1. ## Linear Transformation problem

Let $\displaystyle T:R^2 \rightarrow R^3$ be defined by $\displaystyle T(x,y)=(x+y,0,2x-y)$.
Find bases for N(T) and R(T), compute nullity and rank of T, verify the dimension theorem. Determine if T is one-to-one or onto.

Proof.

T is linear. By a theorem, R(T) = span{T(b)} for a basis b of $\displaystyle R^2$

Let b = {(1,0),(0,1)}, then span{T(b)} = span {(1,0,2),(1,0,-1)} = R(T), so I know that rank of T is 2, then nullity of T must be 0.

But now, I'm working this backward, the problem stated that I should find the bases for N(T) and R(T) first, well, I found the bases for R(T), and I use the theorem to find that N(T) = {0}. So do I need to use another way to prove that N(T) = 0? I meant, it is pretty straight forward that there is no point in $\displaystyle R^2$ that would make T zero.

Now, N(T) = {0}, so T must be one-to-one. But now, how do I know if T is onto or not?

Thank you very much!

2. By definition, a function is onto if the range is the same as the co-domain. So in this case, can the function take every possible value in $\displaystyle R^3$?

3. No, I don't believe so, since dim(R(T)) is 2 and the co-domain is 3.

Now, if the rank (T) is equals to the dimension of the co-domain, can I say it is onto? (Probably not, right?)

4. No, I don't believe so, since dim(R(T)) is 2 and the co-domain is 3.
Excellent. Exactly what I would have done.

Now, if the rank (T) is equals to the dimension of the co-domain, can I say it is onto? (Probably not, right?)
Yes, you can. The only subspace of a vector space V with the same dimension as the vector space is V.