On Tue, Nov 18, 2025 at 11:31:22PM -0800, Alan Grayson wrote: > > > On Monday, November 17, 2025 at 9:54:00 PM UTC-7 Russell Standish wrote: > 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. > > > If they're definitions, there's nothing to be shown or proven. But I have two > questions relating to this subject. First, since uᵀ is a column matrix, is it > OK > to place it on the RHS of M, with the convention that Muᵀ is evaluated first, > followed by the result being evaluated by applying v (a row matrix), so we > we get a real constant as the result? Second, if the function has one > independent variable, say u, I don't see how we can use matrix notation to > evaluate the tensor To get a real value as the result. TY, AG
You mean uᵀv=u.v∈R? In this case u is a vector, and uᵀ is a rank 1 tensor. Cheers -- ---------------------------------------------------------------------------- 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/aR7ZBO3XNlKLlIe0%40zen.
