Succinct data structure: Difference between revisions

Latest revision as of 23:36, 23 November 2014

I'm Yoshiko Oquendo. Arizona is her birth place and she will never move. To maintain birds is 1 of the things he enjoys most. The job I've been occupying for many years is a bookkeeper but I've already applied for an additional one.

Visit my blog: http://muehle-Kruskop.de

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-{{Unreferenced|date=February 2008}}
+I'm Yoshiko Oquendo. Arizona is her birth place and she will never move. To maintain birds is 1 of the things he enjoys most. The job I've been occupying for many years is a bookkeeper but I've already applied for an additional one.<br><br>Visit my blog: [http://muehle-Kruskop.de/index.php?mod=users&action=view&id=22344 http://muehle-Kruskop.de]
-The '''[[tensor product]] of [[quadratic form]]s''' is most easily understood when one views the quadratic forms as ''[[quadratic space]]s''. So, if (''V'', ''q''<sub>1</sub>) and (''W'', ''q''<sub>2</sub>) are quadratic spaces, with ''V'',''W'' [[vector spaces]], then the tensor product is a quadratic form ''q'' on the [[Tensor product#Tensor product of vector spaces|tensor product of vector spaces]] ''V'' ⊗ ''W''.
-It is defined in such a way that for <math>v \otimes w \in V \otimes W</math> we have <math>q(v \otimes w) = q_1(v)q_2(w)</math>. In particular, if we have diagonalizations of our quadratic forms (which is always possible when the [[characteristic (algebra)|characteristic]] is not 2) such that
-:<math>q_1 \cong \langle a_1, ... , a_n \rangle</math>
-:<math>q_2 \cong \langle b_1, ... , b_m \rangle</math>
-then the tensor product has diagonalization
-:<math>q_1 \otimes q_2 = q \cong \langle a_1b_1, a_1b_2, ... a_1b_m, a_2b_1, ... , a_2b_m , ... , a_nb_1, ... a_nb_m \rangle.</math>
-[[Category:Quadratic forms]]
-[[Category:Tensors]]
-{{algebra-stub}}

Succinct data structure: Difference between revisions

Latest revision as of 23:36, 23 November 2014

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