# Linear map over complex vector spaces

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.

## 1 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.

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