If $A$ and $B$ are $3\times 3$ matrices and $A$ is invertible, then can we say that there exist an integer $n$ such that $A+nB$ invertible? I was trying by choosing n such that eigne values of $A+nB$ is non zero. In case of $B = I$ we can find out eigen value of $A+nB$ that would be $\lambda +n B$ (though i am not cleared about its proof). This choosing $n$ such that $\lambda$ not equal to -$n$times eigen value of $B$ will serve the purpose. But i am not sure about general $B$. What if i take any matrix $A$ and $B$.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
Here's how it works:
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
Since $\det(A+xB)$ is a polynomial in $x$, it either has finitely many zeroes or is $0$ for all $x$. Since $A$ is invertible it is not zero when $x=0$, thus all but finitely many integers $n$ are such that $\det(A+nB)\neq 0$ so $A+nB$ is invertible.