Timeline for Walks of odd Lengths in a Matrix
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 22, 2018 at 17:53 | comment | added | Fedor Petrov | Write $x^n=(x^4+a^2 x^2 +1)h(x)+r(x)$, where $\deg r<4$ over $\mathbb{F}_2 [a]$. For even $n$ the polynomial $r$ is even, for odd $n$ it is odd. We have $B^n=r(B)$. Thus $B^n$ is a linear combination either of $1$ and $B^2$ or of $B$ and $B^3$. This yields zeroes in both cases. | |
| Aug 22, 2018 at 17:07 | comment | added | user0410 | A variable over $\mathbb{F} _2$. I upload its power to $6$ in this phto. Thanks | |
| Aug 22, 2018 at 16:59 | comment | added | Fedor Petrov | $a$ is real or variable over $\mathbb{F} _2$ or what? I am confused. | |
| Aug 22, 2018 at 16:57 | comment | added | user0410 | Excuse me I wanted to wrote ${\bf B}^k$ be non-zero over $\mathbb{F}_2$. | |
| Aug 22, 2018 at 16:55 | comment | added | Fedor Petrov | This looks false, the entries of say $B^{10}$ are non-zero polynomials in $a$. | |
| Aug 22, 2018 at 16:47 | comment | added | user0410 | Consider $a\in \mathbb{R}$ is a non-zero element. Consider the matrix $\bf A$ in question such that $A[1,1]=A[3,3]=a$. Please call this new matrix $\bf B$. Based on your comment, how to prove that there is no a positive integer $k$ such that all entries of ${\bf B}^k$ be non-zero over $\mathbb{R}$. The determinant of $\bf B$ is $1$ and its characteristic polynomial is $\left( {\lambda}^{2}+a\lambda+1 \right) ^{2}$ over $\mathbb{F}_2$. I appreciate you taking the time to help me. | |
| Aug 22, 2018 at 13:07 | comment | added | Fedor Petrov | Even if $A$ were singular, the powers of $A$ are eventually periodic modulo 2 (as any map on a finite set is eventually periodic) and once you find $k<n$ such that $A^k$ and $A^n$ coincide modulo 2, you just have to check that all matrices $A,A^2,\dots,A^{n-1}$ contain even entries. | |
| Aug 22, 2018 at 13:03 | vote | accept | user0410 | ||
| Aug 22, 2018 at 13:03 | comment | added | user0410 | Very clear and sweet answer. You know, how much I was thinking about this problem from graph theory point of view. Really, really thanks. | |
| Aug 22, 2018 at 12:59 | history | answered | Fedor Petrov | CC BY-SA 4.0 |