1. ## Linear transformations

I'm stuck with the problem:
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?

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.

2. Originally Posted by omert
I'm stuck with the problem:
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?
Think about the linear transformation T(v)= (0,0,0) for all v.

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.

Given any v in V, there exist vectors u, with u in U, and x such that v= u+ x with u in U. (choose a basis for both U, add vectors to make a basis for V and write v in terms of that basis). Define T(v)= x.

3. 1. the transformation can't be the zero transformation, I need to find another example for these terms.

2. can you elaborate, I don't understand how from this you get to KerT = U?

4. 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.

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 $\{e_1,\ldots,e_k\}$ be a basis for U (where k=dim(U)), and extend it to a basis $\{e_1,\ldots,e_k,e_{k+1},\ldots,e_n\}$ 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 $e_1,\ldots,e_k$ to 0, and takes the vectors $e_{k+1},\ldots,e_n$ to a linearly independent set in W. You can then check that the kernel of T is exactly U.

5. 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.
How do I find the T(1,0,0), T(0,1,0) and T(0,0,1) vectors in terms of only T(1,0,1)???

6. Originally Posted by omert
How do I find the T(1,0,0), T(0,1,0) and T(0,0,1) vectors in terms of only T(1,0,1)???
You have to express (1,0,0), and the other two basis vectors, as linear combinations of (1,0,1), (2,1,2) and (0,1,1). If for example you know that (1,0,0) = –(1,0,1) + (2,1,2) – (0,1,1) then (since T is a linear transformation) T(1,0,0) = –T(1,0,1) + T(2,1,2) – T(0,1,1) = –v+v–v = –v.