Let $H$, the parity check matrix of the Hamming code. $H\in Mat_{m\times 2^{m} - 1}$ where it's columns are all $\mathbb{F}^m \setminus \{o\}$. Prove that the rank of the code is $2^{m} -1 -m$.
let's denote the rank of the code as $k$. Then, $G$ is the generator matrix with the rank of $m+k\times k$. Since $H$ is a mapping from $\mathbb{F}^n$ to $\mathbb{F}^k$ and $G$ is the "opposite" mapping it must be that $m+k = 2^{m}-1 -m$ and therefore $k = 2^{m}-m-1$
Now, as far as I understand $G$ is injective and therefore it's kernel is trivial and so $k = 2^m -1 -m$ is indeed (by the rank-nullity theorem) the rank of the code.
Is that correct? I don't feel that certain about it
Basically you need to check that $H$ has full rank $m$.
The parity check matrix $H$ determines a mapping $\Bbb{F}^n\to\Bbb{F}^m$. If you know that $H$ has full rank, then that mapping is onto, and the code is the kernel which, by rank-nullity, has rank $n-m=2^m-1-m$.
Why has $H$ got full rank? You can surely convince yourself that a subset of columns of $H$ forms an $m\times m$ identity matrix.