Rank of a partitioned matrix 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} &amp; \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.
Write, less confusingly, $A=(A_1~A_2)$, where $A_1$ is a $q\times p$ matrix and $A_2$ a $q\times(n-p)$ matrix, so that in block form
$$
  C=\begin{pmatrix}I&amp;0\\A_1&amp;A_2\end{pmatrix}.
$$
Then row operations will nullify $A_1$, so $\def\rk{\operatorname{rk}}\rk C=\rk I+\rk A_2=p+\rk A_2$, and this is maximal (i.e., equal to the number $n$ of columns of $C$) if and only if $\rk A_2$ is maximal (i.e., equal to the number $n-p$ of columns of $A_2$), in other words if and only if the columns of $A_2$ are linearly independent.
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.