Proving vector is if the vectors w, v2,...,vn is linealy independent then we must have a1 not equal to 0 Image
Let $v_1, . . . ,v_n$ be  linearly  independent  vectors  in $R_m$ and
let $w=a_1v_1+. . .+a_nv_n$, with real numbers $a_1, . . . , a_n$, be a linear combination of these vectors.  Prove the following statements:
(a)  If  the  vectors $w,v_2, . . . ,v_n$ are  linearly  independent,  then  we must have $a_1$ not equal to $0$.
Hint: What happens if $a_1= 0$?
I'm practicing about linearly independent vector, this is the first time I see this problem.
I have thought of a way, so for question a, if $w, v_2,..v_n$ is linearly independent, is it mean $w = 0$ and $a_2,...a_n$ also is $0$?
Can you guys help me solve question a, I appreciate all helps. Thanks a lot
 A: Take the set of vectors $v_{1} ... v_{n}$. A set of vectors is linearly dependent if the equation Ax = 0 does not have only the trivial solution x = 0 (the matrix A is formed with the columns $v_{1} ... v_{n}$). Now say $v_{1}$ = 0. If this is true, Ax = 0 can be written as a linear combination of $v_{1} ... v_{n}$ with not every vector weighted by zero. This looks like: $av_{1} + 0v_{2} + ... + 0v_{n} = 0$, where a is nonzero -- this set is linearly dependent.
In your case, if $a_{1}v_{1}$ = 0, then w is made up of $0 + v_{2} + ... + v_{n}$.
If A is a matrix with columns $w, v_{2}, ... v_{n}$, for the equation Ax = 0 you can write $w - (v_{2} + ... + v_{n}) = 0$ -- this is a nontrivial solution to Ax = 0, because $v_{2} + ... + v_{n}$ are weighted by -1. Since a nontrivial solution to Ax = 0 exists, the vectors that form the columns of A are linearly dependent. For the vectors to be linearly independent, $a_{1}$ therefore must be nonzero.
A: Look at an equation
$\lambda w + \sum_{i=2}^n \mu_i v_i = 0$
for some $\lambda, \mu_i\in \mathbb{R}$ for all $i=2,\dots,n$. Writing out $w$ as linear combination of the $v_i$, we get
$0=\lambda a_1 v_1 + \dots + \lambda a_n v_n + \sum_{i=2}^n \mu_i v_i = \lambda a_1 v_1 + \sum_{i=2}^n (\lambda a_i + \mu_i)v_i $.
Since all the $v_i$ are linearly independent, we must have $\lambda a_1 = \lambda a_i + \mu_i = 0$ for all $i=2,\dots,n$. If $a_1= 0$, we can choose $\lambda = 1$ and $\mu_i = -a_i$, and hence the first equation above has a solution with not all coefficients zero. Therefore, $w,v_1,\dots,v_n$ cannot be linearly independent.
