If A,B are invertible and the set {A<sup>k</sup> - B<sup>k</sup> | k is a natural number} is finite , then there exists a natural number m such that A<sup>m</sup> = B<sup>m</sup>.
Problem
Source:
Tags: algebra, polynomial, induction, superior algebra, superior algebra solved, first college math post
Valentin Vornicu
18.02.2003 01:03
seems to me that you fogot to mention what A and B are? I suppose matrixes with real entries ... ?
iandrei
18.02.2003 06:50
Yes , A and B are square matrices of size n with complex entries. Sorry , I was a bit in a rush last night
andrei negut
02.03.2003 18:51
So, let's denote A<sup>k</sup>-B<sup>k</sup> by C<sub>k</sub>. Thus, we have C<sub>0</sub>=0 and we must find a C<sub>r</sub>, with r>0, that equals 0. Let us consider the sequences (C<sub>i</sub>,C<sub>i+1</sub>,...,C<sub>i+n</sub>), for all i \geq 0. Because the C<sub>k</sub>'s take on finitely many values, therefore there are only finitely many values that the sequences may take on, because n is fixed (being the order of the matrices). Therefore we will be able to find m \geq 0 and r>0 such that: (C<sub>m</sub>,C<sub>m+1</sub>,...,C<sub>m+n</sub>)= (C<sub>m+r</sub>,C<sub>m+r+1</sub>,...,C<sub>m+r+n</sub>). Thus we have C<sub>m</sub>=C<sub>m+r</sub>, C<sub>m+1</sub>=C<sub>m+r+1</sub>, ..., C<sub>m+n</sub>=c<sub>m+n+r</sub>. Now comes the interesting part. Let X<sup>n</sup>+a<sub>n-1</sub>X<sup>n-1</sup>+...+a<sub>0</sub> and Let X<sup>n</sup>+b<sub>n-1</sub>X<sup>n-1</sup>+...+b<sub>0</sub> be the characteristic polynomials of the matrices A and B, respectively. Because A and B are invertible, and (-1)<sup>n</sup>a<sub>0</sub> is the determinant of A, and (-1)<sup>n</sup>b<sub>0</sub> that of B, a<sub>0</sub> and b<sub>0</sub> will be different from 0. According to the Cayley-Hamilton theorem, we have: A<sup>n</sup>+a<sub>n-1</sub>A<sup>n-1</sup>+...+a<sub>0</sub>=0 and B<sup>n</sup>+b<sub>n-1</sub>B<sup>n-1</sup>+...+b<sub>0</sub>=0, so for any k \geq 0 we have A<sup>n+k</sup>+a<sub>n-1</sub>A<sup>n+k-1</sup>+...+a<sub>0</sub>A<sup>k</sup>=0 (1) and B<sup>n+k</sup>+b<sub>n-1</sub>B<sup>n+k-1</sup>+...+b<sub>0</sub>B<sup>k</sup>=0, which leads to B<sup>n+k</sup>+a<sub>n-1</sub>B<sup>n+k-1</sup>+...+a<sub>0</sub>B<sup>k</sup> = (a<sub>n-1</sub>-b<sub>n-1</sub>)B<sup>n+k-1</sup>+...+(a<sub>0</sub>-b<sub>0</sub>)B<sup>k</sup>=B<sup>k</sup>(-X) (2), where -X= (a<sub>n-1</sub>-b<sub>n-1</sub>)B<sup>n-1</sup>+...+(a<sub>0</sub>-b<sub>0</sub>). By substracting (2) from (1) we obtain, for any k \geq 0, the following: C<sub>n+k</sub>+a<sub>n-1</sub>C<sub>n+k-1</sub>+...+a<sub>0</sub>C<sub>k</sub>=B<sup>k</sup>X. Now let's return to the m and r we found at the beginning of the solution. We have C<sub>n+m</sub>+a<sub>n-1</sub>C<sub>n+m-1</sub>+...+a<sub>0</sub>C<sub>m</sub>=B<sup>m</sup>X and C<sub>n+m+r</sub>+a<sub>n-1</sub>C<sub>n+m+r-1</sub>+...+a<sub>0</sub>C<sub>m+r</sub>=B<sup>m+r</sup>X But for any i with 0 \leq i \leq n we have C<sub>m+i</sub>=C<sub>m+r+i</sub>. Therefore by substracting the two equalities above we have B<sup>m</sup>(B<sup>r</sup>-1)X=0. Because B is invertible we have (B<sup>r</sup>-1)X=0. Now, we know that C<sub>n+m-1</sub>+a<sub>n-1</sub>C<sub>n+m-2</sub>+...+a<sub>0</sub>C<sub>m-1</sub>=B<sup>m-1</sup>X and C<sub>n+m+r-1</sub>+a<sub>n-1</sub>C<sub>n+m+r-2</sub>+...+a<sub>0</sub>C<sub>m+r-1</sub>=B<sup>m+r-1</sup>X. By substracting we have a<sub>0</sub>(C<sub>m+r-1</sub>-C<sub>m-1</sub>)=B<sup>m-1</sup>(B<sup>r</sup>-1)X=0, therefore C<sub>m+r-1</sub>=C<sub>m-1</sub>. In the same manner we obtain C<sub>m+r-2</sub>=C<sub>m-2</sub>,..., and by induction, in the end we will have C<sub>r</sub>=C<sub>0</sub>=0, which completes the proof. ______________________________________________________
iandrei
02.03.2003 22:47
Funny thing ! It seems like this is the only solution , in the book (the dreaded Caragea:D) there is a solution in the same manner , slightly simplified.
Valentin Vornicu
02.03.2003 23:02
nevertheless andrei negut's solution look nice to me ... i would have probably cooked up the same solution well done andrei
Z_Math404
22.05.2017 21:33
Valentin Vornicu wrote: nevertheless andrei negut's solution look nice to me ... i would have probably cooked up the same solution well done andrei yeah nice job
mathpro12345
16.03.2021 01:09
iandrei wrote: If A,B are invertible and the set {A<sup>k</sup> - B<sup>k</sup> | k is a natural number} is finite , then there exists a natural number m such that A<sup>m</sup> = B<sup>m</sup>. what does <sup> mean????
jxc516
16.03.2021 03:23
mathpro12345 wrote: iandrei wrote: If A,B are invertible and the set {A<sup>k</sup> - B<sup>k</sup> | k is a natural number} is finite , then there exists a natural number m such that A<sup>m</sup> = B<sup>m</sup>. what does <sup> mean???? its actually [sup][/sup] but it is in AoPS wiki formatting
TheBridgerMC
11.06.2024 21:38
Man, the entire time and history of these replies are all screwed up