quick way to check Linear Independence 
The answer is $V_{3}=2V_{1}-3V_{2}, V_{5}=2V_{1}-2V_{2}-V_{4}$, I understand the coefficients in the dependencies determined by the non-pivot columns, and it looks like the the position of non-pivot column decides dependent vector, so I can easily know the form should be $V_{3}=..., V_{5}=...$, but how can I decide the independent vector on the right hand side? let's take $ V_{5}$ for example, I wish it could be $V_{5}=2V_{1}-2V_{2}-V_{3}$ but it is not.
 A: Let's say you have a matrix $A \in M_{n \times m}(\mathbb{F})$ written as
$$ A = \begin{pmatrix} \mathbf{v}_1 | \mathbf{v}_2 | \dots | \mathbf{v}_m \end{pmatrix} $$
where each $\mathbf{v}_i$ is a column vector in $\mathbb{F}^n$. Any linear relation
$$ a_1 \mathbf{v}_1 + \dots + a_m \mathbf{v}_m = \mathbf{0} $$
between the vectors $\mathbf{v}_i$ can be written as
$$ a_1 \mathbf{v}_1 + \dots + a_m \mathbf{v}_m = A \begin{pmatrix} a_1 \\ \dots \\ a_m \end{pmatrix} = \mathbf{0}. $$
Performing row reduction on $A$ results in replacing $A$ with 
$$PA = \begin{pmatrix} \mathbf{v}'_1 | \mathbf{v}'_2 | \dots | \mathbf{v}'_m \end{pmatrix} $$
where $P \in M_{n \times n}(\mathbb{F})$ is invertible. Since $P$ is invertible, we have
$$ PA \begin{pmatrix} a_1 \\ \dots \\ a_m \end{pmatrix} = \vec{0} \iff A \begin{pmatrix} a_1 \\ \dots \\ a_m \end{pmatrix} = \vec{0}. $$
This means that any linear relation satisfied by the columns $\mathbf{v}_i$ is satisfied by the columns $\mathbf{v}'_i$ and vice versa. When $PA$ is in row-reduced form, it is easy to read the linear relations satisfied by the columns of $PA$.
In your case, you can see that 
$$\mathbf{v}'_3 = \begin{pmatrix} 2 \\ -3 \\ 0 \\ 0 \\ 0 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} - 3 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} =  2\mathbf{v}'_1 - 3\mathbf{v}'_2$$ 
which implies that $\mathbf{v}_3 = 2\mathbf{v}_1 - 3\mathbf{v}_2$. Similarly for $\mathbf{v}'_5$.
