I know these are parity check matrixes of linear codes $C_1$ and $C_2$.
$H_1= \begin{matrix} 1 & 0 & 1 & 0 & 1\\ 0 & 1 & 0 & 1 & 1\\ 1 & 1 & 0 & 0 & 1\\ \end{matrix}$
$H_2= \begin{matrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 1 & 1 & 0 & 1 & 0\\ 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 1\\ \end{matrix}$
I need to find the number of code words belonging to each of the codes $C_1, C_2$. I know that I can solve $Hx=0$ and get all words that way. But I have seen examples were this is done by computing the dimension as $k = cols - rows$ and thereby the words are $2^k$ and saying that this can be done because these parity check matrixes rows are obviously linearly independent.
How can I easily verify that this is the case? I suspect there is a neat trick involved, and not much actual computation.
Many times the $(n-k)\times n$ parity check matrix $H$ of a $[n,k]$ code is written in the form $$H = [P_{(n-k)\times(k)} \mid I_{(n-k)\times (n-k)} ]$$ which makes the independence of the rows obvious. But for your matrices, you can try finding the reduced-echelon matrix with the same row space. For example, what happens if you add the first and second rows of $H_1$ into the third row?