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Let $A,B$ be $n\times n$ complex matrices. If $B$ is invertible then there exists a scalar $c \in \mathbb C$ such that $A+cB$ is not invertible.

Since $\det(A+cB)$ is a polynomial in $\mathbb C$, it must have a root in $\mathbb C$, i.e. there must exist a $c$ such that $\det(A+cB)=0$. Then why is the condition "$B$ is invertible" necessary?

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    $\begingroup$ Your argument is incorrect. You need to know that $\det(A + cB)$ is a nonconstant polynomial in $c$ (or the zero polynomial). $\endgroup$ – user49640 Jun 13 '17 at 7:06
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Let's take the full journey. $B$ is inversible so $B^{-1}$ exists and $\det{B}\neq 0$. So we can rewrite

$$\det(A+cB)=\det(AB^{-1}+cI)\det{B}$$

Now the coefficient of $c^n$ in the $\det(AB^{-1}+cI)$ is $1$ so the polynomial has degree $n\gt 1$ and so it has a root in $\Bbb{C}$.

Now let's find a counterexample. Consider

$$ A=\begin{bmatrix} 1 & 0\\1 & 1\end{bmatrix}$$

$$ B=\begin{bmatrix} 0 & 1\\0 & 1\end{bmatrix}$$

One has

$$A+cB =\begin{bmatrix}1 & c\\1 & 1+c\end{bmatrix}$$

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    $\begingroup$ I dont get the counterexample, even if $B$ is not invertible, we again find $A+cB$ is not invertible, so it doesnot about necessity of invertible $B$. Where am I wrong? $\endgroup$ – Azer BABA fanlari May 5 '19 at 8:46
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    $\begingroup$ Thanks ! I just edited for the right counterexample $\endgroup$ – marwalix May 8 '19 at 7:00
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Hint: Consider the polynomial $\varphi(z) = 1$ in $\mathbb C$. Does it have a root?

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  • $\begingroup$ Isnt the condition "B not equal to 0" sufficient then? $\endgroup$ – timotheechalamet Jun 13 '17 at 6:43
  • $\begingroup$ @AnwitaBhowmik $A$ could be the identity matrix, and $B$ could be any upper-triangular matrix with zeros on the diagonal. $\endgroup$ – user49640 Jun 13 '17 at 7:01
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A necessary and sufficient condition for the existence of $c$ is that either: $A$ is non-invertible; or $A$ is invertible and $BA^{-1}$ is not nilpotent. This answers the question, since if $B$ is invertible, then $BA^{-1}$ is invertible, hence cannot be nilpotent.

If $A$ is non-invertible, choose $c = 0$. Otherwise $\det(A + cB) = 0$ if and only if $c \ne 0$ and $\det(c^{-1}I + BA^{-1}) = 0$. This is feasible if and only if $BA^{-1}$ has a nonzero eigenvalue, which occurs precisely when $BA^{-1}$ is not nilpotent.

Alternative proof. Assume $B$ invertible. Then $\det(A + cB) = 0$ if and only if $\det(AB^{-1} + cI) = 0$. So it's enough to note that since $\mathbf{C}$ is algebraically closed, the matrix $AB^{-1}$ has an eigenvalue.

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