Question on Subspace and Standard Basis. Show that $e_1, e_2, e_3$ are not in a subspace $V$ of $R^3$ that is spanned by \begin{bmatrix}1\\3\\5\end{bmatrix} and \begin{bmatrix}2\\4\\6\end{bmatrix}
I know you would create an augmented matrix with $e_1, e_2, e_3$ and the two vectors and reduce, but I'm not sure where to go from there. Thanks
 A: If $ V $ is spanned by $ (1,3,5) $ and $ (2,4,6) $, then any vector $ \boldsymbol{v} \in V $ can be written as a linear combination $ c_1 (1,3,5) + c_2 (2,4,6) $. 
If $ e_1 = (1,0,0) $ is in $ V $, then there exists some $ c_1 $ and $ c_2 $ such that $ c_1 (1,3,5) + c_2 (2,4,6) = (1,0,0) $. Similarly for $ e_2 $ and $ e_3 $. Now all we need to do is construct an augmented matrix, reduce it and show that there are no solutions. 
$$
\left[
\begin{array}[cc|ccc]
11 & 2 & 1 & 0 & 0 \\
3 & 4 & 0 & 1 & 0 \\
5 & 6 & 0 & 0 & 1 
\end{array}
\right]
\rightarrow
\left[
\begin{array}[cc|ccc]
11 & 0 & 0 & -3 & 2 \\
0 & 1 & 0 & 5/2 & -3/2 \\
0 & 0 & 1 & -2 & 1 
\end{array}
\right]
$$
which is inconsistent. 
A: Personally, I wouldn't use an "augmented matrix" or matrices at all. If $e_1= \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$ is in this subspace then there exist numbers, a and b, such that $\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= a\begin{bmatrix}1 \\ 3 \\ 5 \end{bmatrix}+ b\begin{bmatrix}2\\ 4 \\ 6\end{bmatrix}= \begin{bmatrix}a+ 2b \\ 3a+ 4b \\5a+ 6b\end{bmatrix}$.
That is, do there exist numbers, a and b, that satisfy  a+ 2b= 1, 4a+ b= 3, and 5a+ 6b= 5. Start by solving two of these equations for a and b.  For example if we multiply the first equation by 4 we get 4a+ 8b= 4.  Subtracting 4a+ b= 3 from that, 7b= 1 so b= 1/7.  Then a+ 2/7= 1 so a= 5/7.  
Now, do a= 5/7 and b= 1/7 satisfy the third equation, 5a+ 6b= 5?
