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.


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.

  • $\begingroup$ This makes a lot of sense, thank you. Happy holidays! :) $\endgroup$ – Jens Kramer Dec 28 '18 at 20:46

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