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.


1 Answer 1


$\textbf{Hint:}$ 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.

  • $\begingroup$ 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. $\endgroup$
    – Ankit Seth
    Jun 5, 2021 at 8:36
  • $\begingroup$ 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, $b_1(a_{11},\dots,a_{1n})+b_2(a_{21},\dots,a_{2n})=(b_1a_{11},\dots,b_1a_{1n})+(b_2a_{21},\dots,b_2a_{2n})=(b_1a_{11}+b_2a_{21},\dots,b_1a_{1n}+b_2a_{2n})$. $\endgroup$
    – Stuck
    Jun 5, 2021 at 8:40
  • $\begingroup$ If you write this in general, you will have a vector with each component being one of the sums $\sum b_k a_{kj}$ $\endgroup$
    – Stuck
    Jun 5, 2021 at 8:44

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