Inferring from having a basis of one vector space, and a linear map Let $T:V \to W$ be a linear map and let $\{v_i\}_{i=1}^7$ be a basis of $V$. Which of the following claims are true?
A. If $\{T(v_1),T(v_2),T(v_3),T(v_4)\}$ is a basis for $Image(T)$ then $\{v_5,v_6,v_7\}$ is a basis of $kernel(T)$.
B. If $\{v_1,v_2\}$ span $kernel(T)$ then $\{T(v_3),T(v_4),T(v_5),T(v_6),T(v_7)\}$ is a basis of $Image(T)$.
C. If $\{T(v_1),T(v_2),T(v_3)\}$ span $Image(T)$ then $dim(ker(T))=4$.
I think B is true. The fact that the first two vectors span the kernel, implies that they form a basis of the kernel (because they are independent). Which implies the image must be of dimension 5. My guess is that if they weren't independent, we would be able to find a linear combination of them that is in the kernel, which will contradict the independence of the first seven vectors.
As for A  - is it possible that $T(v_5)$ will be a linear combination of $\{T(v_1),T(v_2),T(v_3),T(v_4)\}$?
As for C - is it possible that for example $T(v_1)=T(v_2)$? so that the $dim(ImageT))<3$?
Thank you!
 A: B is true. It is clear that the list $Tv_{3}, \ldots, Tv_{7}$ spans $\text{im }T$. To prove linear independence, suppose that $\alpha_{3}, \ldots, \alpha_{7}$ are scalars such that
$$ \sum_{i=3}^{7} \alpha_{i} Tv_{i} = 0 .$$
Then
$$ T\left( \sum_{i=3}^{7} \alpha_{i} v_{i} \right) = 0, $$
so $ \sum_{i=3}^{7} \alpha_{i} v_{i} \in \ker T$. Since $v_{1}, v_{2}$ span $\ker T$, there exist scalars $\alpha_{1}$ and $\alpha_{2}$ such that
$$ \sum_{i=3}^{7} \alpha_{i} v_{i} = \alpha_{1}v_{1} + \alpha_{2}v_{2},$$
or, equivalently,
$$-\alpha_{1}v_{1} - \alpha_{2}v_{2} +\sum_{i=3}^{7} \alpha_{i} v_{i} = 0. $$
Using the fact that the $v_{i}$'s are linearly independent, we deduce that all the coefficients in this linear combination are zero. In particular, $\alpha_{3} = \ldots = \alpha_{7} = 0$, which is what we needed to prove.
Edit. A much shorter proof following the observations made in the original post could go as follows:
As noted, $\dim \ker T = 2$, so we must have $\dim \text{im }T = 5$. The list of five vectors $Tv_{3}, \ldots, Tv_{7}$ spans $\text{im }T$, so we can conclude that it forms a basis for $\text{im }T$.
