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Consider a $N \times N$ matrix $A=B_1+B_2$, where $B_1$ is a diagonal matrix with all the diagonal entries between $0$ and $1$ and $B_2$ is a skew symmetric matrix which can be written as

$$B_2= \begin{bmatrix} 0_{N-1} & f \\ -f^T & 0 \end{bmatrix}$$

where $0_{N-1}$ is the $N-1 \times N-1$ zero matrix, $f$ is a $N-1$ by $1$ vector. So $B_2=-B_2^T$.

My question is how many complex eigenvalues (non-zero imaginary part) $A$ has?

I have run some numerical simulations. It seems that $A$ can at most have one pair of complex eigenvalues. And it is plausible to me because the skew symmetric $B_2$ is only rank $2$.

How can we prove this mathematically or come up with a counter example? If needed, we can put an upper bound on $|f|_2$. But I don't know that if it is necessary, as indicated from my simulations.

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  • $\begingroup$ I don't know if it helps, but using the Schur complement, the characteristic polynomial of $A$ is$$p_A(z) = \prod_{k=1}^{n-1}(z_n-z) + \sum_{j=1}^{n-1}\left(\prod_{k=1,k\neq j}^{n-1}(z_k-z)\right)f_j^2,$$where $z_k$ is the $k$-th enty in the diagonal of $B_1$. $\endgroup$ – Friedrich Philipp Mar 18 '16 at 15:43
  • $\begingroup$ I meant $z_k$ instead of $z_n$ in the first product. $\endgroup$ – Friedrich Philipp Mar 18 '16 at 15:58
  • $\begingroup$ Does $p_A(z)$ depend on $z_N$, last entry of $B_1$? $\endgroup$ – user322840 Mar 18 '16 at 16:01
  • $\begingroup$ The first product ranges to $n$. So, it is $\prod_{k=1}^n(z_k-z)$. Sorry for the confusion. $\endgroup$ – Friedrich Philipp Mar 18 '16 at 16:06
  • $\begingroup$ What you could try is the argument principle en.wikipedia.org/wiki/Argument_principle. But this means you would have to derive $p_A$ and consider $p_A'/p_A$. Not that nice... $\endgroup$ – Friedrich Philipp Mar 18 '16 at 16:08
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We only need $B_1$ to be diagonal. Positive definiteness is not needed. Let $A=\pmatrix{D&f\\ -f^T&a}$. If some entries of $f$ is zero, we can reduce the problem to case with a smaller-sized $A$. Assume $f$ is entrywise nonzero. As Friedrich Philipp suggests in his comment, we may make use of Schur complement: \begin{align} \det(tI-A)&=\det(tI-D)\left(t-a+f^T(tI-D)^{-1}f)\right)\\ &=\det(tI-D)\left(t-a+\sum_{j=1}^{n-1} \frac{f_j^2}{t-d_j}\right). \end{align} Hence it suffices to show that $g(t)=t-a+\sum_{j=1}^{n-1} \frac{f_j^2}{t-d_j}$ has at least $n-2$ real roots.

By permutation, we may assume that the diagonal entries are arranged in strictly ascending order. Since $g(d_j^-)=-\infty$ and $g(d_j^+)=+\infty$, if the diagonal elements of $D$ are distinct, $g$ changes sign on every interval $(d_j,d_{j+1})$. Consequently, $g$ changes sign at least $n-2$ times on $\mathbb R\setminus\{d_1,d_2,\ldots,d_{n-1}\}$ and $A$ has at least $n-2$ (distinct) real eigenvalues. By continuity of eigenvalues in the matrix entries, $A$ also has at least $n-2$ real eigenvalues when some diagonal elements of $D$ are the same.

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  • $\begingroup$ Thank you and @FriedrichPhilipp so much. I think your answer solves the problem! If two of the diagonal elements are the same, one eigenvalue of the A matrix will be the same as that diagonal element. So we are back to the distinct eigenvalue case. $\endgroup$ – user322840 Mar 20 '16 at 15:40
  • $\begingroup$ @user322840 Huh? No, that's not what I mean. Although some diagonal entries of $D$ may be equal to each other, $D$ is always the limit of a sequence of diagonal matrices that have distinct diagonal entries. In fact, if we define $D_k=D+\operatorname{diag}(k^{-1},k^{-2},\ldots,k^{-n})$, then $\lim_{k\to\infty}D_k=D$ and $D_k$ has distinct diagonal elements when $k$ is sufficently large. Let $A_k$ be the matrix obtained by replacing the $D$ in $A$ by $D_k$. Then each $A_k$ has at least $n-2$ real eigenvalues. Therefore, $A$, the limit of $A_k$, also has at least $n-2$ real eigenvalues. $\endgroup$ – user1551 Mar 20 '16 at 16:02
  • $\begingroup$ Thanks, @user1551, for your detailed explanation. Here is what I meant: let $d_{1}=d_{2}\neq d_j, \forall j\neq 1,2$. We then have $\det (tI-A) = \Pi_{i=1}^{n-1}(t-d_i)\times (t-a+\sum_{i=1}^2\frac{f_1^2+f_2^2}{t-d_1}+\sum_{j=3}^{n-1}\frac{f_j^2}{t-d_j})=\Pi_{i=2}^{n-1}(t-d_i)\times ((t-a)(t-d_1)+{f_1^2+f_2^2}+(t-d_1)\sum_{j=3}^{n-1}\frac{f_j^2}{t-d_j})$. Then in addition to an eigenvalue at $t=d_1=d_2$, $A$ has at least $n-3$ other real eigenvalues. $\endgroup$ – user322840 Mar 20 '16 at 16:40
  • $\begingroup$ Probably not as simple as your method. $\endgroup$ – user322840 Mar 20 '16 at 16:42

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