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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 $B$.

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Hint: $p(n) := \det(A + nB)$ is a polynomial of degree $\le 3$ in $n$, which isn't zero as $p(0) \ne 0$. –  martini May 10 '12 at 6:15
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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. –  Joel Cohen May 10 '12 at 6:15
    
@martini is it a polynomial of degree 3 or $\leq 3$? –  srijan May 10 '12 at 6:42
    
still i haven't got a complete answer. I need little more help. –  srijan May 10 '12 at 6:46
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up vote 4 down vote accepted

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 $x$. –  srijan May 10 '12 at 6:23
    
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). –  Alex Becker May 10 '12 at 6:30
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