Let $A$ and $B$ be square real matrices such that $A+iB$ is non-singular. Show that there exists $t\in \mathbb{R}$ such that $A+tB$ is non-singular.

My Attempt:

I thought about considering the polynomial over $\mathbb{C}$. Let $$f(t)=|A+tB|.$$ Then $f(i)\not =0.$ This shows that $f\not \equiv 0$ and since $f$ has a finite number of roots of which let $\{t_1,t_2,...,t_k\}$ be the real roots. Then we can easily, find $t\in \mathbb{R}$ such that $f(t)=|A+tB|\not =0$ and hence $A+tB$ will be non-singular. Is this correct reasoning?

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    $\begingroup$ Yes, that's right. $\endgroup$ – David C. Ullrich Jan 5 '18 at 17:28
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    $\begingroup$ The characteristic polynomial $\det(A+tB)$ is an analytic function of $t$. If this function is $0$ for all $t\in\mathbb{R}$, then by analytic continuation, it must be $0$ for all $t\in\mathbb{C}$, which contradicts with that $A+iB$ is non-singular. $\endgroup$ – Zhuoran He Jan 5 '18 at 17:34
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    $\begingroup$ @Zhuoran: that's overkill. It's a polynomial, so it's either zero or it has finitely many roots. $\endgroup$ – Qiaochu Yuan Jan 7 '18 at 0:30

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