I'm currently studying for my Linear Algebra exams coming up in January, thus going through some previous exam questions and I'm at the following.
Let $T:V_1\rightarrow V_2$ denote a linear transformation between two complex vectorspaces with, $\dim V_1=3,\: \dim V_2=4$. Furthermore there are two subspaces $U_1\subseteq V_1$ and $U_2\subseteq V_2$, both of dimension $1$, such that $T(U_1)=U_2$.
- A. Determine the possible values of $\dim null\: T$
- B. Determine the possible values of $\dim ran \: T$
- C. Determine whether $T$ is always injective, always non-injective, or both options can occur.
- D. Determine whether $T$ is always surjective, always non-surjective, or both options can occur.
- E. Define $S:U_1\rightarrow U_2$ by $Su=Tu$ for all $u\in U_1$. Determine whether $S$ is injective, surjective, invertible or not; or if different combinations can occur.
I've tried to get through the questions and my answer are.
A. and B. By the theorem $\dim V=\dim ran \: T + \dim null \: T$, I should be able to answer both A and B by answering just one of them knowing $\dim V_1=3$.
I choose to find $\dim ran \: T$.
Knowing that there are some subspaces of $V_1$ that are mapped to a subspace of $V_2$ with a dimension $>0$, gives that $\dim ran \: T>0$.
By the stated theorem, $\dim ran \: T$ can be either $1,2,3$, thus $\dim null \: T$ can be $0,1,2$.
C. The linear map is injective iff. $ null \: T=\{0\}$, so it can be both.
D. Mapping from a dimension to a higher dimension rejects the possibility of surjectivity. So it is never surjective.
E. I'm really not sure about this answer, but knowing that both subspaces $U_1,U_2$ are of the same dimension, I guess it makes the linear map an operator, which means that it is either surjective and injective thus invertible OR none of the statements. From the assumptions, $Su=Tu$ which implies $T=S$ or that they are isometric atleast? I think this means that $u=u$, thus making it injective, thus surjective and invertible?
I'm really not sure about my answers, and I hope i got some parts of it right. I hoping to get some tips or corrections. Thank you in advance!
Best regards
Jens.
Best Answer
Parts A-D look good, but I think you misunderstood part E.
Part E is saying "let's take $T$ and restrict its domain to $U_1$." In other words, $S(u)=T(u)$ for all $u \in U_1$, but is undefined for any $u \in V$ outside of $U_1$.
Now, we are given that $T(U_1)=U_2$, so since $S(u)=T(u)$ for all $u \in U_1$, this means that $S(U_1)=U_2$, as well. Thus, $S$ is surjective.
Furthermore, this means $S$ has image $U_2$, which has dimension 1, so the rank of $S$ is also $1$. By rank-nullity theorem:
$$\text{rank } S+\text{nullity }S=\text{dim }U_1\rightarrow 1+\text{nullity }S=1\rightarrow\text{nullity }S=0$$
Thus, since $S$ has nullity of $0$, it is also injective.
In short, $S$ is both surjective, because $S(U_1)=U_2$, and injective, because it has a nullity of $0$. Therefore, it is a bijection and thus invertible.