Show that block matrix has full row rank ii and only if $A$ has full row rank. I've read a text, and they say the following lemma,

Lemma: For given matrices $A,B$ , the matrix
$$
\begin{pmatrix}
 A & B\\
 0^T & v^T\\
\end{pmatrix}
$$
has full row rank for all nonzero $v$ iif $A$ has full row rank.

where the matrix, not needed to be squared. can give me som hint, thanks!
 A: If $A$ does not have full row rank, then there exists a nonzero $u$ such that $u^\top A = 0$. Try to come up with a nonzero vector $v$ and scalar $c$ such that
$$\begin{pmatrix} u^\top & c \end{pmatrix} \begin{pmatrix} A & B \\ 0^\top & v^\top \end{pmatrix} = 0.$$
(It will depend on what $u^\top B$ is.)

If there exists nonzero $v$ such that $M := \begin{pmatrix} A & B \\ 0^\top & v^\top \end{pmatrix}$ does not have full row rank, then there exists a vector $u$ and scalar $c$ such that $\begin{pmatrix}u \\ c \end{pmatrix}$ is nonzero and
$$\begin{pmatrix} u^\top & c \end{pmatrix} \begin{pmatrix} A & B \\ 0^\top & v^\top \end{pmatrix} = 0.$$
Show that the above conditions imply that $u$ is nonzero and that $u^\top A = 0$.
A: call your overall matrix $C$ which has $m$ rows  .
1.) $A \text{ is full row rank }\implies C \text{ is full row rank}$
since $A$ has full row rank, it is surjective and has a right inverse.  Thus there is some (block) matrix $W$ such that
$CW = \begin{pmatrix}
 A & B\\
 \mathbf 0^T & \mathbf v^T\\
\end{pmatrix}W = \begin{pmatrix}
 I_{m-1} & *\\
 \mathbf 0^T & \mathbf v^T\\
\end{pmatrix}$
$m = \text{rank}\big(CW\big)\leq \text{rank}\big(C\big)\leq m$
2.) $C \text{ is full row rank} \implies A\text{ is full row rank } $
$C$ is sujective so it has a right inverse $W'$
$CW' = I_m$
This implies that for all non-zero $\mathbf x':=\begin{pmatrix}\mathbf x \\  0 \end{pmatrix}$
$(\mathbf x')^TC \neq\mathbf 0^T\implies \mathbf x^TA\neq \mathbf 0^T$
and hence all rows of $A$ are linearly independent.
