Is the rank $A+\sqrt{2} B$ greater than the rank of $A$?

Question

Let $A$ and $B$ be two matrices of same dimensions with rational coefficients.

1. Is $\text{rank } A+\sqrt{2} B \ge \text{rank } A$ and $\text{rank } A+\sqrt{2} B \ge \text{rank } B$?
2. More generally, is it true for any field extension $\mathbf{Q}[x]$?
3. Can we generalize, is $\text{rank } A_0+\sqrt{2} A_1 + \sqrt{3} A_2 + \sqrt{6} A_3 \ge \max\text{rank } A_i$?

Answer 1

The answer is no, even for the weakest of these statements. As a counterexample, consider the matrices $$ A = \pmatrix{0&1\\2&0}, \quad B = \pmatrix{1&0\\0&1}. $$ We have $\operatorname{rank}(A) = \operatorname{rank}(B) = 2$, but $\operatorname{rank}(A + \sqrt{2}B) = 1$.