Finite field, basis

Question

In $\mathbb{F}_3^3$, I am given:$$U = \text{span}\left(\begin{pmatrix}0\\1\\2\end{pmatrix}, \begin{pmatrix}1\\1\\1\end{pmatrix}\right),\quad W = \text{span}\left(\begin{pmatrix}-1\\0\\3\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}\right)$$ I have to find a basis for $U\cap W$. Do I have to rewrite the vectors that in the spans? Could I say that in $\mathbb{F}_3^3$, $\begin{pmatrix} 0\\1\\2 \end{pmatrix}=\begin{pmatrix} 0\\1\\0 \end{pmatrix}$? And how do I notate that correctly?

Answer 1

Vectors in U are, in general, $a*{(0,1,2)}+b*{(1,1,1)}$ and W are $c*{(-1,0,3)} + d*{(0,1,0)}$. To find $U∩W$ you need to find vectors that allow both of the above. Meaning, for example for the first row $0*a+1*b = -1*c+0*d$. finding all parameters that allow all equations to be true are $U∩W$. Another method is to find all solutions $v$ for $Av=0$ where A is the matrix with all vectors from both U and W(it is essentially the same explanation as before). In finite fields the first might be easier than the second due to $mod3$ calculations.

Answer 2

Let's assume $v$ lies in U∩W

and hence $v$ lies in U and $v$ lies in W

Hence $v$ = $a1$ * [0,1,2] +$a2$ * [1,1,1]

also $v$= $b1$ * [-1,0,3] +$b2$ * [0,1,0]

now equate $v$ and solve for $a1$,$a2$,$b1$,$b2$

$a1$ * [0,1,2] +$a2$ * [1,1,1] = $b1$ * [-1,0,3] +$b2$ * [0,1,0]

$a2$=-$b1$

$a1$+$a2$= $b2$

2$a1$+$a2$=3$b1$

On solving in terms of $b1$ you will get

$a1$=2$b1$

$a2$=-$b1$

$b2$=$b1$

replacing in your initial equation of $v$ you will get

$v$ = $a1$ * [0,1,2] +$a2$ * [1,1,1]

implies,

$v$ = 2$b1$ * [0,1,2] -$b1$ * [1,1,1]

taking $b1$ as common

$v$ = $v$*(2 * [0,1,2] - [1,1,1])

$v$ = $b1$*([-1,1,3])

Hence $v$=span([-1,1,3])

Hope this helps