Column space in linear algebra Looking to get some help with the following problem. Given matrix $A$ and vector $c$. Is vector $c$ in the coumn space of $A$? And if so why? 
Matrix $A$
$\begin{bmatrix}
1 & 2  \\
3 & 4  \\
5 & 6  \\
\end{bmatrix}$
vector c:
$\begin{bmatrix}
1 \\ 
2 \\ 
3 \\
\end{bmatrix}$
 A: Hint:
$$\frac{1}{2}\begin{pmatrix} 2 \\ 4 \\ 6\end{pmatrix}=\begin{pmatrix} 1 \\ 2 \\ 3\end{pmatrix}$$
A: Calculation
You can create an augmented matrix with the columns being the three vectors: 
$$\begin{bmatrix}1&2&1\\3&4&2\\5&6&3\end{bmatrix}$$ and then row reduce, getting
$$\begin{bmatrix}1&0&0\\0&1&\frac{1}{2}\\0&0&0\end{bmatrix}$$
From there, we see $0$ and $\frac{1}{2}$ are the numbers to multiply the vector by; double checking, $$0\begin{bmatrix}1\\3\\5\end{bmatrix}+\frac{1}{2}\begin{bmatrix}2\\4\\6\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$$
$$\frac{1}{2}\begin{bmatrix}2\\4\\6\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$$
Yes, this is correct. So yes, vector $c$ is in the column space of $A$.
Theory
Remember here that we just want to check if there is some scalar that we can multiply by each of the column vectors (remember, these are the individual columns of the matrix $A$) that when these new vectors are added up, it equals the vector we are wondering about; i.e.:
$$x_1(v_1)+x_2(v_2)+...+x_n(v_n)=c$$
This is the general equation, where you are trying to figure out if there is a coefficient that will give $c$. In this case, there is, and so $c$ is in the column space of $A$. 
Further, you can remember that column space is the span of the linear combinations of the column vectors which again can be represented as the equation above, where $x_1$ through $x_n$ can be any number. 
Hope this helps!
A: Column space of $A$ is just the set generated by the linear combinations of columns of $A$, which is equivalent to asking if there is some $a, b \in \mathbb{R}$ such that
$$a\begin{bmatrix}1 \\ 3 \\ 5\end{bmatrix} + b\begin{bmatrix}2 \\ 4 \\ 6\end{bmatrix} = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$$
Notice that if we choose $a = 0, b = \frac{1}{2}$ we get?
