Is C is basis of vector space V If $\{e_1, e_2, e_3\}$ are basis of vector space $V$ over $\Bbb{R}$ then is
$C=\{ 2e_1, 3e_{1}+e_{3}, 6e_{1}+3e_{2}+e_{3} \} $ a basis of $V$?
Can anyone help ? 
$C$ can be reduce to trivial basis given by some column operation so I think it must be basis .
But my textbook says its not!!
The equation
\begin{align}
x_1(2e_1)+x_2(3e_1+e_3)+x_3(6e_1+3e_2+e_3)=0
\end{align}
is equivalent to
\begin{align}
(2x_1+3x_2+6x_3)e_1+3x_3e_2+(x_2+x_3)e_3=0
\end{align}
Now $e_1,e_2,e_3$forms a basis, thus we have
\begin{align}
2x_1+3x_3+6x_3&amp;=0\\
3x_3&amp;=0\\
x_2+x_3&amp;=0
\end{align}
This implies the family $C$ is linear independent, hence a basis.
Hint : 


You can write the matrix of passage from $\{e_1, e_2, e_3\}$ to C. Your problem is then: is that matrix invertible ?
I think you're right. Can you show us exactly how the result is given in your textbook ?