在俄罗斯学到的第一个trick First trick I learnt in Russia

考虑一个$n\times n$级方阵$A$, 每个位置都是$0,1,-1$.如果$A$的特征值$\lambda$取到特征值绝对值的最大值. 假设$\lambda$的几何重数(特征子空间的维数)是$m$. 那么$A$的任意$(n-m+1)\times (n-m+1)$级主子阵一定有一列非零元数目超过$|\lambda|$. 

Consider a matrix $A$ of size $n\times n$ with each indices $0$, $1$ or $-1$. Assume that the eigenvalue $\lambda$ takes the maximum of the absolute value of all eigenvalues. Assume that the geometrically multiplicity (dimension of eigenspace) of $\lambda$ is $m$. Then any $(n-m+1)\times (n-m+1)$ principal submatrix has a row with nonzero indices more than $|\lambda|$.

证明如下.

The proof is as follow.

假设$A=(A_{\alpha\beta})_{n\times n}$. 不妨假设就是前$(n-m+1)$级子式. 取$n\times m$级矩阵$B$是特征子空间的基组成的矩阵. 考虑$B$的末$n-(n-m+1)=m-1$行$B^*$, 因为降秩, 存在非零向量$x$使得$B^*x=0$. 此时$y=Bx$前$n-m+1$行不全为零, 末$m-1$ 行全为$0$. 假设$y=(y_{\alpha})_{n\times 1}$, 那么

$$\begin{array}{rll}|\lambda|\cdot |y_{\alpha}|&=|(\lambda y)_{\alpha}|=|(\lambda Bx)_{\alpha}|= |(ABx)_{\alpha}|=|(Ay)_{\alpha}|\\& \leq \sum_{\beta} |A_{\alpha\beta}| |y_{\beta}|= \sum_{\beta \leq n-m+1} |A_{\alpha\beta}| |y_{\beta}| & \textrm{取$\alpha$使得$|y_{\alpha}|$最大} \\& \leq |y_{\alpha}|\sum_{\beta\leq n-m+1} |A_{\alpha\beta}| = |y_{\alpha}|\cdot \#\{\textrm{$\alpha$行的非零元素}\}\end{array}$$

两边消去$|y_{\alpha}|$即知结果.

Assume that $A=(A_{\alpha\beta})_{n\times n}$. WLOG assume it is exactly the first $n-m+1$ rows and columns. Taking the matrix of basis of eigenspace belonging to $\lambda$ whose size is $n\times m$. Denote $B^*$ the last $n-(n-m+1)=m-1$ rows of $B$. Then there exists nonzero vector $x$ such that $B^* x=0$ by the argument of rank. Now $y=Bx$ has first $n-m+1$ rows not all zero, but last $m-1$ rows vanishing. Assume that $y=(y_{\alpha})_{n\times 1}$, then

$$\begin{array}{rll}
|\lambda|\cdot |y_{\alpha}|&=|(\lambda y)_{\alpha}|=|(\lambda Bx)_{\alpha}|= |(ABx)_{\alpha}|=|(Ay)_{\alpha}|\\& \leq \sum_{\beta} |A_{\alpha\beta}| |y_{\beta}|\\
&= \sum_{\beta \leq n-m+1} |A_{\alpha\beta}| |y_{\beta}| & \textrm{taking $\alpha$ such that $|y_{\alpha}|$ is maximal} \\& \leq |y_{\alpha}|\sum_{\beta\leq n-m+1} |A_{\alpha\beta}| \\& = |y_{\alpha}|\cdot \#\{\textrm{nonzero indices of row $\alpha$}\}\end{array}$$

Cancelling $|y_{\alpha}|$ of both sides completes the proof.

应用: $n$维超立方体$Q_n$的任意顶点数$\geq 2^{n-1}+1$的子图必有顶点$\deg \geq \sqrt{n}$.

Application. Any subgraphs of $n$-dimensional hyper-cube $Q_n$ who has more than $2^{n-1}+1$ vertices possesses a vertex with degree $\geq \sqrt{n}$.

解释: 考虑$n$位的2进制码, 如果两串码只相差1位, 就说是容易弄混的, 那么任意一半以上的码拿出来之中一定有一个码和$\sqrt{n}$个码容易弄混.

Explanation. Consider the $0$-$1$ code with length $n$. If two codes differ only one position, we say them difficult to distinguish. If we pick more than a half of codes from all of them, there always exists a code difficult to distinguish with $\sqrt{n}$ codes.

证明: 考虑$A_0=(0), A_n=\left(\begin{matrix}A_{n-1} & I \\ I & -A_{n-1}\end{matrix}\right)$, 那么归纳可证$A_n^2=nI$.

Proof. Consider $A_0=(0), A_n=\left(\begin{matrix}A_{n-1} & I \\ I & -A_{n-1}\end{matrix}\right)$, now $A_n^2=nI$ by induction.