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If $A$ and $B$ are any $3 \times 3$ matrices and A is any invertible matrix, then there exist an integer $n$ such that $A + nB$ is invertible.

It is easy to check if we take $n = 0$, then the result always holds, But I want to know, when $n$ is non-zero then the result is true or not.

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Why did you accept an incomplete solution? –  Did Sep 30 '12 at 15:22
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1 Answer

up vote 2 down vote accepted

Yes, the determinant of $A+nB$ can be written as a polynomial $f$ in $n$ of degree 3. There are at most 3 real roots of $f$ and any integer $m$ which is not a root gives $A+mB$ which has non-zero determinant, and so invertible.

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The coefficient of n^3 in det(A+nB) is det(B) hence, if B is not invertible, the degree of det(A+nB) is not 3. –  Did Sep 30 '12 at 14:45
You need to assume something on $B$ so that your polynomial is not identically zero though. I don't know what one needs to assume. It would probably have to do with the rank of $A$ and/or $B$. –  Patrick Da Silva Sep 30 '12 at 14:45
The degree of $P(X) = \det(A+XB)$ is indeed the rank of $B$. But we know $P(0) \ne 0$, so $P$ is not identically zero, and there exist $n \ne 0$ such that $P(n) \ne 0$ (because a polynomial of any degree can only have finitely many zeros). –  Joel Cohen Sep 30 '12 at 14:55
@JoelCohen This is a proof. –  Did Sep 30 '12 at 15:21
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