Dimension of nullspace I can understand the dimension of column space of matrix is the no. of independent column vectors
But why is the dimension of nullspace = no. of free variables?


For an $m \times n$ matrix, $A$, the Rank-Nullity theorem says that:
  $$ \text{column rank}(A) + \text{nullity}(A) = n.$$
  where $\text{nullity}(A)$ is the dimension of the null space of $A$.

When you find the reduced row echelon form of a matrix, the max number of independent columns (i.e. the column rank) is the number of pivot columns (columns containing a leading one for some row). Notice now that free variables correspond to the columns without pivots. So the number of free variables is $n - \text{column rank}(A)$.
So 
\begin{align*}
\text{nullity}(A) &= n - \text{column rank}(A) \\
&= \text{number of free variables}.
\end{align*}
