Solving for $F(v,f)$ in Tensor $F$

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SUMMARY

The discussion focuses on calculating the value of the tensor $F(v,f)$ where $F = e^1 \otimes e_2 + e^2 \otimes (e_1 + 3e_3) \in T^1_1(V)$, with $v = e_1 + 5e_2 + 4e_3$ and $f = e^1 + e^2 + e^3$. It is confirmed that $e^1 \otimes e_2$ does not equal zero, as it corresponds to the matrix $E_{12} = \begin{bmatrix}0&1&0\\0&0&0\\0&0&0 \end{bmatrix}$, which is a non-zero matrix. The calculation of $e^i \otimes e_j(v,u^{\ast}) = u^TE_{ij}v$ yields a scalar in the underlying field, affirming the tensor's non-triviality.

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Hello everyone

Here is the problem:

Find the value $F(v,f)$ of the tensor $F=e^1\otimes e_2 +e^2\otimes(e_1+3e_3)\in T^1_1(V)$ where $v=e_1+5e_2+4e_3, f=e^1+e^2+e^3$

Does $e^1\otimes e_2=0$ in this problem?Thanks
 
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I wouldn't think so, it seems to me that $e^1 \otimes e_2$ corresponds to the matrix:

$E_{12} = \begin{bmatrix}0&1&0\\0&0&0\\0&0&0 \end{bmatrix}$

which is not the 0-matrix.

That is, that:

$e^i \otimes e_j (v,u^{\ast}) = u^TE_{ij}v $, a scalar in the underlying field.
 
Deveno said:
I wouldn't think so, it seems to me that $e^1 \otimes e_2$ corresponds to the matrix:

$E_{12} = \begin{bmatrix}0&1&0\\0&0&0\\0&0&0 \end{bmatrix}$

which is not the 0-matrix.

That is, that:

$e^i \otimes e_j (v,u^{\ast}) = u^TE_{ij}v $, a scalar in the underlying field.

Got it, thanks a lot:)
 

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