Question

(4) Let V and W be vector spaces over R: consider the free vector space F(V × W) on the Cartesian product V x W of V and W. Given an element (v, w) of V x W, we view (v, w) as an element of F(V x W) via the inclusion map i : V x W F(V x W) Any element of F(V x W) is a finite linear combination of such elements (v, w) Warning. F(V × W) disregards the vector space structures on V and W, and just treats VxW as the set of ordered pairs (v, w) wherev E V and w E W. For example, if vメ0 in V, then (v.0) and (2 .0) are linearly independent in F(V x W), even though v and 2v are not linearly independent in V Let S be the subset of F(V x W) consisting of the following elements: . (v, w) + (v', w) - (v +v', w) for all v, v' EV and wEW ·(v, w) + (v, w,) _ (v. w+ w,) for all v E V and w, w, E W · o(v, w)-(cv.w) for all v E V, w E W, and c E R Definition. Define F(V × W) span(S) Given an element (v, w) of F(V x W), its class [(v, w)| in the quotient space V & W warning. In general, not every element of V W can be written as vⓧw for some element of VⓧW is a linear combination of pure tensors, but it might not be a pure will be denoted by v& w V and w E W. Elements like vⓧw are called "pure tensors." In general, an tensor itself. In quantum mechanics, impure tensors correspond to entangled states Recall that if V, W, Z are vector spaces over R, a map f :Vx W -Z is said to be bilinear if: Problem. Show that the map π : V × W → VⓧW sending (v,w) to v u is bilinear. Hint: To see that π(v ay, w) πίν, w) CT(v,,w), show that (v+ cv, ,w) (v, w) - c(v', w) is in the span of the set S. The other equation is similar (you don't need to write out the proof again)

Answer 1

. T(vw) + c. Cr'@w gan ㅠ is bilinear