0

Let $F=\mathbb Z_3, V=F^4$.

Let $U=sp\{(1,0,0,0),(1,0,1,0),(0,1,1,1) \} \\W=sp\{(0,0,1,0),(-1,1,0,1),(1,1,1,1) \}$

Find $dim (U\cap W)$

we have $v\in U \text{ and } v\in W$ so $v=v$ therefore: $au_1+bu_2+cu_3=xw_1+yw_2+zw_3$ and after some algebra trying to find $a,b,c$ I get: $a=z-x\\ b=x-y\\c=y+z$

So $U\cap W = \{(a,b,c)|x(-1,1,0)+y(0,-1,1)+z(1,0,1) \}=sp\{(-1,1,0),(0,-1,1)(1,0,1)\}$

But this is in $F^3$! Why did I lose a dimension in this process?

Community
  • 1
shinzou
  • 3,981

1 Answers1

1

Your last step isn't correct; $U\cap W$ consists of all vectors which can be written in the form $(a+b,c,b+c,c)$ and also in the form $(-y+z,y+z,x+z,y+z)$.

As pointed out by ajotatxe, $U=W=\{(r,s,t,s): r, s, t \in F_3\}$, since

$\hspace{.3 in}(a+b,c,b+c,c)=(r,s,t,s)$ for $a=r-t+s,\;b=t-s,\; c=s$ and

$\hspace{.3 in}(-y+z,y+z,x+z,y+z)=(r,s,t,s)$ for $x=t+r+s, \;y=2s+r, \;z=2r+2s$;

and it follows that $\dim(U\cap W)=\dim U=3$.

user84413
  • 27,211
  • In this approach, shouldn't we find $a,b,c$ or $x,y,z$? – shinzou Jun 24 '15 at 18:30
  • I mean $(a+b,c,b+c,c)$ is just a linear combination of $U$, it doesn't mean that this is the intersection... – shinzou Jun 24 '15 at 18:39
  • @kuhaku You don't need to find $a,b,c$ in terms of $x,y,z$, although you're correct that a vector must be able to be expressed in both forms to be in the intersection. I will add some more information to my answer. – user84413 Jun 24 '15 at 20:07