Rank of sum of matrix and arbitrarily small matrix Suppose I have an $n\times n$ matrix $A$ with full rank. Let $B$ be another $n\times n$ matrix and let $c>0$ be a constant. Can we always find a constant $c>0$ sufficiently small such that
$$
A + cB
$$
also has full rank?
 A: Yes, because the determinant is a continuous function of the matrix entries. If it's not zero then it's not zero for nearby matrices.
Edit in response to comment.
This is true for rectangular matrices too. To prove it, note that Gaussian elimination produces the reduced row echelon form $R$ of $M$ by multiplication on the right by an invertible matrix $A$. Now $M$ is full rank if and only if $R$ has a full rank (identity) matrix in its upper left corner. You can perturb that corner and keep it full rank. Since multiplication by $A$ is a continuous bijection, matrices near $M$ will be full rank.
A: Another approach, valid over any field with more than $n+1$ elements:
Since $A$ is full rank and square, it has an inverse. Then
$$\det(A+cB)=\det(A(I+cA^{-1}B))=\det(A)\det(I+cA^{-1}B).$$
Since $\det(A)\neq0$, $A+cB$ is full rank iff $\det(I+cA^{-1}B)\neq0$.
Now, if $c\neq0$ then $\det(I+cA^{-1}B)=c^n\det(c^{-1}I+A^{-1}B)$, so $A+cB$ is full rank iff $$\det(c^{-1}I+A^{-1}B)\neq0.$$
This happens iff $-c^{-1}\not\in$Spec$(A^{-1}B)$. So there are at most $n$ values of $c$ which do not give full rank.
