# 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.
• You mean I should get something like $b_1(a_{11}, a_{12},....,a_{1n}) + .... + b_m(a_{m1}, a_{m2},....,a_{mn}) = 0$ and then use independence of the vectors? I thought of this but the problem is getting this equation. Jun 5, 2021 at 8:36
• That equation is equivalent to what you have. One way to see this is how I mentioned, another is to simplify the equation you just wrote. For example, $b_1(a_{11},\dots,a_{1n})+b_2(a_{21},\dots,a_{2n})=(b_1a_{11},\dots,b_1a_{1n})+(b_2a_{21},\dots,b_2a_{2n})=(b_1a_{11}+b_2a_{21},\dots,b_1a_{1n}+b_2a_{2n})$. Jun 5, 2021 at 8:40
• If you write this in general, you will have a vector with each component being one of the sums $\sum b_k a_{kj}$ Jun 5, 2021 at 8:44