Tensor Operation on Vector a: Is T a Tensor?

[DougFleming]

A tensor T operates on a vector a and gives:

Ta=a/(magnitude)a

Is this a tensor?

given previous examples of such questions I know that if Ta=3a then T(a+b)=3(a+b)=3a+3b. So I have an idea of how to solve these

Well by looking at what I have if I substituted just like the examples I come up with

T(a+b)=(a+b)/(magnitude) (a+b)

Given this I believe that it isn't a tensor, but should I solve further for the magnitude which will be the square root of vector "a+b". Somehow I am not certain though, should I set T(b) equal to an equation? I am just confused please help

 

[HallsofIvy]

Well, what definition of "tensor" are you using? And what would Ta be if the magnitude of a were 0?

 

[DougFleming]

A tensor transforms a vector into another vector, just like a
dyad. T assigns to an arbitrary vector a to another vector, denoted by Ta and it must satisfy T(au + bv) = aTu + bTv

I believe it would also be for all values other than zero

 

[HallsofIvy]

So far, you have just said that a tensor is a linear transformation- and not all linear transformations are tensors. The crucial point is in that phrase "just like a dyad". What does that mean?

 

[qlinsey]

HI, dear all,

I am very sorry to post my question here since I cannot find where to start a new thread ...sorry ..

Currently I am reading the paper "spatial scan statistics: approximation and performance study " and confused with section 4.1. I read many times , but still cannot get the meaning how the formula of (4.2) come from and why hi=p /4 can give a very good approximation for the convex function? could you guide me which mathematic theoretic background I should read or...how to understand this part? Futhermore, in fact,I cannot relate lemma 4.2 to the part of explanation using Figure 3 (a),(b).

This paper --spatial scan statistics: approximation and performance study (authors: Deepal Agarwal, Andrew McGregor, etc.) , can be downloaded on website from google searching...


Hope can helpe me to figure out !

thanks !
linsey

 

[HallsofIvy]

There is a "New Topic" button just below the title of each category.

 

[DougFleming]

I GOT IT! I was looking in way to deep into this equation. In a way you check if it is linear and substitute in another variable. Very easy! Thanks for your help!