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Let $[\theta_1,\theta_2, \dots, \theta_N]^\mathrm{T} \, \in \mathbb{R}^N$. The angles are not all identical (on the circle), i.e. $[\theta_1,\theta_2, \dots, \theta_N] \not \equiv c [1,1,\dots, 1]^\mathrm{T}\,\, \mathrm{mod}\,\, 2\pi$. Define matrices $C$ and $S$ as:

\begin{align} \begin{split} [C]_{jl}&= 1 \,\, \mathrm{if}\,\, j =l \\ [C]_{jl}&= \cos(\theta_j-\theta_l)\,\, \mathrm{if} \,\, j \neq l \end{split} \end{align}

and

\begin{align} \begin{split} [S]_{jl}&= 0 \,\, \mathrm{if}\,\, j =l \\ [S]_{jl}&= \sin(\theta_j-\theta_l)\,\,\mathrm{if} \,\, j \neq l\,. \end{split} \end{align}

Is it true that any $v$ that belongs to the nullspace of S also belongs to the nullspace of C?

My repeated simulations in MATLAB with randomly generated $[\theta_1,\theta_2, \dots, \theta_N]^\mathrm{T}$ seems to suggest that $C$ and $S$ are rank $2$ and any eigenvectors with $0$ eigenvalue for one is an eigenvector with $0$ eigenvalue for the other.

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    $\begingroup$ Could you explain the motivation behind the question? This looks like it could make for a really nice problem on a take-home linear algebra final; how did you stumble across it "in the wild"? $\endgroup$ – Elena Yudovina Dec 13 '16 at 23:47
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    $\begingroup$ With the updated version of the problem, you don't need to separate the $j=l$ and $j \neq l$ cases any more, since they are both subsumed by the $j \neq l$ formula. Also, I'm wondering whether all the matrices in the family $\left(S_\phi\right)_{\phi\in\mathbb{R}}$, where the $\left(j,l\right)$-th entry of $S_\phi$ is $\sin\left(\theta_j-\theta_l+\phi\right)$, have the same nullspace. $\endgroup$ – darij grinberg Dec 14 '16 at 1:15
  • $\begingroup$ @ElenaYudovina. I am trying to solve a nonlinear dynamics problem. Lyapunov analysis lead me to such kind of matrices. $\endgroup$ – Mohit Dec 14 '16 at 4:17
  • $\begingroup$ What happens if $\theta_1=\pi$ and all other $\theta_i=0$? In that case, $S=0$ and $C\neq 0$, if I do not misunderstand. $\endgroup$ – Federico Poloni Dec 14 '16 at 7:26
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You must have made a mistake.

$E = -C + 2I + i S$ has entries $\exp(i (\theta_j - \theta_l)) = \exp(i\theta_j) \exp(-i \theta_l)$ and thus is the product $V V^*$ where $V$ is the column vector with entries $\exp(i \theta_j)$, so it has rank $1$.
Then $S = (E - \overline{E})/(2i)$ has rank at most $2$ (it would be $0$ e.g. if all $\theta_j$ are multiples of $\pi$). Similarly, $-C+2I$ has rank at most $2$, so $C$ has rank at least $N-2$ (that part at least you got right). But "generically" $C$ will have rank $N$.

EDIT: OK, now with the corrected question we have $E = C + iS$. Again $S = (E - \overline{E})/(2i)$ and now $C = (E + \overline{E})/(2i)$ have rank at most $2$. Moreover, $\ker(E) \cap \ker(\overline{E}) = (V^*)^\perp \cap (\overline{V^*})^\perp$ is contained in both $\ker(S)$ and $\ker(C)$. Note that the following are equivalent:

  1. $\theta_j - \theta_l \not \equiv 0 \mod \pi$ for some $j,l$.
  2. $V$ and $\overline{V}$ are linearly independent.
  3. $(V^*)^\perp$ and $(\overline{V^*})^\perp$ are distinct linear subspaces.
  4. $\ker(E) \cap \ker(\overline{E})$ has dimension exactly $N-2$.
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  • $\begingroup$ Thanks for your answer. I realized I made a mistake in writing my question. Actually my matrix $C$ had $\cos (\theta_j -\theta_l)$ as entries, without the negative sign. Also, I saw they were rank $2$ with $N-2$ zero eigenvalues as your argument points out. I have edited my question. I apologize for my sloppiness. $\endgroup$ – Mohit Dec 13 '16 at 22:16
  • $\begingroup$ Your answer can be used to solve the corrected problem. Let me write $n$ for your $N$, out of habit. Let us use the notation $\operatorname{Ker}_{\mathbb{R}} U$ for "the intersection of the kernel of an $n\times n$-matrix $U \in \mathbb{C}^{n\times n}$ with the real vector space $\mathbb{R}^n$". With the correction in place, we have $E = C + iS$. But from $E = VV^\ast$, we obtain $\operatorname{Ker}E = \operatorname{Ker} \left(V^\ast\right)$ (by a standard argument using the positivity of the inner product). Hence, ... $\endgroup$ – darij grinberg Dec 14 '16 at 1:59
  • $\begingroup$ ... we get $\operatorname{Ker}_{\mathbb{R}} E = \operatorname{Ker}_{\mathbb{R}} \left(V^\ast\right) = \left\{\left(x_1, x_2, \ldots, x_n\right)^T \in \mathbb{R}^n \mid \sum\limits_{k=1}^n e^{-i\theta_k}x_k = 0\right\}$ $ = \left\{\left(x_1, x_2, \ldots, x_n\right)^T \in \mathbb{R}^n \mid \sum\limits_{k=1}^n \left(\sin\theta_k\right) x_k = 0 \text{ and } \sum\limits_{k=1}^n \left(\cos\theta_k\right) x_k = 0\right\} $ (because the sum $\sum\limits_{k=1}^n e^{-i\theta_k}x_k$ has real part $\sum\limits_{k=1}^n \left(\cos\theta_k\right) x_k = 0 $ and ... $\endgroup$ – darij grinberg Dec 14 '16 at 2:01
  • $\begingroup$ ... imaginary part $\sum\limits_{k=1}^n \left(\sin\theta_k\right) x_k = 0 $). This is the intersection of two hyperplanes in $\mathbb{R}^n$. Okay, now I am seeing that the problem is missing an assumption: We want the two hyperplanes to be distinct. But they will coincide if we have $\theta_1 \equiv \theta_2 \equiv \cdots \equiv \theta_n \mod \pi$ (not just $2 \pi$). That said, I hope that if we require this not to happen, the hyperplanes will be distinct, and thus $E$ (being their intersection) will be an $\left(n-2\right)$-dimensional real subspace of $\mathbb{R}^n$. This ... $\endgroup$ – darij grinberg Dec 14 '16 at 2:05
  • $\begingroup$ ... said, I have no idea how to finish this off now, but a few minutes ago I saw it clearly that $\operatorname{Ker}_{\mathbb{R}} E = \operatorname{Ker}_{\mathbb{R}} C = \operatorname{Ker}_{\mathbb{R}} S$, at least assuming that the latter two kernels also have dimension $n-2$. $\endgroup$ – darij grinberg Dec 14 '16 at 2:13

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