Given $u_1,u_2,u_3,u_4$ independent vectors in $R^5$. $v_1,v_2,v_3$ in $R^5$ too. Given $u_1,u_2,u_3,u_4$ linearly independent vectors in $R^5$. $v_1,v_2,v_3$ $\in R^5$ .
They are defined like that:
$v_1 = u_1 + u_2 - u_4$
$v_2 = au_1 - u_3 + u_4$
$v_3 = u_2 + au_3 - 6u_4$
How can I find for which $a$ , vectors $v_1,v_2,v_3$ will be linearly dependent?
 A: We show that $v_1,v_2,v_3\in \mathbb{R}^5$ are always linearly independent. First suppose $a=0$. Now let $\lambda_1,\lambda_2,\lambda_3\in \mathbb{R}$ such that $\lambda_1v_1+\lambda_2v_2+\lambda_3v_3=0$. Then $$\lambda_1u_1+2(\lambda_1+\lambda_3)u_2-\lambda_2u_3-6(\lambda_1+\lambda_2+\lambda_3)u_4=0  \implies \lambda_1=\lambda_2=\lambda_3=0$$
Now suppose $a\not=0$. We express the $v_i$ as column vectors now and perform elementary matrix operations on 
\begin{bmatrix}
1 & a & 0 \\
1 & 0 & 1 \\
0 & -1 & a \\
-1 & 1 & -6 \\
0 & 0 & 0
\end{bmatrix}
to obtain
\begin{bmatrix} 
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & 1-\frac{1}{a^2} \\
0 & 0 & -1 -\frac{1}{a}+\frac{6}{a^2} \\
0 & 0 & 0 
\end{bmatrix}
Note that since $a=1,-1$ both do not yield the zero vector in the third column, we can conclude that $v_1,v_2,v_3\in \mathbb{R}^5$ are always linearly independent 
A: It is immaterial that the vectors are in $\mathbb{R}^5$ or any other vector space. The vectors $v_1$, $v_2$ and $v_3$ belong to the span of $\{u_1,u_2,u_3,u_4\}$ of which the set is a basis.
A set of vectors is linearly independent if and only if their coordinate vectors with respect to a basis are linearly independent. Thus consider the matrix having as columns the coordinate vectors with respect to the given basis, that is
$$
\begin{bmatrix}
1 & a & 0 \\
2 & 0 & 1 \\
0 & -1 & a \\
-1 & 1 & -6 \\
\end{bmatrix}
$$
Let's try Gaussian elimination:
\begin{align}
\begin{bmatrix}
1 & a & 0 \\
2 & 0 & 1 \\
0 & -1 & a \\
-1 & 1 & -6 \\
\end{bmatrix}
&\to
\begin{bmatrix}
1 & a & 0 \\
0 & -2a & 1 \\
0 & -1 & a \\
0 & 1+a & -6 \\
\end{bmatrix}
&&\begin{aligned}R_2&\gets R_2-2R_1\\ R_4&\gets R_4+R_1\end{aligned}
\\[6px]
&\to
\begin{bmatrix}
1 & a & 0 \\
0 & 1 & -a \\
0 & -2a & 1 \\
0 & 1+a & -6 \\
\end{bmatrix}
&&\begin{aligned}R_2&\leftrightarrow R_3\\ R_2&\gets -R_2\end{aligned}
\\[6px]
&\to
\begin{bmatrix}
1 & a & 0 \\
0 & 1 & -a \\
0 & 0 & 1-2a^2 \\
0 & 0 & -6+a+a^2 \\
\end{bmatrix}
&&\begin{aligned}R_3&\gets R_3+2aR_2\\ R_4&\gets R_4-(1+a)R_2\end{aligned}
\end{align}
For no value of $a$ we have both $1-2a^2=0$ and $a^2+a-6=0$, so the rank of this matrix is $3$ and therefore the vectors $v_1$, $v_2$ and $v_3$ are linearly independent for any value of $a$.
A: $v_1, v_2, v_3$ will be linearly dependent if there are $\alpha_1, \alpha_2, \alpha_3$ not all $0$ such that $\alpha_1v_1 + \alpha_2v_2 + \alpha_3v_3 = 0$
For your $v_1, v_2$ and $v_3$ we have:
$0 = \alpha_1v_1 + \alpha_2v_2 + \alpha_3v_3 = \alpha_1(u_1 + u_2 - u_4) + \alpha_2(au_1 - u_3 + u_4) + \alpha_3(u_2 + au_3 - 6u_4) = u_1(\alpha_1 + a\alpha_2) + u_2(\alpha_1 + \alpha_3) + u_3(-\alpha_2 + a\alpha_3) + u_4(-\alpha_1 + \alpha_2 -6\alpha_3)$
As you have that $u_1, u_2, u_3, u_4$ are linearly independent, you have that
\begin{cases}
\alpha_1 + a\alpha_2 = 0 \\
\alpha_1 + \alpha_3 = 0\\
-\alpha_2 + a\alpha_3 = 0 \\
-\alpha_1 + \alpha_2 -6\alpha_3 = 0
\end{cases}
$\implies$
\begin{cases}
a = \frac{-\alpha_1}{\alpha_2} \\
\alpha_1 = -\alpha_3\\
a = \frac{\alpha_2}{\alpha_3} \\
-\alpha_1 + \alpha_2 -6\alpha_3 = 0
\end{cases}
