positive definite matrices + kronecker product

alekk

Let $A = \left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ and $B = \left(b_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ be two $n\times n$ positive definite matrices. Show that the matrix $C = \left(c_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ defined by $c_{i,j} = a_{i,j}b_{i,j}$ is also positive definite.

Remark (inserted by moderator). Under a positive definite matrix, we understand a symmetric (or Hermitean, if we work over the complex numbers) matrix $U$ such that $x^TUx > 0$ (or $x^*Ux > 0$ , respectively) for any nonzero vector $x$ .

**Posted:** Fri Jan 14, 2005 5:49 am

_________________
Please, could you rephrase your question in the form of the answer?

fredbel6

is that correct?
the product of two positive definite matrices isn't even always a symmetric (or hermitian if you work complex)

take A = [ 1 1/2
1/2 1

B = 3 1
1 2

the product is not symmetric!

**Posted:** Fri Jan 14, 2005 6:23 am

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

? he just claimed it's positive definite, not symmetric...

**Posted:** Fri Jan 14, 2005 6:33 am

harazi

Use Schur decomposition theorem to prove that any positive semidefinite matrix can be written as the sum of k matrices of rank 1 of the form $x\cdot x^{*}$ where k is the rank of A and the vectors x are ortogonal. Then use the fact that the sum of positive semidefinite matrices is semidefinite.

**Posted:** Fri Jan 14, 2005 7:24 am

fredbel6

whenever i think i know something i make a fool of myself
indead, i am so used at working with symmetric positive definite that i thought other positive definite matrices
weren't considered
sorry!

**Posted:** Fri Jan 14, 2005 7:29 am

alekk

**Posted:** Fri Jan 14, 2005 1:06 pm

_________________
Please, could you rephrase your question in the form of the answer?

alekk

OK, now it's clear. Very nice Mr. Green

**Posted:** Fri Jan 14, 2005 4:10 pm

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Please, could you rephrase your question in the form of the answer?

liyi

**Posted:** Sun Jan 16, 2005 5:47 pm

alekk

I think here that to prove the statement of Harazi you just have to diagonalize the matrices in an orthogonal basis.
Conclusion follows.

**Posted:** Mon Jan 17, 2005 7:07 am

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Please, could you rephrase your question in the form of the answer?

liyi

I still don't understand this...
What is the next step after obtaining the spectrum decomposition of the positive definite matrices?

**Posted:** Mon Apr 04, 2005 3:40 am

harazi

Well, it suffices to write that $A=U^{*}DU$ with D diagonal and to identify the coefficients in the equality.

**Posted:** Mon Apr 04, 2005 6:02 am

liyi

Can you give a detailed solution...
I still don't understand.

**Posted:** Mon Apr 04, 2005 6:27 am

harazi

I will post a complete solution tomorow. Now I have a discussion... guess with whom.

**Posted:** Mon Apr 04, 2005 6:32 am

harazi

So, any positive semidefinite matrix A of rank k can be written in the form $v_1 v_1^{*}+...+v_k v_k^{*}$ with $v_i$ ortogonal vectors. Indeed, write by spectral theorem $A=UDU^{*}$ with U ortogonal and D diagonal with entries $d_i\geq 0$ and take $v_i$ the vector $\sqrt{d_i}\cdot M_i$ where $M_i$ is the column i vector in the matrix U.

**Posted:** Wed Apr 06, 2005 6:58 am

rucarden · New Member Offline Joined: 01 Jul 2005 Posts: 1

**Posted:** Sat Jul 02, 2005 10:12 pm

nguyenkims · New Member Offline Joined: 23 Jan 2005 Posts: 13

I don't think it's true.
The positive defination is symmetric?

**Posted:** Tue Jul 12, 2005 10:18 am

yassinus

hi ,wath are you talking about , a matrix witch is defined positivly isn't necesserly symetric ,take for exemple (1 -1) ( of course in IR-space )
(0 1)

**Posted:** Sun Aug 07, 2005 4:40 pm

darij grinberg

Do you know what I find neat about this topic? The matrix $C$ defined above is not the Kronecker product of the matrices $A$ and $B$ , but throwing in the words "Kronecker product" points to another (in my opinion, better) solution of the problem:

The matrix $C$ is a principal minor of the Kronecker product of the matrices $A$ and $B$ (in fact, we can think of the rows of the Kronecker product indexed by pairs of numbers $\left(i,j\right)\in\left\{1,2,...,n\right\}^2$ , and of the columns indexed the same way; now, the matrix $C$ is the principal minor of this Kronecker product obtained by taking only the rows $\left(i,i\right)$ and the columns $\left(k,k\right)$ ). Now, the Kronecker product of the matrices $A$ and $B$ is positive definite (says the Wikipedia, and I guess this is not hard to prove), and a principal minor of a positive definite matrix is positive definite again (trivial by the definition). Thus we are done.

By the way, for me, a positive definite matrix is supposed to be symmetric by the definition of "positive definite". The Wikipedia agrees with me...

darij

**Posted:** Fri Feb 15, 2008 7:34 pm

_________________
lift this prayer to heaven's heights, feed my mind with the most glorious thoughts,
sacrificing all my dreams...
for this i will serve and honour to death... or do i just fear the tyrant's wrath?

blahblahblah

the proper phrase is 'hadamard product', i believe.

**Posted:** Sat Feb 16, 2008 12:25 am

alekk

as Darij says, it is not hard to see that if $A,B$ are positive, then so is their Kronecker product. For the simple case, $dim (A) = dim(B)$ : the Kronecker product of $A,B$ can be seen as the matrix of the application $\phi_{A,B}: M_n \to M_n$ defined by $\phi_{A,B}(M) = AMB$ . One has to show that for every matrix $M$ , the following holds: $Tr(M^tAMB) = Tr(AMBM^t) \geq 0$ . Write $A = O^t D O$ where $D$ is diagonal with positive elements to see that $Tr(M^t AMB) = Tr(D (OM) B (OM)^t)$ . The conclusion then follows since $[(OM) B (OM)^t]_{i,i} \geq 0$ (because $B$ is positive).

**Posted:** Sat Feb 16, 2008 7:14 am

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Please, could you rephrase your question in the form of the answer?