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positive definite matrices + kronecker product
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drapt@
Poincare Conjecture

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#21

here is a nice application of the result , prove that if A is symetrique positif matrix then so is the matrix (exp(ai,j))

Posted: Fri Nov 07, 2008 7:26 am

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QuyBac
Riemann Hypothesis

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#22

Re: positive definite matrices + kronecker product
my teacher

alekk wrote:

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 $x$ .

By $A \in S_n^{ + }(\mathbb{R})$ then the exist $\omega \in O_n{\mathbb{(R)}}$ such that : $A = \omega D \omega^{T}$ ("*")
Here $\omega = [\omega_{ij}]_{1 \leq i,j \leq n}$ and $\omega^T = [\omega_{ji}]_{1 \leq i,j \leq n}$ , $D = diag(\lambda_1,\lambda_2,..,\lambda_n)$ , $\lambda_k \geq 0 ,\forall k \in \{1,2,..,n\}$
Then easy see from "*" : $a_{ij} = \sum_{k = 1}^{n}\sum_{m = 1}^{n}{\omega_{ik}\lambda_{ik}\omega_{jm}} = \sum_{k = 1}^{n}{\lambda_k\omega_{ik}\omega_{...$ (1)
You known that $A \in S_n^{ + }(\mathbb{R})$ then $\forall x \in M_{n1}{\mathbb{(R)}}$ then $x^TAx \geq 0$
Now put $y_{ik} = \omega_{ik}x_i,1 \leq i ,j \leq n$ and $y_k = \begin{bmatrix}\omega_{1k}x_1 \\ \omega_{2k}x_2 \\ .... \\ \omega_{nk}x_n \end{bmatrix}=\begin{bmatrix}y_{1k}\\y_{2k}\\...$ , $1 \leq k \leq n$
We have :
$x^TCx = \sum_{i = 1}^{n}\sum_{j = 1}^{n}{c_{ij}x_ix_j} = \sum_{i = 1}^{n}\sum_{j = 1}^{n}{b_{ij}(a_{ij}x_ix_j)} = \sum_{i = 1...$
By $B \in S_n^{ + }(\mathbb{R}),y_k \in M_{n1}{\mathbb{(R)}}$ .....
You can read a different proof of Elizabeth Million about it but I don't find proof Oppenheim's Inequality can post it Embarassed

With $\forall A,B,C \in M_n{\mathbb{(C)}}$ , $C = [a_{ij}b_{ij}]_{1 \leq i,j \leq n}$ then $rank(C) \leq rank(A \otimes{B})$ (easy see small matrix)
About Hadamard's Inequality , $A \in S_n^{ + }{\mathbb{(R)}}$
Then $\det(A) \leq a_{11}a_{22}...a_{nn}$ You can use lemma $A = M^TM,M \in M_n{\mathbb{(R)}}$ then :
$|\det(M)| = |\det(m_1,..,m_n)| \leq ||m_1||...||m_n||$ (Hadamard's Inequality ) then easy $a_{ii} = \sqrt {||m_i||} , 1 \leq i \leq n$ ....
You can talk more ? Rotfl

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Posted: Wed Jul 29, 2009 7:27 pm

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