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Given matrices $A, B \in \Bbb R^{3 \times 3}$ whose ranks satisfy $\mbox{rank} (A), \mbox{rank} (B) \geq 2$, I would like to prove that for large (or small) enough scalar $\alpha \in \mathbb{R} \setminus \{0\}$ the following does hold.

$$\mbox{rank} (A+\alpha B) \geq 2$$

This seems to be true by hand waving argument, but I would like to find some short proof or reference that formally proves it. Thanks.

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Since rank$(A)\geq 2$, the matrix $A$ has a $2\times 2$ submatrix $a$ with nonzero determinant; the determinant is a continuous function of the matrix elements, so adding a sufficiently small perturbation $\alpha B$ to $A$ will leave $\det a\neq 0$ and hence the rank of $A+\alpha B$ remains $\geq 2$.

If instead of small $\alpha$ you wish to take large $\alpha$, define $B'=\alpha B$, $A'=\alpha A$ and work with $B'+(1/\alpha)A'$, where $B'$ has rank $\geq 2$.

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    $\begingroup$ By the way, this argument shows that the rank function is lower-semicontinuous. $\endgroup$ – HenrikRüping Jul 29 at 11:24

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