Quote:
Originally Posted by omert  1. does a linear transformations (T:R3 to R3) exists that follwing these terms:
T(1,0,1)=T(2,1,2)=T(0,1,1)=T(2,3,3)
if there is give an example for such transformation, if not explain why? |
Look at the set of vectors {(1,0,1), (2,1,2), (0,1,1), (2,3,3)}. The first three of these vectors are linearly independent. Thus they form a basis for R^3, and you can express the vectors (1,0,0), (0,1,0), (0,0,1) in terms of them. Then use the linearity of T to find T(1,0,0), T(0,1,0) and T(0,0,1) in terms of the vector v=T(1,0,1). Finally, check whether v=T(2,3,3) is equal to 2*T(1,0,0) + 3*T(0,1,0) + 3*T(0,0,1). If so (and assuming that v is not the zero vector) then you have constructed a nonzero T with the required properties.
Quote:
Originally Posted by omert 2. Let V,W be vector spaces over the same field, both from a finite dimension.
Let U sub space of V and dimU>=dimv-dimw
prove that there is a transformation T:V to W, that sustains kerT = U. |
Let
be a basis for U (where k=dim(U)), and extend it to a basis
of V. The condition on dim(U) tells you that dim(W)≥n–k, so you can define T to be the linear transformation that takes each of
to 0, and takes the vectors
to a linearly independent set in W. You can then check that the kernel of T is exactly U.