Find basis of intersection of 2 spans with unequal dimensions

Question

I've been stuck on this question for quite a while:

Given $U =$ span $\left\{ \begin{pmatrix} 0\\ 2\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 2\\ 1\\ 3\\ 7 \end{pmatrix} \right\} $ and $W =$ span $\left\{ \begin{pmatrix} 1\\ 0\\ 3\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 1\\ -3\\ 7 \end{pmatrix} \right\} $, find a basis for $U \cap W$.

Now, I've tried writing general $u \in U, w \in W$ and then setting $u=w$ and finding a general solution using Gaussian elimination. However, because the linear equation system $u=w$ has 5 variables and not 4, my attempts were unsuccessful. I also tried looking for similar questions on this site, however every question I found had dim$U$=dim$W$ which is not the case here. Any help will be appreciated, thanks in advance!

Answer 1

I would solve this problem by finding a homogenous system of linear equations describing $W$, and the same for $U$. Then $U \cap W$ would be described by the homogenous system of linear equations consisting of both the equations describing $W$ and $U$. Then you could go about and solve this system of linear equations, thus obtaining a basis for the desired vector space.

By converting the span to a homogenous system of linear equations I mean doing the following:

Let $u \in U$, denote $u = \begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}$ then of course $u \in U$ if and only if the system $$\begin{pmatrix}0\;\;\;\;1\;\;\;\;2\\2\;\;\;\;0\;\;\;\;1\\0\;\;\;\;0\;\;\;\;3\\0\;\;\;\;0\;\;\;\;7\end{pmatrix} \begin{pmatrix}\lambda_1\\\lambda_2\\\lambda_3\\\end{pmatrix}=\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}$$

Has a solution.

This logic will lead you towards getting a homogenous system a linear equation describing $U$.

Hope this helps!

Answer 2

The most comfortable situation when you want to compute an intersection of two subspaces is when one of them is given by linear equation(s) and the other one by parametrization, and when there are the least possible equations and parameters.

Here, the equation of the hyperplane $U$ is $$\det\begin{pmatrix}0&1&2&x\\2&0&1&y\\0&0&3&z\\0&0&7&t\end{pmatrix}=0,$$i.e. $$7z-3t=0.$$ So, an arbitrary vector of $W,$ $$\begin{pmatrix} a\\ b\\ 3a-3b\\ 7b \end{pmatrix},$$ also belongs to $U$ iff $$7(3a-3b)-3(7b)=0,$$ i.e. $a=2b.$ This proves that $U\cap W$ is the set of vectors of the form $$\begin{pmatrix} 2b\\ b\\ 3(2b)-3b\\ 7b \end{pmatrix}.$$ A basis for this line is made of the single vector $$\begin{pmatrix} 2\\ 1\\3\\ 7 \end{pmatrix}.$$ If you chose to flip the roles of the two subspaces, you would have 2 equations for $W$ and 3 parameters for $U,$ which would be less tractable.