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| In [[mathematics]], particularly in [[linear algebra]], the '''Schur product theorem''' states that the [[Hadamard product (matrices)|Hadamard product]] of two [[positive definite matrices]] is also a positive definite matrix. The result is named after [[Issai Schur]]<ref name="Sch1911">{{Cite doi|10.1515/crll.1911.140.1}}</ref> (Schur 1911, p. 14, Theorem VII) (note that Schur signed as J. Schur in ''Journal für die reine und angewandte Mathematik''.<ref>{{Cite doi|10.1007/b105056}}, page 9, Ch. 0.6 ''Publication under J. Schur''</ref><ref>{{Cite doi|10.1112/blms/15.2.97}}</ref>)
| | <br><br>dawonls.dothome.co.kr [http://dawonls.dothome.co.kr/db/?document_srl=1017665 http://dawonls.dothome.co.kr/db/?document_srl=1017665]. <br><br>There are more inexpensive services out there, but I am comfortable with these ones. I didn't think so either. When you are looking to purchase something specially on the net you always desire to make positive you might have the very best deal. If you are pondering of organising your individual website, then perhaps one of the first issues that you want to take into account is finding a great web hosting service that may cater to your on-line needs. They offer so much for so little, this truly is one of the diamonds in the ruff sites that sound to good to be true, but deliver on everything they claim. They are a domain name and web hosting. It is such good value for capital that it is tough to say no.For more info check out HostGator. The green (stoplight) was the aim. I also use adsense and kontera on my sites. Hostgator Quick Install Feature Is Better Than [http://imgur.com/hot?q=Fantastico+Hostgator Fantastico Hostgator] is one of the best webhosting provider in the world.It is the only webhosting which has constantly strived to its top position without considering any of the competition provided by other webhosting companies.<br><br><br><br>This program can show you if the niche you chose will be able to make you money. Achievements in modern small business calls for an powerful net existence. So, you need to choose the best one that match with your requirement. Depending on your own individual taste and the all round theme of your site, you can pick templates from a selection or classes like trend, holiday, design and style studio and hotel. Soon after I signed up, as a end result of an error on my element, I had an matter with a credit card on my [http://Dict.Leo.org/?search=account account] when a new card was issued and had to change cards. Are you going to use the first coupon you can find or are you willing to search high and low for the best? With wordpress, learning to create or edit content is generally fairly simple. But there seems to be a relative low number of Brokers using wordpress to build web entities at the company level. |
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| == Proof == | |
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| === Proof using the trace formula ===
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| It is easy to show that for matrices <math>M</math> and <math>N</math>, the Hadamard product <math>M \circ N</math> considered as a bilinear form acts on vectors <math>a, b</math> as
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| : <math>a^T (M \circ N) b = \operatorname{Tr}(M \operatorname{diag}(a) N \operatorname{diag}(b))</math>
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| where <math>\operatorname{Tr}</math> is the matrix [[Trace (linear algebra)|trace]] and <math>\operatorname{diag}(a)</math> is the [[diagonal matrix]] having as diagonal entries the elements of <math>a</math>.
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| Since <math>M</math> and <math>N</math> are positive definite, we can consider their square-roots <math>M^{1/2}</math> and <math>N^{1/2}</math> and write
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| : <math>\operatorname{Tr}(M \operatorname{diag}(a) N \operatorname{diag}(b)) = \operatorname{Tr}(M^{1/2} M^{1/2} \operatorname{diag}(a) N^{1/2} N^{1/2} \operatorname{diag}(b)) = \operatorname{Tr}(M^{1/2} \operatorname{diag}(a) N^{1/2} N^{1/2} \operatorname{diag}(b) M^{1/2})</math>
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| Then, for <math>a=b</math>, this is written as <math>\operatorname{Tr}(A^T A)</math> for <math>A = N^{1/2} \operatorname{diag}(a) M^{1/2}</math>
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| and thus is positive. This shows that <math>(M \circ N)</math> is a positive definite matrix. | |
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| === Proof using Gaussian integration ===
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| ==== Case of ''M'' = ''N'' ====
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| Let <math>X</math> be an <math>n</math>-dimensional centered [[Gaussian random variable]] with [[covariance]] <math>\langle X_i X_j \rangle = M_{ij}</math>.
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| Then the covariance matrix of <math>X_i^2</math> and <math>X_j^2</math> is
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| : <math>\operatorname{Cov}(X_i^2, X_j^2) = \langle X_i^2 X_j^2 \rangle - \langle X_i^2 \rangle \langle X_j^2 \rangle</math>
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| Using [[Wick's theorem]] to develop <math>\langle X_i^2 X_j^2 \rangle = 2 \langle X_i X_j \rangle^2 + \langle X_i^2 \rangle \langle X_j^2 \rangle</math> we have
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| : <math>\operatorname{Cov}(X_i^2, X_j^2) = 2 \langle X_i X_j \rangle^2 = 2 M_{ij}^2</math>
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| Since a covariance matrix is positive definite, this proves that the matrix with elements <math>M_{ij}^2</math> is a positive definite matrix.
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| ==== General case ====
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| Let <math>X</math> and <math>Y</math> be <math>n</math>-dimensional centered [[Gaussian random variable]]s with [[covariance]]s <math>\langle X_i X_j \rangle = M_{ij}</math>, <math>\langle Y_i Y_j \rangle = N_{ij}</math> and independt from each other so that we have
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| : <math>\langle X_i Y_j \rangle = 0</math> for any <math>i, j</math> | |
| Then the covariance matrix of <math>X_i Y_i</math> and <math>X_j Y_j</math> is
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| : <math>\operatorname{Cov}(X_i Y_i, X_j Y_j) = \langle X_i Y_i X_j Y_j \rangle - \langle X_i Y_i \rangle \langle X_j Y_j \rangle</math>
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| Using [[Wick's theorem]] to develop
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| : <math>\langle X_i Y_i X_j Y_j \rangle = \langle X_i X_j \rangle \langle Y_i Y_j \rangle + \langle X_i Y_i \rangle \langle X_i Y_j \rangle + \langle X_i Y_j \rangle \langle X_j Y_i \rangle</math>
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| and also using the independence of <math>X</math> and <math>Y</math>, we have
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| : <math>\operatorname{Cov}(X_i Y_i, X_j Y_j) = \langle X_i X_j \rangle \langle Y_i Y_j \rangle = M_{ij} N_{ij}</math>
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| Since a covariance matrix is positive definite, this proves that the matrix with elements <math>M_{ij} N_{ij}</math> is a positive definite matrix.
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| === Proof using eigendecomposition ===
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| ==== Proof of positivity ====
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| Let <math>M = \sum \mu_i m_i m_i^T</math> and <math>N = \sum \nu_i n_i n_i^T</math>. Then
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| : <math>M \circ N = \sum_{ij} \mu_i \nu_j (m_i m_i^T) \circ (n_j n_j^T) = \sum_{ij} \mu_i \nu_j (m_i \circ n_j) (m_i \circ n_j)^T</math>
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| Each <math>(m_i \circ n_j) (m_i \circ n_j)^T</math> is positive (but, except in the 1-dimensional case, not positive definite, since they are [[Rank (linear algebra)|rank]] 1 matrices) and <math>\mu_i \nu_j > 0</math>, thus the sum giving <math>M \circ N</math> is also positive.
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| ==== Complete proof ====
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| To show that the result is positive definite requires further proof. We shall show that for any vector <math>a \neq 0</math>, we have <math>a^T (M \circ N) a > 0</math>. Continuing as above, each <math>a^T (m_i \circ n_j) (m_i \circ n_j)^T a \ge 0</math>, so it remains to show that there exist <math>i</math> and <math>j</math> for which the inequality is strict. For this we observe that
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| : <math>a^T (m_i \circ n_j) (m_i \circ n_j)^T a = \left(\sum_k m_{i,k} n_{j,k} a_k\right)^2</math>
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| Since <math>N</math> is positive definite, there is a <math>j</math> for which <math>n_{j,k} a_k</math> is not 0 for all <math>k</math>, and then, since <math>M</math> is positive definite, there is an <math>i</math> for which <math>m_{i,k} n_{j,k} a_k</math> is not 0 for all <math>k</math>. Then for this <math>i</math>and <math>j</math> we have <math>\left(\sum_k m_{i,k} n_{j,k} a_k\right)^2 > 0</math>. This completes the proof.
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| == References ==
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| {{reflist}}
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| == External links ==
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| * [https://eudml.org/doc/149352 Bemerkungen zur Theorie der beschränkten Bilinearformen mit unendlich vielen Veränderlichen] at [https://eudml.org EUDML]
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| [[Category:Linear algebra]]
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| [[Category:Matrix theory]]
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