# How to find the intersection of $W$ and $Z$? [duplicate]

Subspaces$$W$$ and $$Z$$ of $$\mathbb R^4$$ are generated by $$\{(1,1,0,-1),(1,2,3,0),(2,3,3,-1)\}$$ and $$\{(1,1,0,-1),(1,2,3,4),(0,1,3,5)\}$$, respesctively. Find a basis for $$W\capZ$$.

I already know how to find the basis for $$W+Z$$, but I am confused on how to find the basis of $$W\capZ$$.

• What is the word “matrix” doing in the title? – José Carlos Santos Oct 14 '18 at 13:47
• – Chinnapparaj R Oct 14 '18 at 14:12
• It might be expeditious to work out $\dim(W\cap Z)$ from $\dim(W),\dim(Z),$ and $\dim(W+Z)$, if you already have a basis for the latter. Note that the "generated by" vectors (spanning sets) for $W,Z$ may turn out not to be resp. bases for them (linearly independent as well as spanning)! – hardmath Oct 14 '18 at 15:39

Here are some steps you can take to solve this problem. First, find bases of the orthogonal complements $$W^\perp$$ and $$Z^\perp$$ (of course, with respect to the usual nondegenerate bilinear form of $$\mathbb{R}^4$$). Then, we have $$(W\cap Z)^\perp = W^\perp + Z^\perp.$$ That is, $$W\cap Z=(W^\perp+Z^\perp)^\perp.$$
Now, to find a basis of $$W^\perp$$, write down the matrix $$w=\begin{pmatrix}1&1&0&-1\\1&2&3&0\\2&3&3&-1\end{pmatrix}$$ whose rows are the given vectors of $$W$$ that span $$W$$. Solve for the (right) null space of $$w$$. Similarly, write down the matrix $$z=\begin{pmatrix}1&1&0&-1\\1&2&3&4\\0&1&3&5\end{pmatrix}$$ whose rows are the given vectors of $$Z$$ that span $$Z$$. Solve for the null space of $$z$$.
If you did the job properly, you should see that $$\operatorname{null}(w)$$ is spanned by $$(3,-3,1,0)$$ and $$(2,-1,0,1)$$, while $$\operatorname{null}(z)$$ is spanned by $$(3,-3,1,0)$$ and $$(6,-5,0,1)$$. That is, $$V=W^\perp +Z^\perp$$ is spanned by $$(3,-3,1,0)$$, $$(2,-1,0,1)$$, $$(3,-3,1,0)$$, and $$(6,-5,0,1)$$ (the repetition of $$(3,-3,1,0)$$ can be removed). We now try to find $$W\cap Z=V^\perp$$.
Define the matrix $$v$$ by stacking up the known spanning elements of $$V$$: $$v=\begin{pmatrix}3&-3&1&0\\2&-1&0&1\\3&-3&1&0\\6&-5&0&1\end{pmatrix}.$$ Determine the null space of $$v$$, and the work is now yours.