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Let $V$ is a finite dimensional vector space over a field $\mathbb F.$ Let $\rm dim (V)=n>0$ and $\mathcal {B}=\{v_1,\ldots,v_n\}$ be a basis of $V.$ Now we know dimension of $V \otimes_\mathbb {F}V$ is $n^2$ as $V \otimes_\mathbb {F}V \cong \mathbb {F^{n^2}}.$ Now since the set $\mathcal {A}=\{v_i\otimes v_j:1 \leq i,j\leq n\}$ spans $V \otimes_\mathbb {F}V$ and the number of elements in $\mathcal {A}$ is $n^2$, $\mathcal {A}$ forms a basis for $V \otimes_\mathbb {F}V$ as a vector space over $\mathbb F.$ In this way every element in $\mathcal {A}$ is non zero.

Now my question is if I don't want to use the above arguments how can I show that for any $i$ and $j$ the element $v_i\otimes v_j$ is non zero in $V \otimes_\mathbb {F}V$. By the construction of tensor product if it can be shown then it will help me.

Help me. Many Thanks.

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2 Answers 2

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Whatever your definition of tensor is, it should be true that any bilinear function $$ V\times V\longrightarrow \mathbb{F} $$ must factor uniquely through $$ V\times V\longrightarrow V\otimes_\mathbb{F} V\longrightarrow \mathbb{F} $$ Now take any linear functional $\ell :V\to\mathbb{F}$ such that $\ell(v_i)\neq0 $ and $\ell(v_j)\neq 0$. Define a bilinear function $f:V\times V\to \mathbb{F}$ as $$ f(u,v) = \ell(u)\ell(v) $$ Since $f$ is clearly bilinear, $f$ must factor through a homomorphism $g:V\otimes_\mathbb{F} V\to \mathbb{F}$, i.e. $g(u\otimes v) = f(u,v)$. Since $g(v_i\otimes v_j) = f(v_i,v_j) = \ell(v_i)\ell(v_j)\neq 0$, $v_i\otimes v_j$ is nonzero.

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  • $\begingroup$ you were faster :) $\endgroup$
    – user171326
    Jan 26, 2017 at 7:07
  • $\begingroup$ Thank you for writing in details. $\endgroup$
    – user371231
    Jan 26, 2017 at 7:14
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If $(v_i,v_j)$ are non-zero there is a bilinear application $\beta : V \times V \to k$ with $\beta(v_i,v_j) = 1$ and by the universal property of tensor product we get a map $b : V \otimes V \to k $ with $b(v_i \otimes v_j) = 1$ so it cannot be zero.

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