Defining Bilinear, Multilinear Maps (Tensor Prod., maybe)

[Bacle]

Hi, Everyone:

We know that for a fin.dim V.Space , given a basis {v1,..,vn}, then a linear map
T is uniquely defined/specified once we know the values t(v1),T(v2),..,T(vn).

Now, let's consider a bilinear map on VxW (with W not nec. different from V),
both fin. dim. V.spaces over the same field F, with respective bases
{v1,..,vn} and {w1,..,wm}.

I am trying to see what info re the basis vectors of V,W to uniquely determine
a bilinear map defined on VxW. ( we turn VxW into a V.Space over F in the standard
way: basis is {(v1,0),..,(vn,0), (0,w1),..,(0,wm)} , addition is done pairwise, etc.

I know that defining a bilinear map B on the basis alone is not enough to determine
B. What else do we need? I think this has to see with the tensor product V(x)W.

Thanks.

 

[Fredrik]

T is determined by its action on basis vectors because $Tx=T(x_iv_i)=x_iTv_i$. B is determined by its action on basis vectors because $B(x,y)=x_iy_jB(v_i,w_j)$.

 

[Bacle]

But the pairs (vi,wj) are not basis vectors of VxW; the basis vectors of VxW
are :{(v1,0),..,(vn,0),(0,w1),...,(0,wm)}, so B is determined by its action on
the pairs (vi,wj), not on basis vectors for VxW.

 

[Bacle]

But now the goal is, I think, to show that every bilinear map in VxW can be
represented by a linear map L in V(x)W, by:

L(v(x)w)= B(v,w)

And then I think we need to show that L is actually linear on V(x)W .

 

[Landau]

But the pairs (vi,wj) are not basis vectors of VxW; the basis vectors of VxW
are :{(v1,0),..,(vn,0),(0,w1),...,(0,wm)}, so B is determined by its action on
the pairs (vi,wj), not on basis vectors for VxW.

But

$$B(v_i,w_j)=B((v_i,0)+(0,w_j))=B(v_i,0)+B(0,w_j)$$

so B ís uniquely determined by its action on basis vectors.

 

[mathwonk]

my lucid explanation of tensor products and bilinear maps is in my notes at:

http://www.math.uga.edu/~roy/and a google search will certainly find better sources. but i hope you like mine.