Inner product of linearly independent rows of $m \times n$ matrix with $n$ linearly independent vectors gives $m$ independent vectors This question is a supplementary problem of Schaum's outlines Linear Algebra. 
The question says:

Suppose $(a_{11},..,a_{1n}), \dots \dots , (a_{m1},..,a_{mn}) $ are
  linearly independent vectors in $K^n$, and suppose $v_1, v_2, \dots,
 v_n$ are linearly independent vectors in a vector space $V$ over $K$.
  Show that the vectors 
$w_1 = a_{11}v_1 + \dots +a_{1n}v_n, \dots , w_m = a_{m1}v_1 + \dots +a_{mn}v_n$
are also linearly independent.

Being a beginner, I am unable to understand how to start the proof. I tried using the definition of independence of vectors and i know a bit about rowspace but I really don't know what works here. 
 A: To test if $w_1,\ldots,w_m$ are linearly independent, we set a linear combination equal to zero, and set to show that the coefficients have to be zero. So we assume
$$
0=\sum_{j=1}^mb_jw_j.
$$
Then
$$
0=\sum_{j=1}^mb_jw_j=\sum_{j=1}^mb_j\sum_{k=1}^na_{jk}v_k=\sum_{j=1}^m\sum_{k=1}^nb_ja_{jk}v_k=\sum_{k=1}^n\left(\sum_{j=1}^mb_ja_{jk}\right)v_k.
$$
Because of the linear independence of $v_1,\ldots,v_n$, we get that 
$$
\sum_{j=1}^mb_ja_{jk}=0,\ \ \ k=1,\ldots,n.
$$
These are precisely the $n$ entries of 
$$
\sum_{j=1}^mb_j(a_{j1},\ldots,a_{jn}),
$$
so
$$
\sum_{j=1}^mb_j(a_{j1},\ldots,a_{jn})=0.
$$
Now the linear independence of the vectors $(a_{j1},\ldots,a_{jn})$ gives you that $b_1=\cdots=b_m=0$. So $w_1,\ldots,w_m$ are linearly independent. 
A: Let $$\sum_{i=1}^m\lambda_iw_i=0$$
with $\lambda_i\in K $. You want to prove that all $\lambda_i$ must be $0$.
Expand:$$\sum_{i=1}^m\lambda_i\sum_{j=1}^na_{ij}v_j=0$$
Rewrite:$$\sum_{j=1}^n\left(\sum_{i=1}^m\lambda_ia_{ij}\right)v_j=0$$
Now use the fact that the $v_j$ are independent and then the fact that the $(a_{ij})$ are independent.
