Show that $(B(E\times E), \otimes)$ is a tensor product of $E^*\times E^*$ , where $E$ is a real finite dimensional vector space , $E^*$ is its dual space, $B(E\times E)$ is the space of all bilinears from $E\times E$ to $\mathbb{R}$ and $\otimes (f,g)=b$ such that $b(u,v)=f(u)g(v)$. The concept of tensor product is fairly new to me but I know that I need to show that for every other bilinear $\varphi$ that maps $E\times E$ onto an space $G$ there exists a unique linear transformation $h$ such that $h \circ \otimes = \varphi $. The problem I have is constructing that linear transformation for any given $\varphi$, any clues as to where to start?

  • $\begingroup$ Is $E$ finite-dimensional? $\endgroup$ – wj32 Jun 27 '13 at 3:49
  • $\begingroup$ Yes it is, just edited the question to avoid confusion. $\endgroup$ – Julio Cáceres Jun 27 '13 at 3:51
  • $\begingroup$ Does there really not occur any $g$ in the definition of $b(u,v)$? $\endgroup$ – HSN Jun 27 '13 at 7:56
  • $\begingroup$ Mistyped, just changed that bit. $\endgroup$ – Julio Cáceres Jun 27 '13 at 11:21

Be $\{e_i\}$ a basis of $E$ and $\{e^i\}$ its dual basis in $E^*$, that is, $e^i(e_j)=\delta^i_j$. Then every vector $u$ can be written as $\sum_i u^i e_i$ and every covector $f$ can be written as $f = \sum_i f_i e^i$. Note that $e^i(u) = u^i$.

Now $$\otimes(f,g) = \otimes(\sum_i f_i e^i,\sum_j g_j e^j) = \sum_{i,j} f_i g_j\, {\otimes(e^i,e^j)}$$ and thus $$(\otimes(f,g))(u,v) = \sum_{i,j} f_i g_j\, {\otimes(e^i,e^j)}(u,v)$$ Now for $\phi$, we have $$\phi(u,v) = \phi(\sum_i u^i e_i, \sum_j v^j e_j) = \sum_{i,j} u^i v^j \phi(e_i,e_j) = \sum_{i,j} \phi(e_i,e_j) (\otimes(e^i,e^j))(u,v)$$ or short $$\phi = \sum_{i,j} \phi(e_i,e_j) ( \otimes(e^i,e^j))$$ If we define $e^{ij}:=\otimes(e^i,e^j)$, then it is easy to see that the $e^{ij}$ form a basis of $B(E\times E)$. Therefore the function $h$ has the form $$h(\sum_{ij}\lambda_{ij} e^{ij}) = \sum_{ij}\lambda_{ij} \phi(e_i,e_j)$$

  • $\begingroup$ Nice construction, my actual problem was to show that $\otimes(e^i,e^j)$ was a basis of $B(E \times E)$ which I already did. I knew that if demonstrated that $(B(E \times E),\otimes)$ was the tensor product the basis thing would follow, thanks a ton! $\endgroup$ – Julio Cáceres Jun 28 '13 at 0:52

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