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Let $A$,$B$ $\in M_n(\mathbb{R})$ and $B$ is invertible, then prove that there exists a $c \in \mathbb{R}$ such that $A+cB$ is not invertible.

My attempt: We need to show that $\det(A+cB)=0$. So $\det(A+cB)=\det(ABB^{-1}+cB)=\det(B(AB^{-1}+cI))=\det(B)\det(AB^{-1}+cI)$. Now all I need to show that $\det(AB^{-1}+cI)=0$, that is to prove $-c$ is the eigenvalue of $AB^{-1}$. But I don't know how to proceed. Help me out.

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    $\begingroup$ Well, you just need the existence of some $c$, but $AB^{-1}$ is not guaranteed to have an eigenvalue over the real numbers. So this doesn't seem to actually be true, i.e. take A = [ [0, -1], [1, 0] ], $B = I$, so $\det(A + cB) = c^2 + 1$ which is not zero. $\endgroup$ – user148177 May 1 '16 at 1:25
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    $\begingroup$ the question seems to be true only when n is odd. $\endgroup$ – nivekgnay May 1 '16 at 1:31
  • $\begingroup$ Yes, this is true only for $n$ odd (in the complex case this is true for all $n$). The matrices $A=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ and $B=I_2$ are a counter-example for $n=2$. $\endgroup$ – Luiz Cordeiro May 1 '16 at 1:40
  • $\begingroup$ Hint: You can reformulate the problem as to "Find a solution for the equation $\det(A+xB)=0$". The LHS has a nice form. $\endgroup$ – Luiz Cordeiro May 1 '16 at 1:41
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    $\begingroup$ @Quintic It is polynomial. $\endgroup$ – Luiz Cordeiro May 1 '16 at 2:46

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