Given two subspaces $U$ and $W$ in $\mathbb{R^n}$ then is it right to say that:
$\dim(U \cap W) = \dim(\operatorname{nullSpace}(U \cap W))$, because $U\cap W$ is equal to number of vectors which are both in $U$ and $W$ and if we apply row operations then all linearly dependent vectors will be cancelled?
and
$\dim(U+W) = n -\operatorname{Nullspace}(U+W)$ ?
There are already a few answers but all of them make use of this theorem:
$\dim(U+W) + \dim(U \cap W) = \dim (U) + \dim(W)$.
which is too long to compute if we want only $\dim(U+W)$ or $\dim(U \cap W)$
Example:
$U = (1,1,1)$ and $W= {(1,-1,2),(3,1,0)}$ then we can make a matrix
$\lambda_1(1,1,1)-\lambda_2(1,-1,2)-\lambda_3(3,1,0)$
and we get
$U\cap W = \begin{bmatrix}1 &&1 && 3 \\ 1 && 1 && -1 \\ 1 && -2 && 0 \end{bmatrix}$