Is this proof sufficient for this problem (linear independence of vectors) Problem:
If $v_1, v_2, v_3$ are linearly independent then prove that the vectors $2v_1 + 3v_2 + v_3, v_1 - v_2 - v_3, 3v_1 + 5v_2 - v_3$ are also linearly independent.
My solution:
Since $v_1,v_2,v_3$ are lin. ind. we know that $c_1v_1 + c_2v_2 + c_3v_3=0$ only if $c_1,c_2,c_3=0$
We suppose that
$c_4(2v_1 + 3v_2 + v_3) + c_5(v_1 - v_2 - v_3) + c_6(3v_1 + 5v_2 - v_3) = 0$
So
$v_1(2c_4 + c_5 + 3c_6) + v_2(3c_4 - c_5 + 5c_6) + v_3(c_4 - c_5 - c_6) = 0$
We set from the first definition that
$c1 = 2c_4 + c_5 + 3c_6, c_2 = 3c_4 - c_5 + 5c_6, c_3 = c_4 - c_5 - c_6$
and so we see that the vectors in question are lin. ind.
Is this a sufficient proof for this problem? I saw a solution in another problem that involved finding a determinant at the end and i cant wrap my head why
 A: A)
As you say, from equations:
$$c_1v_1 + c_2v_2 + c_3v_3=0 \textit{ only if } c_1,c_2,c_3=0$$
and
$$v_1(2c_4 + c_5 + 3c_6) + v_2(3c_4 - c_5 + 5c_6) + v_3(c_4 - c_5 - c_6) = 0$$
we reach:
$$2c_4 + c_5 + 3c_6 = 3c_4 - c_5 + 5c_6 = c_4 - c_5 - c_6 = 0$$
and solving the system we found $c_4 = c_5 = c_6 = 0$, finishing the proof.
B)
Note the system:
$$2c_4 + c_5 + 3c_6 = 3c_4 - c_5 + 5c_6 = c_4 - c_5 - c_6 = 0$$
can be expressed in matrix form as:
$$\left(\begin{matrix}
2 & 1 & 3\\
3 & -1 & 5\\
1 & -1 & -1\\
\end{matrix}\right) 
\left(\begin{matrix}
c_4\\
c_5\\
c_6
\end{matrix}\right) = \left(\begin{matrix}
0\\
0\\
0
\end{matrix}\right)$$
multiplying by inverse (assuming it exists):
$$\left(\begin{matrix}
c_4\\
c_5\\
c_6
\end{matrix}\right) =\left(\begin{matrix}
2 & 1 & 3\\
3 & -1 & 5\\
1 & -1 & -1\\
\end{matrix}\right) ^ {-1}
\left( \begin{matrix}
0\\
0\\
0
\end{matrix}\right)$$
you can calculate the inverse, multiply by (0,0,0) and find $c_4=c_5=c_6=0$. However, note that the values in the inverse are not important, always $c_4=c_5=c_6=0$ after multiply by (0,0,0). The only important thing is "EXIST the inverse?". If it exists, $c_4=c_5=c_6=0$, no matter the exact value of the inverse.
Thus, the problem now is answer "has this matrix an inverse?", that is "is this matrix invertible". And a matrix has inverse only if its determinant is non-zero.
In conclusion, just evaluate the determinant of this matrix. If it is non-zero, the matrix has inverse. If the matrix has inverse, $c_4=c_5=c_6=0$, finishing the proof.
A: As an alternative by

*

*$w_1=2v_1 + 3v_2 + v_3$

*$w_2=v_1 - v_2 - v_3$

*$w_3=3v_1 + 5v_2 - v_3$
we have

*

*$w_1-2w_2=5v_2+3v_3$

*$3w_2-w_3=-8v_2-2v_3$
and then

*

*$2(w_1-2w_2)+3(3w_2-w_3)=2w_1+7w_2-3w_3=-14v_2$

*$8(w_1-2w_2)+5(3w_2-w_3)=8w_1+13w_2-5w_3=14v_3$
and since $w_2=v_1 - v_2 - v_3\implies v_1=v_2+v_3+w_2$ we have

*

*$6w_1+20w_2-2w_3=14v_1$
therefore $w_1$, $w_2$, $w_3$ span the same space and therefore they are linerly independent.
A: Let $A = \begin{pmatrix}2 & 1 & 3 \\ 3 & -1 & 5 \\ 1 & -1 & -1\end{pmatrix}$ and let $P \colon \mathbb{R}^3 \to \text{span}(v_1, v_2, v_3)$ be the linear map given by $Pe_j = v_j$. The question can be phrased as are $PAe_1, PAe_2, PAe_3$ linearly independent? This happens if and only if $PA$ is invertible. Since $\{v_1, v_2, v_3\}$ are a basis of $\text{span}(v_1, v_2, v_3)$, $P$ is invertible, so $PA$ is invertible if and only if $A$ is invertible. Now you just have to compute whether $A$ has an inverse or not.
