Let $n \in \mathbb{N},n \ge 2,$ and $A,B \in \mathcal{M}_n(\mathbb{R}).$ Prove that there exists a complex number $z,$ such that $|z|=1$ and $$\Re \left( {\det(A+zB)} \right) \ge \det(A)+\det(B),$$where $\Re(w)$ is the real part of the complex number $w$.
We know that $f(z)=\det(A+zB)$ is a polynomial in $z$. Can we use this idea?
On
Here is the answer I sketched in the comments, in more detail.
I will prove a more general statement:
Theorem 1. Let $n$ be a positive integer. Let $A,B\in\mathbb{C}^{n\times n}$ be two complex matrices. Then, there exists a complex number $z$ on the unit circle (actually, an $n$-th root of unity) such that \begin{equation} \operatorname*{Re}\left( \det\left( A+zB\right) \right) \geq \operatorname*{Re}\left( \det A+\det B\right) . \label{darij1.eq.t1.eq} \tag{1} \end{equation}
Here, I am using the notation $\operatorname*{Re}w$ for the real part of a complex number. (You are using $\mathfrak{R}\left( w\right) $ for this.)
Theorem 1 generalizes your question, because if $A,B$ are real matrices, then $\operatorname*{Re}\left( \det A+\det B\right) =\det A+\det B$.
In order to prove Theorem 1, we need some notations regarding polynomials. We set $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $.
Definition. Let $\mathbb{K}$ be a commutative ring, and let $f\in \mathbb{K}\left[ t\right] $ be a polynomial. For any $n\in\mathbb{N}$, we let $\left[ t^{n}\right] f$ denote the coefficient of $t^{n}$ in the polynomial $f$.
Thus, any polynomial $f\in\mathbb{K}\left[ t\right] $ satisfies $f=\sum_{k\in\mathbb{N}}\left( \left[ t^{k}\right] f\right) \cdot t^{k}$.
We recall a basic property of determinants:
Proposition 2. Let $\mathbb{K}$ be a commutative ring. Let $n\in \mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$ and $B\in\mathbb{K}^{n\times n}$ be two $n\times n$-matrices. Consider the matrix $tA+B\in\left( \mathbb{K}\left[ t\right] \right) ^{n\times n}$.
(a) Then, $\det\left( tA+B\right) \in\mathbb{K}\left[ t\right] $ is a polynomial of degree $\leq n$ in $t$.
(b) We have $\left[ t^{0}\right] \left( \det\left( tA+B\right) \right) =\det B$.
(c) We have $\left[ t^{n}\right] \left( \det\left( tA+B\right) \right) =\det A$.
Proposition 2 appears as Proposition 2.2 in my note The trace Cayley-Hamilton theorem (which has gradually become a grab-bag for proofs about characteristic polynomials). It also follows from "the final result" in Giuseppe Negro's answer https://math.stackexchange.com/a/2189752/ . You probably already know why it is true by the time you have finished reading the preceding two sentences.
On the other hand, we recall a basic property of roots of unity:
Proposition 3. Let $n$ be a positive integer. Let $\zeta$ be a primitive $n$-th root of unity in $\mathbb{C}$. Let $k\in\mathbb{Z}$. Then:
(a) If $n\nmid k$, then $\sum\limits_{j=0}^{n-1}\zeta^{jk}=0$.
(b) If $n\mid k$, then $\sum\limits_{j=0}^{n-1}\zeta^{jk}=n$.
Proposition 3 is the engine behind the Discrete Fourier Transform. For the sake of completeness, let me prove it:
Proof of Proposition 3. We have $\zeta^{n}=1$ (since $\zeta$ is an $n$-th root of unity), hence $\zeta^{nk}=\left( \underbrace{\zeta^{n}}_{=1}\right) ^{k}=1^{k}=1$.
(a) Assume that $n\nmid k$. Then, $\zeta^{k}\neq1$ (since $\zeta$ is a primitive $n$-th root of unity). Hence, $\zeta^{k}-1\neq0$. Now, set $\alpha=\sum\limits_{j=0}^{n-1}\zeta^{jk}$. Then, \begin{align*} \zeta^{k}\underbrace{\alpha}_{=\sum\limits_{j=0}^{n-1}\zeta^{jk}} & =\zeta^{k}\sum\limits_{j=0}^{n-1}\zeta^{jk}=\sum\limits_{j=0}^{n-1} \underbrace{\zeta^{k}\zeta^{jk}}_{=\zeta^{k+jk}=\zeta^{\left( j+1\right) k} }=\sum\limits_{j=0}^{n-1}\zeta^{\left( j+1\right) k}=\sum_{j=1}^{n} \zeta^{jk}\\ & \qquad\left( \text{here, we have substituted }j\text{ for }j+1\text{ in the sum}\right) \\ & =\sum_{j=1}^{n-1}\zeta^{jk}+\underbrace{\zeta^{nk}}_{=1}=\sum_{j=1} ^{n-1}\zeta^{jk}+1. \end{align*} Comparing this with \begin{equation} \alpha=\sum\limits_{j=0}^{n-1}\zeta^{jk}=\underbrace{\zeta^{0k}}_{=\zeta ^{0}=1}+\sum_{j=1}^{n-1}\zeta^{jk}=1+\sum_{j=1}^{n-1}\zeta^{jk}=\sum _{j=1}^{n-1}\zeta^{jk}+1, \end{equation} we obtain $\zeta^{k}\alpha=\alpha$. Hence, $\zeta^{k}\alpha-\alpha=0$, so that $\left( \zeta^{k}-1\right) \alpha=\zeta^{k}\alpha-\alpha=0$. We can divide this equality by $\zeta^{k}-1$ (since $\zeta^{k}-1\neq0$), and thus obtain $\alpha=0$. Hence, $\sum\limits_{j=0}^{n-1}\zeta^{jk}=\alpha=0$. This proves Proposition 3 (a).
(b) Assume that $n\mid k$. Then, $\zeta^{k}=1$ (since $\zeta$ is an $n$-th root of unity). Now, \begin{equation} \sum\limits_{j=0}^{n-1}\underbrace{\zeta^{jk}}_{=\zeta^{kj}=\left( \zeta ^{k}\right) ^{j}}=\sum\limits_{j=0}^{n-1}\left( \underbrace{\zeta^{k}} _{=1}\right) ^{j}=\sum\limits_{j=0}^{n-1}\underbrace{1^{j}}_{=1} =\sum\limits_{j=0}^{n-1}1=n. \end{equation} This proves Proposition 3 (b). $\blacksquare$
Next, we put Proposition 3 to use in establishing the so-called roots-of-unity filter (in one of its simplest forms):
Proposition 4. Let $n$ be a positive integer. Let $\zeta$ be a primitive $n$-th root of unity in $\mathbb{C}$. Let $f\in\mathbb{C}\left[ t\right] $ be a polynomial. Then, \begin{equation} \sum\limits_{j=0}^{n-1}f\left( \zeta^{j}\right) =n\sum_{\substack{k\in \mathbb{N};\\n\mid k}}\left[ t^{k}\right] f. \end{equation}
Proof of Proposition 4. Recall that $f=\sum_{k\in\mathbb{N}}\left( \left[ t^{k}\right] f\right) \cdot t^{k}$ (since $\left[ t^{0}\right] f,\left[ t^{1}\right] f,\left[ t^{2}\right] f,\ldots$ are the coefficients of the polynomial $f$). Hence, for each $j\in\mathbb{Z}$, we have \begin{equation} f\left( \zeta^{j}\right) =\sum_{k\in\mathbb{N}}\left( \left[ t^{k}\right] f\right) \cdot\underbrace{\left( \zeta^{j}\right) ^{k}}_{=\zeta^{jk}} =\sum_{k\in\mathbb{N}}\left( \left[ t^{k}\right] f\right) \cdot\zeta^{jk}. \end{equation} Summing this equality over all $j\in\left\{ 0,1,\ldots,n-1\right\} $, we find \begin{align*} \sum\limits_{j=0}^{n-1}f\left( \zeta^{j}\right) & =\underbrace{\sum \limits_{j=0}^{n-1}\sum_{k\in\mathbb{N}}}_{=\sum\limits_{k\in\mathbb{N}} \sum\limits_{j=0}^{n-1}}\left( \left[ t^{k}\right] f\right) \cdot \zeta^{jk}=\sum\limits_{k\in\mathbb{N}}\sum\limits_{j=0}^{n-1}\left( \left[ t^{k}\right] f\right) \cdot\zeta^{jk}\\ & =\sum\limits_{k\in\mathbb{N}}\left( \left[ t^{k}\right] f\right) \cdot\sum\limits_{j=0}^{n-1}\zeta^{jk}\\ & =\sum\limits_{\substack{k\in\mathbb{N};\\n\nmid k}}\left( \left[ t^{k}\right] f\right) \cdot\underbrace{\sum\limits_{j=0}^{n-1}\zeta^{jk} }_{\substack{=0\\\text{(by Proposition 3 (a))}}}+\sum\limits_{\substack{k\in \mathbb{N};\\n\mid k}}\left( \left[ t^{k}\right] f\right) \cdot \underbrace{\sum\limits_{j=0}^{n-1}\zeta^{jk}}_{\substack{=n\\\text{(by Proposition 3 (b))}}}\\ & \qquad\left( \begin{array} [c]{c} \text{since each }k\in\mathbb{N}\text{ satisfies either }n\nmid k\\ \text{or }n\mid k\text{, but not both at the same time} \end{array} \right) \\ & =\underbrace{\sum\limits_{\substack{k\in\mathbb{N};\\n\nmid k}}\left( \left[ t^{k}\right] f\right) \cdot0}_{=0}+\sum\limits_{\substack{k\in \mathbb{N};\\n\mid k}}\left( \left[ t^{k}\right] f\right) \cdot n=\sum\limits_{\substack{k\in\mathbb{N};\\n\mid k}}\left( \left[ t^{k}\right] f\right) \cdot n\\ & =n\sum_{\substack{k\in\mathbb{N};\\n\mid k}}\left[ t^{k}\right] f. \end{align*} This proves Proposition 4. $\blacksquare$
Combining Proposition 2 with Proposition 4, we obtain the following:
Corollary 5. Let $n$ be a positive integer. Let $\zeta$ be a primitive $n$-th root of unity in $\mathbb{C}$. Let $A\in\mathbb{C}^{n\times n}$ and $B\in\mathbb{C}^{n\times n}$ be two $n\times n$-matrices. Then, \begin{equation} \sum_{j=0}^{n-1}\det\left( \zeta^{j}A+B\right) =n\left( \det A+\det B\right) . \end{equation}
Proof of Corollary 5. Define a polynomial $f\in\mathbb{C}\left[ t\right] $ by $f=\det\left( tA+B\right) $. Then, Proposition 2 (a) (applied to $\mathbb{K}=\mathbb{C}$) shows that $\det\left( tA+B\right) \in \mathbb{C}\left[ t\right] $ is a polynomial of degree $\leq n$ in $t$. In other words, $f\in\mathbb{C}\left[ t\right] $ is a polynomial of degree $\leq n$ in $t$ (since $f=\det\left( tA+B\right) $). Thus, $\left[ t^{u}\right] f=0$ for all integers $u>n$. Thus, the coefficients $\left[ t^{2n}\right] f,\left[ t^{3n}\right] f,\left[ t^{4n}\right] f,\ldots$ are all $0$ (since the integers $2n,3n,4n,\ldots$ are all $>n$).
Proposition 2 (b) (applied to $\mathbb{K}=\mathbb{C}$) yields $\left[ t^{0}\right] \left( \det\left( tA+B\right) \right) =\det B$. In view of $f=\det\left( tA+B\right) $, this rewrites as $\left[ t^{0}\right] f=\det B$.
Proposition 2 (c) (applied to $\mathbb{K}=\mathbb{C}$) yields $\left[ t^{n}\right] \left( \det\left( tA+B\right) \right) =\det A$. In view of $f=\det\left( tA+B\right) $, this rewrites as $\left[ t^{n}\right] f=\det A$.
But we have $f=\det\left( tA+B\right) $. Hence, for any $z\in\mathbb{C}$, we have \begin{equation} f\left( z\right) =\det\left( zA+B\right) \label{darij1.pf.cor5.1} \tag{2} \end{equation} (because if we plug $z$ for $t$ into the matrix $tA+B$ and then we take the determinant, then we obtain the same result as if we first take the determinant and then plug $z$ for $t$ in it).
Proposition 4 yields \begin{align*} \sum\limits_{j=0}^{n-1}f\left( \zeta^{j}\right) & =n\underbrace{\sum _{\substack{k\in\mathbb{N};\\n\mid k}}\left[ t^{k}\right] f} _{\substack{=\left[ t^{0}\right] f+\left[ t^{n}\right] f+\left[ t^{2n}\right] f+\left[ t^{3n}\right] f+\cdots\\\text{(since the } k\in\mathbb{N}\text{ satisfying }n\mid k\\\text{are }0,n,2n,3n,\ldots\text{)} }}\\ & =n\underbrace{\left( \left[ t^{0}\right] f+\left[ t^{n}\right] f+\left[ t^{2n}\right] f+\left[ t^{3n}\right] f+\cdots\right) }_{\substack{=\left[ t^{0}\right] f+\left[ t^{n}\right] f\\\text{(since the coefficients }\left[ t^{2n}\right] f,\left[ t^{3n}\right] f,\left[ t^{4n}\right] f,\ldots\text{ are all }0\text{)}}}\\ & =n\left( \underbrace{\left[ t^{0}\right] f}_{=\det B}+\underbrace{\left[ t^{n}\right] f}_{=\det A}\right) =n\left( \det B+\det A\right) \\ & =n\left( \det A+\det B\right) . \end{align*} Comparing this with \begin{equation} \sum\limits_{j=0}^{n-1}\underbrace{f\left( \zeta^{j}\right) } _{\substack{=\det\left( \zeta^{j}A+B\right) \\\text{(by \eqref{darij1.pf.cor5.1}, applied to }z=\zeta^{j}\text{)}}}=\sum_{j=0} ^{n-1}\det\left( \zeta^{j}A+B\right) , \end{equation} we obtain \begin{equation} \sum_{j=0}^{n-1}\det\left( \zeta^{j}A+B\right) =n\left( \det A+\det B\right) . \end{equation} This proves Corollary 5. $\blacksquare$
Now we can prove Theorem 1:
Proof of Theorem 1. Choose any primitive $n$-th root of unity in $\mathbb{C}$ (for example, $e^{2\pi i/n}$), and denote it by $\zeta$. We claim that there exists some $j\in\left\{ 0,1,\ldots,n-1\right\} $ such that \begin{equation} \operatorname*{Re}\left( \det\left( A+\zeta^{j}B\right) \right) \geq\operatorname*{Re}\left( \det A+\det B\right) . \label{darij1.pf.t1.1} \tag{3} \end{equation}
[Proof: Assume the contrary. Thus, no $j\in\left\{ 0,1,\ldots,n-1\right\} $ satisfies \eqref{darij1.pf.t1.1}. In other words, each $j\in\left\{ 0,1,\ldots ,n-1\right\} $ satisfies \begin{equation} \operatorname*{Re}\left( \det\left( A+\zeta^{j}B\right) \right) <\operatorname*{Re}\left( \det A+\det B\right) . \end{equation} Summing up these inequalities over all $j\in\left\{ 0,1,\ldots,n-1\right\} $, we obtain \begin{align} \sum_{j=0}^{n-1}\operatorname*{Re}\left( \det\left( A+\zeta^{j}B\right) \right) & <\sum_{j=0}^{n-1}\operatorname*{Re}\left( \det A+\det B\right) =n\operatorname*{Re}\left( \det A+\det B\right) \nonumber\\ & =n\operatorname*{Re}\left( \det B+\det A\right) . \label{darij1.pf.t1.1.pf.1} \tag{4} \end{align} But Corollary 5 (applied to $B$ and $A$ instead of $A$ and $B$) yields \begin{equation} \sum_{j=0}^{n-1}\det\left( \zeta^{j}B+A\right) =n\left( \det B+\det A\right) . \end{equation} Taking real parts on both sides of this equality, we find \begin{align} \operatorname*{Re}\left( \sum_{j=0}^{n-1}\det\left( \zeta^{j}B+A\right) \right) =\operatorname*{Re}\left( n\left( \det B+\det A\right) \right) =n\operatorname*{Re}\left( \det B+\det A\right) , \end{align} so that \begin{align*} n\operatorname*{Re}\left( \det B+\det A\right) & =\operatorname*{Re} \left( \sum_{j=0}^{n-1}\det\left( \zeta^{j}B+A\right) \right) \\ & =\sum_{j=0}^{n-1}\operatorname*{Re}\left( \det\underbrace{\left( \zeta ^{j}B+A\right) }_{=A+\zeta^{j}B}\right) =\sum_{j=0}^{n-1}\operatorname*{Re} \left( \det\left( A+\zeta^{j}B\right) \right) \\ & <n\operatorname*{Re}\left( \det B+\det A\right) \end{align*} (by \eqref{darij1.pf.t1.1.pf.1}). This is clearly absurd. This contradiction shows that our assumption was wrong. Hence, we have proven our claim that there exists some $j\in\left\{ 0,1,\ldots,n-1\right\} $ satisfying \eqref{darij1.pf.t1.1}.]
Now, consider a $j\in\left\{ 0,1,\ldots,n-1\right\} $ satisfying \eqref{darij1.pf.t1.1}. (We have just proven that such a $j$ exists.) Then, $\zeta^{j}$ is an $n$-th root of unity (since $\zeta$ is an $n$-th root of unity) and thus lies on the unit circle. Moreover, \eqref{darij1.pf.t1.1} shows that $\zeta^{j}$ is a complex number $z$ on the unit circle (actually, an $n$-th root of unity) satisfying \eqref{darij1.eq.t1.eq}. Hence, such a $z$ exists. This proves Theorem 1. $\blacksquare$
Incomplete proof
Assume first that both $A$ and $B$ are singular matrices. We have to prove that $\mathfrak R(f(z))\ge 0$ for some $z \in \partial D$ where $\partial D$ is the unit complex circle.
$f(z)$ is a complex polynomial whose integral on $\partial D$ vanishes. If $\mathfrak R(f(z))$ is always vanishing on $\partial D$, we’re done. If not it must take positive and negative values (if the integral of a non always vanishing continuous function is equal to zero, the function must take positive and negative values). Therefore we’re also done.
Now, removing our initial assumption, suppose for example that $\det B \neq 0$. We have to prove that $$\mathfrak R(\det(C+zI)) \ge \det(C)+1$$ for some $z \in \partial D$ where $C =B^{-1}A$.
As $f(z) = \det(C+zI)=z^n +a_{n-1} z^{n-1} +\dots +a_1 z+\det(C),$ we’re left to prove that any real trigonometric polynomial $$\cos n\theta + a_{n-1} \cos (n-1)\theta + \dots a_1 \cos \theta -1$$
takes at least a non negative value for $\theta \in \mathbb R$.
and here I don’t know how to follow on...