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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?

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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.

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  • $\begingroup$ Great, thank you for the response. Do you have a similar argument for non square matrices? $\endgroup$ – Marc Oct 23 '17 at 12:10
  • $\begingroup$ Yes. See my edit. $\endgroup$ – Ethan Bolker Oct 23 '17 at 13:00
  • $\begingroup$ Does it hold even if $B$ is not full rank? $\endgroup$ – Cloud JR Sep 18 '18 at 18:25
  • $\begingroup$ @CloudJR Yes. The rank of $B$ is irrelevant. Read the proof sketch in the answer. $\endgroup$ – Ethan Bolker Sep 18 '18 at 19:41
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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.

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