# Depressing exam question on linearly independent set $C$ and the proof behind it

Let $V$ and $W$ be vector spaces of field $F$.

$T : V \to W$ is a surjective linear transformation. Let $D$ be a linearly independent subset of $W$, for each $\mathbf{d}$ in $D$, fix a vector $\mathbf{c}_d$ in $V$ such that $T(\mathbf{c}_d) = \mathbf{d}$ and denote $$C = \{\mathbf{c}_d~|~ \mathbf{d} \in D\}$$

a) Show $C$ is linearly independent subset of $V$.

Due to insufficient time, i think i penned down a wrong answer, i think i will get some partial marks because i believe the idea is there. Hopefully someone can help me correct the proof.

My proof, I define a "restriction map" as such $$T|_{V_{1}} : V_1 = \text{span}(C) \to T(\text{span}(C)) = \text{span}(D)$$

In a hurry, i did not have time to think and i claimed that this map is surjective by construction (of course $T$ is surjective too) and i claim that this map is injective. And hence $D = T(C)$ being linearly independent will imply $D$ being linearly independent by isomorphism between maps. After i got home, i think the injective claim is a bit too fast. Anyone can help me finish it?

Let $c_1,...,c_n \in C$, $f_1,...,f_n \in F$ and $0=f_1c_1+...+f_nc_n$.

For $j \in \{1,...,n\}$ there is $d_j \in D$ with $c_j=c_{d_j}$

Then $0=T(f_1c_1+...+f_nc_n)=f_1T(c_1)+...+f_nT(c_n)=f_1d_1+...f_nd_n$.

Since $D$ is linearly independent, we get $f_1=...=f_n=0.$

• oh shuts that's an easier solution i did not think of, that said, is my solution somewhat correct ? – nan Dec 6 '17 at 12:51
• In general $T|_{V_{1}}$ is not injective. – Fred Dec 6 '17 at 12:54
• @ilovewt I think your solution just changed the given statement to be proved into an equivalent one. Proving that $T$ is injective is pretty much the same thing as proving $C$ is linearly independent. – Matthew Leingang Dec 6 '17 at 12:54
• @Fred yes that i know, however in the way i defined it, it is slightly different from the normal restriction map – nan Dec 6 '17 at 12:55
• @MatthewLeingang I see, is the new map i defined really injective? – nan Dec 6 '17 at 12:57