(Generic) rank of an augmented matrix Let $A\in\mathbb{R}^{n\times m}$ be a full column rank matrix and $D_1$, $D_2\in\mathbb{R}^{n\times n}$ be diagonal matrices. Consider the augmented matrix
$$
\bar A=\left[D_1A \ \ D_2A\right]\in\mathbb{R}^{n\times 2m}
$$
If $m\le n/2$ and $D_1$, $D_2$ take generic (nonzero) diagonal values, it is expected that $\bar A$ will still have full column rank.
However, is there a (simple) way to formalize this intuition? (For instance, in some kind of probabilistic/measure-theoretic way)
Thanks in advance for any help/suggestion.
 A: Here's an algebraic sort of approach. Fix your matrix $A$. Let $x_1,\dots,x_n$ denote the diagonal entries of $D_1$, and $y_1,\dots,y_n$ the diagonal entries of $D_2$.
Your augmented matrix will have full column-rank if and only if the determinant
$$
\det(\bar A^T \bar A) = p(x_1,\dots,x_n,y_1,\dots,y_n)
$$
is non-zero. Notably, this is the zero-set of a (multivariate) polynomial. It follows that one of the two following situations must occur: either $p = 0$ for all $x_1,\dots,x_n,y_1,\dots,y_n$, or the set of solutions has measure $0$ in $\Bbb R^{2n}$.
In other words, as soon as we know that there exists one pair of diagonal matrices $D_1,D_2$ such that $\bar A$ has full column-rank, it immediately follows that this occurs for almost all such diagonal matrices.

Other thoughts: it would be nice to be able to show that if $A$ has full column rank, then $\bar A$ is indeed of full column rank for at least one choice of $D_1,D_2$.
We can write this matrix as the product
$$
\bar A = \pmatrix{D_1 & D_2} \pmatrix{A & 0\\0 & A} \implies\\
\bar A^T = \pmatrix{A^T & 0\\0 & A^T} \pmatrix{D_1 \\ D_2}.
$$
Let $T$ denote the transformation over $\Bbb R^{2m}$ given by
$$
T(x) = \pmatrix{A^T & 0\\0 & A^T}x.
$$
Let $D$ denote the matrix
$$
D = \pmatrix{D_1\\D_2}.
$$
The rank of $\bar A^T$ is equal to the rank of $T|_{\mathcal C(D)}$, where $\mathcal C(D)$ denotes the column space of $D$. By the rank nullity theorem, this rank is equal to
$$
\dim(\mathcal C) - \dim [\mathcal C(D) \cap \ker(T)] = n - \dim[\mathcal C(D) \cap \ker(T)].
$$
So, $\bar A$ will have maximal rank when $\dim [\mathcal C(D) \cap \ker(T)]$ is minimal. If $\bar A$ has full column-rank, then this minimal dimension must be equal to $m - 2n$. Notably, $\dim \ker A^T = m - n$.
Now, write
$$
\bar A^T x = 0 \implies \pmatrix{A^TD_1 x\\ A^TD_2 x} = 0 \implies \pmatrix{A^Ty\\ A^TBy} = 0,
$$
where $y = D_1x$ and $B = D_2D_1^{-1}$.
