Linear independence of images by $A$ of vectors whose span trivially intersects $\ker(A)$ 
I know that any subset of V with more than k vectors is always linearly dependent, but how do I include matrix A into the vectors given to prove the statement? 
Also, for part (b), does the statement mean that Ax=0 only has the trivial solution since V intersects span{u1, u2, ..., uk}={0}?
Any help would be appreciated! Thank you!
 A: Part $(a)$, To prove that $\pmb{A}\pmb{u}_i$ for $i = 1 \ldots k$ are linearly dependent, it sufficies to stack them into one matrix $[\pmb{A}\pmb{u}_1 \ldots \pmb{A}\pmb{u}_k]$, which is a fat matrix since $k > m$ and hence it is rank deficient, i.e. the rank of this matrix is at most $m$. Therefore there exists at least $k - m$ vectors $\pmb{n}_j$ such that $[\pmb{A}\pmb{u}_1 \ldots \pmb{A}\pmb{u}_k]\pmb{n}_j = \pmb{0}$.
Part $(b)$, let $\alpha_j$ for $j = 1 \ldots k$ be arbitrary scalars, then
\begin{equation}
\begin{split}
&\rightarrow\sum\limits_{i=1}^k \alpha_i \pmb{A}\pmb{u}_i
=
\pmb{0}
\\&\rightarrow
\pmb{A}\sum\limits_{i=1}^k (\alpha_i \pmb{u}_i)
=
\pmb{0}
\\&\rightarrow
\pmb{A}\pmb{v}
=
\pmb{0}
\end{split}
\end{equation}
where $\pmb{v} \in \text{span}\lbrace \pmb{u}_1 \ldots \pmb{u}_k \rbrace$. But $\pmb{v} \in V$ since it nulls $\pmb{A}$, which leaves you with two cases:
-either $\text{span}\lbrace \pmb{u}_1 \ldots \pmb{u}_k \rbrace \cap V \neq \lbrace \pmb{0} \rbrace$ and $\pmb{u}_1 \ldots \pmb{u}_k $ are independent so that $\pmb{v} \neq \pmb{0}$.
-or  $\text{span}\lbrace \pmb{u}_1 \ldots \pmb{u}_k \rbrace \cap V =\lbrace \pmb{0} \rbrace$ and $\pmb{u}_1 \ldots \pmb{u}_k $ must be linearly dependent so that $\pmb{v} = \pmb{0}$.(i.e. the trivial solution as you state).
A: (a) If $u \in \mathbb R^n$ then you have $Au \in \mathbb R^m$ (by rules of matrix multiplication). So you get $Au_1, Au_2, \dots, Au_k \in \mathbb R^m$ and if $k > m$ they have to be dependent because you got more vectors than dimensions in $\mathbb R^m$.
(b) Suppose $\sum_{i=1}^k b_i Au_i = 0$. We have to show that $b_i = 0$ for all $i$. It holds $$0 = \sum_{i=1}^k b_i Au_i = \sum_{i=1}^k b_i \sum_{j=1}^n a_{i,j} u_i^{(j)} = \sum_{j=1}^n a_{i,j} \sum_{i=1}^k b_i u_i^{(j)}.$$
Set $x^{(j)} = \sum_{i=1}^k b_i u_i^{(j)}$ then we have $x \in \mathbb R^n$ and the equation above can be written as $Ax = 0$. Since $x \in span\{u_1, u_2, \dots, u_k\}$ we get $x = 0$ and since $u_1, u_2, \dots, u_k$ are linearly independent it follows that $b_i = 0$ for all $i$.
(c) No. (just compare (b))
I hope that it helps you :)
