Find the dimension of a subspace and the orthogonality complement of another

Question

I would like to the fastest way to find $\dim(V \cap W^\bot)$ where $V$ and $W$ are two subspaces of $\Re^4$. $V$ is defined as the span of three vectors and $W$ the span of one vector.For example as in here - $$ V = \left\langle\begin{pmatrix}2\\2\\2\\1\end{pmatrix}, \begin{pmatrix}5\\4\\1\\1\end{pmatrix}, \begin{pmatrix}2\\1\\1\\0\end{pmatrix}\right\rangle,\quad W = \left\langle\begin{pmatrix}1\\-3\\2\\1\end{pmatrix}\right\rangle $$

Answer 1

Consider the only vector $u_1$ which spans $W$.You can use gram-schmidt process to continue this vector and make an orthonormal basis $\{u_1,u_2,u_3,u_4\}$ for $\mathbb R^4$. The set $\{u_2,u_3,u_4\}$ is a basis for $W^T$. Now we know that $dim(V\cup W^\bot)=dim(V)+dim(W^\bot)-dim(V\cap W^\bot )$. If $V=span\{v_1,v_2,v_3\}$ then $(V\cup W^\bot )=span\{u_2,u_3,u_4,v_1,v_2,v_3\}$ and here you can easily find $dim(V\cup W^\bot)$.

If $u_1=0$, it means zero vector, then $(V\cup W^\bot)=V$ because $W^\bot=\mathbb R^4$