Rank of a partitioned matrix

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Let $p$, $q$ and $n$ be positive integers such that $p+q \geq n$ and $p \leq n$. Denote by $\mathbf{I}$ the identity matrix of size $p$, by $\mathbf{0}$ the zero matrix of size $(p\times (n-p))$ and by $\mathbf{A}=(a_{ij})_{i=1,\ldots,q,j=1,\ldots,n}$ a matrix over $\mathbb{R}$.

Can one formulate simple conditions on $\mathbf{A}$, which guarantee that the partitioned matrix

\begin{equation} \mathbf{C}:= \left( \begin{array}{cc} \mathbf{I} & \mathbf{0}\\ \mathbf{A} \end{array} \right) \end{equation}

has full rank (i.e. $\text{rank}(\mathbf{C}) =n$)?

CLARIFICATION: Note that $\mathbf{A}$ has $n$ columns, so the matrix $\mathbf{C}$ does not necessarily have any columns that are all zero.

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By the fact that these are equivalent conditions;

1) Rank$(C)=n$

2) The $n$ column vectors of C are linearly independent,

you can not achieve your goal.