grime]

if soemthing states "there is an R" for example, note it is just a label; labelling two things with the same letter here was exactly the problem. lectuer notes won't teach you that since they assume that you won't do that.