Symmetric vector to tensor representation?

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The discussion focuses on converting a discrete 2D vector field with two opposite vectors at each point into a symmetric tensor representation. The user seeks to represent these vectors in a 2x2 tensor format, specifically T(point)=[T11(x,y) T12(x,y); T12(x,y) T22(x,y)]. They realize that using the dyadic product of the two vectors can generate the desired tensor. An example is provided where the vectors v1 and v2 yield a tensor T(point)=[-1 -2; -2 -4]. The conversation highlights the relationship between the dyadic product and the eigenvector properties of the resulting tensor.
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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