On Sunday, November 9, 2025 at 3:06:45 PM UTC-7 Russell Standish wrote:

On Sun, Nov 09, 2025 at 01:11:47AM -0800, Alan Grayson wrote: 
> 
> 
> On Saturday, November 8, 2025 at 6:25:17 AM UTC-7 Alan Grayson wrote: 
> 
> In some treatments of tensors, they're described as linear maps. So, in 
GR, 
> if we have a linear map described as a 4x4 matrix of real numbers, which 
> operates on a 4-vector described as a column matrix with entries (ct, x, 
y, 
> z), which transforms to another 4-vector, what must be added in this 
> description to claim that the linear transformation satisfies the 
> definition of a tensor? TY, AG 
> 
> 
> Let's call the linear transformation T, then the answer to my question 
might be 
> that T is a tensor iff it has a continuous inverse.  I'm not sure if this 
is 
> correct, but I seem to recall this claim in a video about tensors I 
viewed in 
> another life. But even if it's true, it seems to conflict with the claim 
that 
> an ordinary vector in Euclidean space is a tensor because it's invariant 
under 
> linear (?) transformations. In this formulation, it is the argument of T, 
which 
> we can call V, which is invariant, not the map T. I'd appreciate it if 
someone 
> here could clarity my confusion. TY, AG 

It's got nothing to do with being invertible (which is the conjunction 
of being 1:1 and onto). 

Rather a tensor is a multilinear map -


*If it's a map, how can an ordinary vector in Euclidean space be a tensor?*
*Such vectors are NOT maps! See my problem? AG*
 

is a map with multiple 
arguments, and linear in each. Obviously a standard linear map R^n -> R^n 
is a 
rank 2 tensor. We recognise them generally as matrices. Vectors 
correspond to linear maps by means of transposing them and forming the 
inner product, ie a linear map from R^n->R, and are rank 1 tensors as a 
result. 



> 

Dr Russell Standish Phone 0425 253119 (mobile) 
Principal, High Performance Coders [email protected] 
http://www.hpcoders.com.au 
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