Prove that vectors formed by linear combination of independent vectors are independent I am learning vector space for the first time from the book Schaum's outlines Linear Algebra. There I stumbled upon a question, which is as follows-

So, I started solving in this way-
If $w_1,....,w_m$ are independent, then the equation
$b_1w_1 + b_2w_2 + .... + b_mw_m = 0$
should have the solution $b_1 = b_2 = .... = b_m = 0$, where all $b_i$'s are scalers.
After substituting the values of $w_i$'s and grouping together by vectors $v_j$'s, I got an equation whose left side is a linear combination of vector $v_j$'s and coefficient of each $v_j$ is $\sum b_ka_{kj}$. Since the vectors $v_j$'s are independent, that means
$\sum b_ka_{kj} = 0$ for all j
From here I am not able to proceed on how to show that all $b_k$'s are 0.

$\textbf{Hint:}$ You have a system of equations which you can write as $Ab=0$ where $b=(b_1,\dots,b_m)^T$ and $A$ is the matrix whose $j^{th}$ column is $a_j=(a_{j1},\dots,a_{jn})^T$. Now apply the hypothesis you haven't yet used to get the conclusion.
