Symmetric vector to tensor representation?

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SUMMARY

This discussion focuses on converting a discrete 2D vector field with symmetric vectors into a tensor representation. The vectors at each point are opposites, such as v1(1,1) and v2(-1,-1), leading to a tensor field defined as T(point)=[T11(x,y) T12(x,y); T12(x,y) T22(x,y)]. The dyadic product of the vectors is utilized to generate the tensor, confirming that the tensor remains consistent regardless of the order of the vectors. The concept of "degenerate points" is referenced, drawing from Delmarcelle and Hasselink's 1993 work.

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  • Knowledge of dyadic products in vector mathematics
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  • Basic concepts of discrete vector fields in two dimensions
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hiroman
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Hi all!

I have a discrete 2D vector field with a particular characteristic: At every point, instead of having a single vector, I have two vectors which are in the opposite direction. For example, at point p(x,y)=p(0,0) I have two vectors: v1(1,1) and v2(-1,-1). And so on for all points.

I understand this becomes an "eigenvector field" situation, or a "tensor field", or "symmetric tensor field" situation.

At the end, I wish to find the so called "degenerate points" (refer to Delmarcelle, Hasselink 1993).

But first, I wish to translate these two vectors on a single point to a 2x2 tensor representation, such that T(point)=[T11(x,y) T12(x,y) ; T12(x,y) T22(x,y)].

Summarizing,

Given, two symmetric vectors at a point, v1(point)=[x1;y1] and v2(point)=[x2;y2], such that v2(point)=(-1)*v1(point) represent them in tensor form T(point)=[T11(x,y) T12(x,y) ; T12(x,y) T22(x,y)].

Much appreciated!
 
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Just realized that I can use the dyadic product of two vectors to generate my tensor.

Thus, v1(point)=[1; 2]; v2(point)=[-1;-2] can give T(point)=[(1)(-1) (1)(-2); (2)(-1) (2)(-2)]
T(point)=[-1 -2;-2 -4].

Thus I can have the same eigenvectors if [dyadic product(v1,v2)] or [dyadic product(v2,v1)] since they give me the same tensor.

If anyone is interested.
 

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