# Equivalent conditions for hermitian matrix

1. Jun 20, 2005

### twoflower

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

1) A is positively definite

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

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

$$A = U^{H}U$$

Proof:

$$2) \Rightarrow 3)$$

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

$$A = R^{H}DR$$
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

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

$$AR = RD$$

$$A = RDR^{-1}$$

And because R is unitary

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

$$A = RDR^{H}$$

This isn't imho equivalent to

$$A = R^{H}DR$$

is it?

Thank you.

Last edited: Jun 20, 2005
2. Jun 20, 2005

### 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?

3. Jun 20, 2005

### twoflower

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

$$R^{-1}AR\mbox{ is diagonal matrix}$$

From this I can get:

$$R^{-1}AR = D$$

$$AR = RD$$

$$A = RDR^{-1} = RDR^{H}$$

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

A is hermitian => there exists R unitary such that

$$A = R^{H}DR$$

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

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

$$R^{-1}AR\mbox{ is diagonal matrix}$$

which I know is true.

4. Jun 20, 2005

### 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?

5. Jun 20, 2005

### twoflower

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

6. Jun 20, 2005

### 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.