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If $A$ and $B$ are $3\times 3$ matrices and $A$ is invertible, then can we say that there exists an integer $n$ such that $A+nB$ invertible?

I was trying to show this by choosing $n$ such that eignevalues of $A+nB$ are non-zero. In the case where $B = I$ we can find the eigenvalues of $A+nB$ that would be $\lambda + nB$ (though I am not certain about its proof). This choosing of $n$ such that $\lambda$ is not equal to $-n$ times an eigenvalue of $B$ will serve the purpose. But I am not sure about general $B$. What if I take arbitrary matrices $A$ and $B$.

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    Hint: $p(n) := \det(A + nB)$ is a polynomial of degree $\le 3$ in $n$, which isn't zero as $p(0) \ne 0$.2012-05-10
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    If $-n$ is not an eigenvalue of $A^{-1}B$ then $I+nA^{-1}B$ is invertible so $A+nB$ is also invertible.2012-05-10
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    @martini is it a polynomial of degree 3 or $\leq 3$?2012-05-10
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    still i haven't got a complete answer. I need little more help.2012-05-10

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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.

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    I need explanation for $\det(A+xB)$ has finitely many zeros or is 0 for all $x$.2012-05-10
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    It's a polynomial (when $A$,$B$ are $3\times 3$ it has degree $3$). All nonzero polynomials have only finitely many zeroes (in fact, no more than their degree).2012-05-10