On Sat, Nov 22, 2025 at 03:59:00AM -0800, Alan Grayson wrote: > > > On Friday, November 21, 2025 at 3:27:17 PM UTC-7 Russell Standish wrote: > > On Thu, Nov 20, 2025 at 07:02:53PM -0800, Alan Grayson wrote: > > > > > > I studied linear algebra, but my questions involve tensors. If a tensor > > T is defined as a linear function whose domain is a vector space, and > > maps to a real number, how does one get a real number from T(u), if we > > do the calculation using matrices? Here there is no v, just u. AG > > A matrix corresponds to a rank 2 tensor, ie T(u,v)∈R. T(u)∈R > corresponds to a rank 1 tensor. In matrix notation, a rank 1 tensor is > a transposed vector, ie vᵀ for some vector v∈Rⁿ. vᵀu in matrix > notation corresponds to v.u (ie dot or inner product of two vectors). > > > I'm seeking an unambiguous definition of a TENSOR. You wrote earlier > that a tensor is a MAP whose arguments are VECTORS in a vector space, > which MAP to real numbers, and is INVARIANT under changes in coordinate > systems. Your definition seems OK, but upon more analysis I find it > kind-of vacuous. Firstly, any function which depends on elements in a > vector space which are invariant under changes in coordinate systems, > will necessarily be invariant under changes in coordinate systems, and > it doesn't matter if that function is linear or not in its arguments. So, is > a tensor just limited by the condition of linearity of its arguments? The > invariance under coordinate transformations is a direct result of what > its arguments are, and since vectors are invariant, so the tensor T must > also have this property. That is, the invariance property of T is totally > dependent on the invariance property of its domain, the invariant vectors > in some vector space. TY, AG
Correct so far. Yes - it seems trivial so far, but when you work out how the components of the tensor change with changes of the coordinate system you get the concept of covariance, and when you apply the tensors to tangent spaces on Riemann manifolds, it become decidedly non-trivial. When you're ready, you'll need to work through coveriant differentiation, where there is an additional term coming from curvature of spacetime. > > Moreover, you claim an invariant vector is in fact a tensor of rank 1. It is actually the transpose of a vector. > What is its MAP? The transpose uᵀ. Using tensor multiplication, uᵀv === u.v, where . is the familiar inner product. > Why do you need to introduce v to evaluate T(u), v is the vector corresponding to the tensor T. > which is just a function of u? Using matrices, there's no way to get > a constant as the result of T(u) (which I assume is a row matrix). > I suppose a constant is a tensor of rank 0. What is its MAP, your entity > that DEFINES a tensor? TY, AG Yes - a scalar is considered to be a rank 0 tensor. As you point out, it is not really a map of anything, though, so I suppose it is the exception to the rule that tensors are multilinear maps. > > > > ---------------------------------------------------------------------------- > > Dr Russell Standish Phone 0425 253119 (mobile) > Principal, High Performance Coders [email protected] > http://www.hpcoders.com.au > > ---------------------------------------------------------------------------- > > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email > to [email protected]. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list > /5d526512-286f-4d53-b598-c790f7b6c347n%40googlegroups.com. -- ---------------------------------------------------------------------------- Dr Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders [email protected] http://www.hpcoders.com.au ---------------------------------------------------------------------------- -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/aSOWCrtFKPcFqm1I%40zen.
