Determine whether or not a set is linearly independent Prove or give a counter example:
if $v_1, ..., v_n$ is linearly independent, is $5v_1-4v_2, v_2,...v_m$ also linearly independent.
I'm not sure how to go about this. I tried a couple ways to prove that this is true by writing out the lineal combination of the set equal to zero and solving for $v_1$ or $v_2$ and plugging back in but that was getting messy so I figured this is probably linearly dependent then? I just can't figure this one out, any help?
 A: Let $a_1\dots a_n$ be numbers such that
$$
a_1(5v_1-4v_2) + \sum_{i=2}^n a_iv_i=0.
$$
Then
$$
(5a_1)v_1 + (a_2-4a_1)v_2 + \sum_{i=3}^n a_iv_i=0.
$$
By linear independence of $v_1\dots v_n$ it follows that the coefficients in the above equation are zero:
$$
5a_1 = a_2-4a_1 = a_3 = \dots = a_n =0,
$$
which implies that all the $a_i$ are zero. Hence, the vectors under question are linearly independent.
A: Let $a_1, \ldots, a_n$ be such that $$0=a_1(5v_1-4v_2)+\sum_{i=2}^n a_iv_i= 5a_1v_1+(a_2-4a_1)v_2 +\sum_{i=3}^n a_iv_i,$$
Since $v_1,\ldots,v_n$ are linearly independent we must have $$5a_1= (a_2-4a_1)=a_3=\ldots =a_n=0,$$ in particular $a_1 = 0$ and thus $a_2 = 4a_1 = 0$. This shows that $5v_1-4v_2,v_2,v_3,\ldots,v_n$ are linearly independent.
A: Another way of looking at it: Since $v_1,\ldots,v_n$ are linearly independent, we know $$\mathrm{rank}\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix}=n.$$
Applying row operations to a matrix does not affect its rank, so
\begin{align*}
\mathrm{rank}\begin{bmatrix} 5v_1-4v_2 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} &=\mathrm{rank}\begin{bmatrix} 5v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} & R_1 \gets R_1+4R_2 \\
& =\mathrm{rank}\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} & R_1 \gets \tfrac{1}{5} R_1 \\
& =n.
\end{align*}
So $5v_1-4v_2,v_2,\ldots,v_n$ are also linearly independent.
