The dimension of the intersection of two vector subspaces

Question

We’re given: $$ V = \operatorname{span}\left\{ \begin{bmatrix} 2 \\ 2 \\ 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 5 \\ 4 \\ 1 \\ 1 \end{bmatrix} \right\}$$ and $$W = \operatorname{span} \left\{\begin{bmatrix}1\\-3\\2\\1 \end{bmatrix} \right\}.$$

And we’re asked to find $\dim(V\cap W^\bot)$.

Here’s my approach. First, by inspecting the basis of $W$, I managed to construct a basis for $W^\bot$, which is the following: $$W^\bot = \operatorname{span} \left\{\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\right\}.$$ Then, for each vector $\textbf w$ in the basis of $W^\bot$, I tried to see if the system $A\textbf x=\textbf w$ was compatible or not. In this case, I found out that the system was only compatible with two of the vectors in $W^\bot$, thus indicating me that $\dim(V \cap W^\bot) = 2$ (which is correct).

What I do not get, however, is that since the dimension of the intersection is $2$, why aren’t two of the vectors in the basis of $V$ orthogonal to the vector which spans $W$? That was my initial approach, i.e., try to see which vectors of the basis of $V$ are orthogonal to the vector that spans $W$.

Also, I was wondering if there was a simpler/quicker way of doing this.

Answer 1

At the first find the $\mathrm{dim}(V\cup W^{\bot})$.

And we know that $\mathrm{dim}(V\cup W^{\bot})=\mathrm{dim}(V)+\mathrm{dim}(W^{\bot})-\mathrm{dim}(V\cap W^{\bot})$

Finding the dimension of $(V\cup W^{\bot})$ is also an easy work.

For above one form a matrix $A$ like following one:

$$\begin{bmatrix} 2 \\ 2 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \\ 1 \\ 0 \end{bmatrix}\begin{bmatrix} 5 \\ 4 \\ 1 \\ 1 \end{bmatrix}\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}\begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \end{bmatrix}\begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$

Now find the rank of this matrix via using Gaussian Elimination process.

Answer 2

Imagine an anologous situation in three dimensions, which is easier to visualize: Let $V$ be a plane through the origin and $W$ a nonzero vector that’s not normal to the plane. The orthogonal complement of $W$ is another plane that intersects $V$ in a line that’s orthogonal to $W$. However, by construction there are elements of $V$ that aren’t orthogonal to $W$, and we can find a pair of them that span $V$.

As to a simpler method of solving this, first verify that the generating vectors of $V$ are linearly independent (they are). You’re looking for elements of $V$ that are orthogonal to $W$, which can be expressed by the equation $$[1,-3,2,1](a[2,2,2,1]^T+b[2,1,1,0]^T+c[5,4,1,1]^T)=0.$$ Expanding and simplifying this equation produces $a+b-4c=0$. This has two free variables, so the intersection is two-dimensional.