Let $F= \mathbb{Z_5}$ and $H$ be the following matrix:
\begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ \end{bmatrix}
a) Determinate one base of the space $C \subset F^4$ of the solutions $x \in F^4$ with $Hx^t=0$
b) Verify that $C$ is it's own orthogonal complement.
I have done this so far:
$Hx^t=0 \Leftrightarrow \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ w \\ \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix}$
$\Leftrightarrow \begin{cases} x+2y=0 \\ z+2w=0 \end{cases} \Leftrightarrow \begin{cases} x=-2y \\ z=-2w \end{cases}$
So I can write as $(-2y,y,-2w,w)$.
So, I can have the following $<(0,0,-2,1),(-2,1,0,0),(-2,1,-2,1)>$
Did I calculated $C$ or $C^\bot?$ I don't understand it.
Also, to verify should I use the fact that $dim(C)+dim(C^\bot)=n$?
EDIT:
So, $C=<(0,0,-2,1),(-2,1,0,0),(-2,1,-2,1)>=<(0,0,-2,1),(-2,1,0,0)>$. Now to calculate the orthogonal complement of $C$ I do $\begin{cases} (0,0,-2,1) \cdot(x,y,z,w) \\ (-2,1,0,0) \cdot (x,y,z,w)\\ \end{cases} \Leftrightarrow \begin{cases} -2z+w=0 \\ -2x+y=0\\ \end{cases} \Leftrightarrow \begin{cases} w=2z \\ y=2x\\ \end{cases} \Leftrightarrow (x,2x,z,2z)$
Therefore we have $C^\bot=<(1,2,0,0),(0,0,1,2)> \neq C=<(0,0,-2,1),(-2,1,0,0)>$
I got to $C \neq C^\bot$ , and they should be the same. What am I doing wrong?
The position of the zero entries clearly show that second row is not a scalar multiple of the first, hence the rank of the matrix is $2$. Having 4 columns, the nullity is $4-2=2$, by the rank-nullity theorem. In your computation you show 3 vectors as basis, which should be corrected. The first two are linearly independent and hence form a basis (for what?). Multiplying this vectors with $H$ we see that $Hx^t=0$ (your notation), so they form a base for the solution space.