$z$ is a vector in $R^n$. Given the following functions:
$c_i(z): R^n\to R$
$c(z) = (c_1(z),\dots,c_m(z))^T$
$A(z)^T = [\nabla c_i (z)]_{i\in [m]}$, where $[m] = \{1,2,\dots,m\}$
And $A(z)^T$ has full column rank.
We construct a matrix $Z$, whose columns are a basis of $Nul\ A(z)$.
Please prove $$ \left[\begin{array}{c}{A\left(z\right)} \\\\{Z^{T}}\end{array}\right] $$ is invertible (nonsingular).
Let me know if you have any puzzles about the problem itself.
For any subspace $V$ of $\Bbb R^n$, you have a decomposition $$\Bbb R^n=V\oplus V^\perp$$ In the present case, consider $V$ to be the row space of $A(z)$. Then by definition $V^\perp=\operatorname{Ker}A(z)$.
By construction, the rows of $A(z)$ are a basis of $V$ and the rows of $Z^T$ are a basis of $V^\perp$. Since $\Bbb R^n=V\oplus V^\perp$, it follows that the rows of the full matrix form a basis of $\Bbb R^n$, so that the matrix is non-singular.