# Finding a common matrix of several column full rank matrices to make them invertible

Let $p\ge q$, let $A_1,\dots,A_n\in\mathbb R^{p\times q}$ be matrices such that any nonzero linear combination $k_1A_1+\dots+k_nA_n$ is of rank $q$ (column full rank).

Then can we find a matrix $B\in\mathbb R^{q\times p}$ such that $B(k_1A_1+\dots+k_nA_n)$ is invertible for all $(k_1,\dots,k_n)\neq0$?

For a single matrix we can find $q$ linear independent rows so that a submatrix is invertible. But for several matrices we may not find a common linear independent rows.

My original question is to find a way to reduce a over-determined linear elliptic PDE system (the symbol is injective) to a determined linear elliptic PDE system (the symbol is invertible), just by "deleting some rows in the same time". The condition of first order PDE being elliptic is exactly the assumption of first paragraph.

## 1 Answer

Take $p=q=3$ and $n=2$. Then $\det(A_1+k_2A_2)$ is (when $A_2$ is invertible) a polynomial in $k_2$ of degree $3$; consequently, there is $k_2\in\mathbb{R}$ s.t. $rank(A_1+k_2A_2)<3$. Thus your hypothesis (in your second line) does not work for $p=q$.

Thus we must assume that $p>q$.

For example $p=4,q=3$. If we choose $B\in M_{3,4}$ (with $rank(B)=3$ as required), then $\det(B(A_1+k_2A_2))$ is again a polynomial in $k_2$ of degree $3$ and there is $k_2\in\mathbb{R}$ s.t. $B(A_1+k_2A_2)$ is not invertible.