国外论坛好题
Sylvester equation https://en.wikipedia.org/wiki/Sylvester_equation
To seek the maximum value of $S=x_1x_2+x_2x_3+\cdots+x_nx_1$ on this domain:
补充：这是Ky Fan-Tausski-Todd inequality，参考Converses of two inequalities of Ky Fan, O. Taussky, and J. Todd以及这里。
$x_1+x_2+\cdots+x_n=0$ and
$x_1^2+x_2^2+\cdots+x_n^2=1$.
I have made some trivial observations:
1) $S\in[-1,1)$ by the rearrangement inequality.
2) We can make $S$ arbitrarily close to $1$ by increasing $n$.
3) An equivalent problem is to minimise $(x_1-x_2)^2+\cdots+(x_n-x_1)^2$.
But does the maximum have a meaningful closed form for each $n$?
I propose to do a discrete Fourier transform. To this end put $\omega:=e^{2\pi i/n}$. In the following all sums are over ${\mathbb Z}_n$, unless indicated otherwise. Let
$$y_k:={1\over n}\sum_l x_l\omega^{-kl}\ .$$
Since the $x_l$ are real we have $$y_{-k}=\overline{y_k}\tag{1}$$ for all $k$, furthermore $y_0=\sum_lx_l=0$. One has Parseval's formula
$$n\sum_k y_k\overline{ y_k}=\sum_l x_l^2=1\tag{2}$$
and the inversion formula
$$x_j=\sum_k y_k\omega^{jk}\ .\tag{3}$$
Using $(3)$ one computes
$$S=\sum_j x_jx_{j+1}=\ldots=n\sum_k|y_k|^2\omega^k\ .\tag{4}$$
At this point we have to distinguish the cases (a) $n=2m$, and (b) $n=2m+1$.
(a) If $n=2m$ then $\omega^m=-1$, and $(4)$ gives
$$\eqalign{S&=n\left(\sum_{k=1}^{m-1}|y_k|^2(\omega^k+\omega^{-k}) \ +|y_m|^2\omega^m\right)\cr
&=n\left(\sum_{k=1}^{m-1}|y_k|^2\>2\cos{2k\pi\over n} \ -|y_m|^2\right)\ .\cr}$$
Given the conditions $(1)$ and $(2)$ it is easily seen that the optimal admissible choice of the $y_k$ is $$y_1=y_{-1}={1\over\sqrt{2n}}\>,\qquad y_k=0\quad(k\ne\pm1)\ .\tag{5}
$$ This leads to
$$S_{\rm opt}=\cos{2\pi\over n}\ .$$
In particular when $n=4$ one obtains $S_{\rm opt}=0$, as indicated in Zubzub's answer.
(b) If $n=2m+1$ then $(4)$ gives
$$S=2n\sum_{k=1}^m |y_k|^2\cos{2k\pi\over n}\ ,$$
and the choice $(5)$ leads again to
$$S_{\rm opt}=\cos{2\pi\over n}\ .$$
In particular when $n=3$ one obtains $S_{\rm opt}=-{1\over2}$, and when $n=5$ one obtains $S_{\rm opt}=\cos{2\pi\over5}\doteq0.309$, as indicated in Zubzub's answer.
Let $$
A=\left[ \begin{matrix}
A_1& A_3\\
0& A_2\\
\end{matrix} \right]\quad \text{and} \quad B=\left[ \begin{array}{c}
B_1\\
B_2\\
\end{array} \right],
$$and for any eigenvalue $s$ of $A$, we have
$$
\mathrm{rank}\,\left[ A-sI_n,B \right] =n.
$$
Prove there are a real matrix $K\in \mathbb{R}^{r\times n}$ and invertible matrix $T\in \mathbb{R}^{n\times n}$, such that
$$
T\left( A+BK \right) T^{-1}=\left[ \begin{matrix}
A_1& 0\\
0& \bar{A}_2\\
\end{matrix} \right] ,\qquad TB=\left[ \begin{array}{c}
\bar{B}_1\\
B_2\\
\end{array} \right],
$$
where $A\in \mathbb{R}^{n\times n},B\in \mathbb{R}^{n\times r}$. Meanwhile, $\bar{A}_2$ and $\bar{B}_1$ is real matrix of the proper dimension, $\bar{A}_2$ and $A_1$ Have eigenvalues that are not identical to each other.
This problem is about controllability of linear system and pole assignment, so I think we may try controllability canonical form, or let $K=(K_1,K_2)$, then determine the $K_1,K_2$ and $T$, but it seems so difficult.
The statement about the rank of $\left[A - s\,I, B\right]$ implies that the pair $(A,B)$ is [controllable](https://en.wikipedia.org/wiki/Hautus_lemma). Therefore the poles of the resulting matrix should be able to be placed anywhere. The similarity transformation should allow us to separate the eigenvalues/modes and therefore achieve the stated goal
$$
T \left(A + B\,K\right) T^{-1} =
\begin{bmatrix}
A_1 & 0 \\ 0 & \bar{A}_2
\end{bmatrix}, \quad
T\,B =
\begin{bmatrix}
\bar{B}_1 \\ B_2
\end{bmatrix},
$$
given that
$$
A =
\begin{bmatrix}
A_1 & A_3 \\ 0 & A_2
\end{bmatrix}, \quad
B =
\begin{bmatrix}
B_1 \\ B_2
\end{bmatrix}.
$$
For solving this it is easier to separate the problem in smaller problems. For this I will define $T$ and $K$ as
$$
T =
\begin{bmatrix}
T_1 & T_2 \\ T_3 & T_4
\end{bmatrix}, \quad
K =
\begin{bmatrix}
K_1 & K_2
\end{bmatrix}.
$$
The last goal requires
$$
T\,B =
\begin{bmatrix}
T_1\,B_1 + T_2\,B_2 \\ T_3\,B_1 + T_4\,B_2
\end{bmatrix} =
\begin{bmatrix}
\bar{B}_1 \\ B_2
\end{bmatrix}.
$$
The bottom half of this goal might have infinitely many solution if there is any overlap in the span of $B_1$ and $B_2$. But this problem should be solvable in general, in which case $T_3=0$ and $T_4=I$ should always solve it. Using this then the inverse of $T$ can shown to be
$$
T^{-1} =
\begin{bmatrix}
T_1 & T_2 \\ 0 & I
\end{bmatrix}^{-1} =
\begin{bmatrix}
T_1^{-1} & -T_1^{-1}\,T_2 \\ 0 & I
\end{bmatrix}.
$$
Since nothing is specified about $\bar{B}_1$ then $T_1$ and $T_2$ could be anything for now as long as $T_1$ is invertible. The left hand side of the first goal can now be written as
$$
T \left(A + B\,K\right) T^{-1} =
\begin{bmatrix}
T_1\,A_1\,T_1^{-1} + \bar{B}_1\,K_1\,T_1^{-1} & T_1\,A_3 + T_2\,A_2 - T_1\,A_1\,T_1^{-1}\,T_2 + \bar{B}_1\left(K_2 - K_1\,T_1^{-1}\,T_2\right) \\
B_2\,K_1\,T_1^{-1} & A_2 + B_2\left(K_2 - K_1\,T_1^{-1}\,T_2\right)
\end{bmatrix}.
$$
It can be shown that the top and bottom left half can be set equal to the goal by using $T_1=I$ and $K_1=0$. This allows the top and bottom right of the first goal equation to be simplified to
$$
\begin{bmatrix}
A_3 + T_2\,A_2 - A_1\,T_2 + \bar{B}_1\,K_2 \\
A_2 + B_2\,K_2
\end{bmatrix} =
\begin{bmatrix}
0 \\
\bar{A}_2
\end{bmatrix}.
$$
For the bottom half of this equation you could just use a pole placement algorithm to find $K_2$, such that none of the poles match those of $A_1$. This should be possible since the pair $(A_2,B_2)$ should be controllable. The top half can then be rewritten as
$$
A_3 + T_2\,\bar{A}_2 - A_1\,T_2 + B_1\,K_2 = 0,
$$
which can be transformed into a [Sylvester equation](https://en.wikipedia.org/wiki/Sylvester_equation)
$$
A\,X + X\,B = C,
$$
with $X = T_2$, $A = -A_1$, $B = \bar{A}_2$ and $C = -A_3 - B_1\,K_2$. This equation has an unique solution for $X$ when $A$ and $-B$ do not have a common eigenvalue. This is an identical constraint as mentioned by your problem statement.
So to solve this problem you can first do a pole placement with the pair $(A_2,B_2)$ to find $K_2$, avoiding the eigenvalues of $A_1$. And then solve a Sylvester equation after substituting in this obtained $K_2$ in order to find $T_2$. Using these values then the final solution can then be expressed using
$$
T =
\begin{bmatrix}
I & T_2 \\ 0 & I
\end{bmatrix}, \quad
K =
\begin{bmatrix}
0 & K_2
\end{bmatrix}.
$$
2022年9月06日 03:04
Minister of Education, Bangladesh Mr. Nahid Hasan has announced this year PSC Result 2022 will be announced in the last week of December 2022 with the total or full mark sheet of the Primary School Certificate Exam, Grade 5 Result dhaka Board based on previous years schedule, once the official date is confirmed we will update here with timings.Previously this Prathomik Somaponi Result 2022 was announced on 30th or 31st December, and this year also announced same for the class 5th grade result with merit or toppers list asper regular GPA Marking schedule, and Directorate of Primary Education (DPE) has announced this year is very tough to conduct evaluation process to announce subject wise marks for Grade 5 terminal exams.