Why are the rows of a parity-check matrix linearly independent? I've been reading about error-correcting codes, and I came across the following definition for a parity-check matrix: 
''There is an $(n-k) \times  n$ matrix $H$, called a parity check matrix for an $[n,k]$ linear code $\mathcal{C}$, defined by
$$\mathcal{C} = \{x \in \mathbb{F}^{n}_{q} | Hx^{T} = 0 \}.'' $$
Then, the book states that the rows of $H$ will also be linearly independent, but I am having trouble seeing why this is true. 
As of now, I've noticed that the map $x \mapsto Hx^{T}$ is a linear transformation from $\mathbb{F}^{n}_{q}$ to $\mathbb{F}^{n-k}_{q}$ with $\mathcal{C}$ being the kernel of this linear transformation, but I can't make the connection to linear independence. 
Can anyone help?

For $(n,k,d_{min})$ channel codes there is a parity check matrix $H$ such that a given code is  valid if and only if $Hx^T=\overline{0}$ where $\overline{0}$ is a vector of zeros.
The mapping from information bits to the code bits is bijective which you can have when $H$ is of rank ($n-k$). If $H$ is not of rank $n-k$ then I also think that one can not get $d_{min}=3$, for example for (7,4,3) Hamming codes.
In other words when we think $Hx^T=y$ as a linear system of equations in $GF^2$. For all $x^T$ The solution exists and one to one from $x$ to $y$ if $H$ is of full rank (rank=$n-k$).
