How to Prove matrix $m \times m$ is invertible is equivalent to be $LI$ over $\mathbb{Z_{2}}$?

I came up with one problem that says that a $m \times m$ matrix is invertible is the same as to say that its rows are LI (linearly independent) over $\mathbb{Z_{2}}$.

First of all, I'd like to know how to prove it, in order to prove the following:

Supposing $$z_{m+i} = \sum_{j = 0}^{m-1} c_jz_{i+j} \text{ mod 2}$$

where $(z_1, z_2, ..., z_m)$ comprises the initialization vector. For $i \geq 1$, we define:

$$v_i = (z_i, ..., z_{i+m-1})$$, observing that $v_1,... ,v_m$ are the rows of the $m \times m$ matrix.

It asks to prove with these above informations:

For any $i \geq 1$, $$v_{m+1} = \sum_{j = 0}^{m-1} c_jv_{i+j} \text{ mod 2}$$

I've calculated $v_1 = (0, c_0z_1, ..., ), v_2 = (0, c_0z_2, c_0z_2 +c_1z_3, ...)$, however I don't see a way to know how to prove that for $\alpha_i \in \mathbb{R^*}$ that:

$$\alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3 + ... + \alpha_m v_m = 0$$