How many columns of this matrix are linearly independent? \begin{matrix}
    1 & 2 & 0 & 1 \\
    2 & 4 & 1 & 4 \\
    3 & 6 & 3 & 9 \\
    \end{matrix}
I have tried to transpose it and then reduce it by row echelon form and i get zeros on the last two rows. But i can't grasp if i should be doing that or doing it another way.

If you look at your matrix 
$$ A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 &4 \\ 3 & 6 & 3 & 9 \end{bmatrix} $$
The rank can be no more than $3$. Do the following
Subtract $2 R_{1} $ from $R_{2}$
$$ R_{2} - 2 R_{1} =  \begin{bmatrix} 2 & 4 & 1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 4 & 0 & 2\end{bmatrix} = \begin{bmatrix} 0 & 0& 1  & 2\end{bmatrix} $$
$$  A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 &2 \\ 3 & 6 & 3 & 9 \end{bmatrix} $$ 
Now we need to zero the third row. So subtract $3 R_{1} $ from $R_{3}$ 
$$ R_{3} - 3R_{1} =  \begin{bmatrix} 3 & 6 & 3 & 9 \end{bmatrix} - \begin{bmatrix} 3 & 6 & 0 & 3\end{bmatrix} = \begin{bmatrix} 0 & 0& 3  & 6\end{bmatrix} $$
$$  A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 &2 \\ 0 & 0 & 3 & 6 \end{bmatrix} $$
Now we need to zero that $3$ in row $3$ . So we subtract $3R_{2}$ from $R_{3}$ 
$$ R_{3} - 3R_{2} =  \begin{bmatrix} 0 & 0 & 3 & 6 \end{bmatrix} - \begin{bmatrix} 0 & 0 & 3 & 6\end{bmatrix} = \begin{bmatrix} 0 & 0& 0  & 0\end{bmatrix} $$
You're left with this matrix
$$  A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 &2 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$
There are only $2$ non-zero rows and they aren't multiples of eachother. The rank is $2$
If you want to look at column-wise then $2C_{1} = C_{2}$ and $C_{4} = C_{1}+2C_{3}$.
