On Sat, Nov 15, 2025 at 08:22:23PM -0800, Alan Grayson wrote:
> 
> 
> On Friday, November 14, 2025 at 8:30:29 PM UTC-7 Alan Grayson wrote:
> 
>     On Friday, November 14, 2025 at 7:42:23 PM UTC-7 Russell Standish wrote:
> 
>         T(u,v) = uᵀMv where M is the matrix representation of T, and ᵀ is the
>         transpose operator.
> 
> 
>     This is too succinct for me to understand your explanation. AG 
> 
> 
> Is the above how a tensor is generally evaluated? How would it be evaluated if
> T has three independent variables, u,v,w? AG
> 
> If u and v are modeled as row matrices, then u transposed is a colum vector,
> and
> the total result is a real number. But I don't think it's easy to show that
> this is the
> same value obtained by modeling the tensor as a linear function of u and v.
> Offhand.
> do you have a link for showing the equivalence?  Further, in the case of T(u),
> how is 
> the result a real number when using matrix notation? It looks like a vector
> when u
> is modeled as a column vector. OTOH, I don't think we can model u as a row
> vector
> to do the calculation in this situation (or can we?). AG 
> 

This is Linear Algebra 101.

To convince yourself, try it with a 2x2 matrix to make the
calculations easier. Extending the result to Rⁿis not difficult.

Exercise:

 Show that

(auᵀ₁+buᵀ₂)Mv = auᵀ₁Mv + buᵀ₂Mv

and

uᵀM(av₁+bv₂)=auᵀMv₁+ buᵀMv₂

for a,b∈R, uᵢ,vᵢ ∈ Rⁿ, and M a real valued matrix.

The above two lines are the definition of a bilinear function from Rⁿ⟶ R.

For a trilinear function, it is more convenient to use the Einstein
summation convention, but basically it works the same way.



-- 

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