TITLE: Linear Transformation on $\mathbb{R}^6$
QUESTION [5 upvotes]: Let $W$ be a vector space over $\mathbb R$ and let $T:\mathbb R^6 \to W$ be a linear transformation such that $S = \{Te_2, Te_4, Te_6\}$ spans $W$. Wich one of the following must be true?

(A) $S$ is a basis of $W$
(B) $T(\mathbb R^6) \ne W$
(C) $\{Te_1, Te_3, Te_5\}$ spans $W$
(D) $\ker T$ contains more than one element


I'm having trouble starting this problem.
Here are my findings so far:
I tried Dimension theorem and found that $\ker(T)$ contains more than three elements so (D) is incorrect.
And taking $\dim(T(\mathbb R^6))=\dim(W)$ we get $T(\mathbb R^6)=W$ so (B) is incorrect.
I think (C) option is correct but still not able to prove it.

REPLY [3 votes]: Such a problem is most effectively tackled by looking at each point and asking oneself How could this become wrong?. Let me get you started:
Since $W$ is spanned by a set of three vectors, it has dimension at most $3$. Let's thus assume $W = \mathbb R^d$ with $d\in\{0,1,2,3\}$ for the tasks (this will actually be enough).
For (A) to go wrong, $S$ must be linearly dependent. This is the case if our $d$ is smaller than $3$, wich is certainly possible. A concrete example for $d = 2$ would be
$$T = \pmatrix{1&2&3&0&0&0\\0&0&0&1&2&3}\\
S = \{Te_2, Te_4, Te_6\} = \{(2,0)^T, (0,1)^T, (0,3)^T\}$$
Is there a way to make (B) wrong?
and so on...
I will be happy to provide further help given that you actually show some work. Consider this a "kick-off" ;)