MHB What is the annihilator of a tensor in vector space V?

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The discussion revolves around understanding the concept of the annihilator of a tensor in the context of vector space V, specifically for the tensor \(e_1\wedge e_2+e_3\wedge e_4\). Participants express confusion about the definition and calculation of the annihilator. One user mentions receiving an answer from Stack Exchange but struggles with the complexity due to limited knowledge of tensor algebra. The conversation highlights the need for clearer explanations and examples to grasp the concept effectively. Overall, the thread seeks to clarify the mathematical principles behind the annihilator of tensors.
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Hi everyone, :)

This is a question I don't understand at all. What is the annihilator in this context? Hope you can help me out with this.

Problem: Find the annihilator of the tensor \(e_1\wedge e_2+e_3\wedge e_4\) in \(V=\left<e_1,\,e_2,\,e_3,\,e_4\right>\).
 
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Sudharaka said:
Hi everyone, :)

This is a question I don't understand at all. What is the annihilator in this context? Hope you can help me out with this.

Problem: Find the annihilator of the tensor \(e_1\wedge e_2+e_3\wedge e_4\) in \(V=\left<e_1,\,e_2,\,e_3,\,e_4\right>\).

Hi everyone, :)

I have received an answer to this question on Stack Exchange. I am trying to understand it but due to my limited knowledge of tensor algebra it will take a long time. :p

linear algebra - Annihilator of a Tensor - Mathematics Stack Exchange
 
Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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