Calculate the dimension of the vector subspace $U = \text{span}\left\{v_{1},v_{2},v_{3} \right\}$ 
Given are the vectors $v_{1}=\begin{pmatrix} 1\\  3\\  5
\end{pmatrix}, v_{2}=\begin{pmatrix} 4\\  5\\  6 \end{pmatrix},
v_{3}=\begin{pmatrix} 6\\  4\\  2 \end{pmatrix}$ from
  $\mathbb{R}^{3}$.
What's the dimension of the vector subspace $U =
\text{span}\left\{v_{1},v_{2},v_{3} \right\}$

I was looking for a task about dimensions and I'm glad I could find one. However, I'm not sure if I did it correctly. Also, if I did it right at all, I'd like to know if there is an easier / faster way to do it.
I wrote these vectors as a matrix:
$$\begin{pmatrix}
1 & 4 & 6\\ 
3 & 5 & 4\\ 
5 & 6 & 2
\end{pmatrix}$$
I transposed the matrix:
$$\begin{pmatrix}
1 & 3 & 5\\ 
4 & 5 & 6\\ 
6 & 4 & 2
\end{pmatrix}$$
Then I tried to create as many zero lines as possible by using Gauss. I ended up with (skipped the single steps to keep it short):
$$\begin{pmatrix}
-24 & -72 & -120\\ 
  0 &  56 &  112\\ 
  0 &   0 &    0
\end{pmatrix}$$
Transpose this matrix:
$$\begin{pmatrix}
 -24 &   0 & 0\\ 
 -72 &  56 & 0\\ 
-120 & 112 & 0
\end{pmatrix}$$
Thus the dimension of the vector subspace $U$ is $2$.

Is it correct (I'm pretty sure I did the Gauss part I skipped correctly)?
 A: It is correct. A (perhaps, faster) way to answer is to find the rank of the matrix, defined as the size of the largest square submatrix with a non-zero determinant. The determinant of the matrix is zero, and it has $2 \times 2$ submatrices with non-zero determinant. Then the rank is 2.
A: What you did is a correct way to do it. By definition, rank of a matrix is dimension of row/column space (it is a theorem that these are equal), i.e. for matrix $(v_1\ v_2\ \ldots\ v_n)$, where $\{v_1,\ldots,v_n\}$ are column vectors, $\operatorname{rank}(v_1\ v_2\ \ldots\ v_n) = \dim\operatorname{span}\{v_1,\ldots,v_n\}$.
However, there is no need to transpose matrix back and forth. Rank of matrix is invariant under Gaussian elimination of either rows or columns. In practice, it is enough to produce upper triangular matrix using Gaussian elimination and count non-zero rows. Thus, you could have done something like this:
$$\begin{pmatrix}
1 & 4 & 6\\
3 & 5 & 4\\
5 & 6 & 2
\end{pmatrix}\sim \begin{pmatrix}
1 & 4 & 6\\
0 & -7 & -14\\
0 & -14 & -28
\end{pmatrix}\sim \begin{pmatrix}
1 & 4 & 6\\
0 & -7 & -14\\
0 & 0 & 0
\end{pmatrix}$$
