given-A-a-square-matrix-non-singular-A-I-find-A-such-that-A-3-I- Tinku Tara June 4, 2023 Matrices and Determinants 0 Comments FacebookTweetPin Question Number 80830 by jagoll last updated on 07/Feb/20 givenAasquarematrixnonsingularA≠I.findAsuchthatA3=I Answered by ~blr237~ last updated on 07/Feb/20 M=diag(1,j,j−)withj=[1,−2π3]theproblemwillbealittledifficultifwewantamatrixwithrealsfactorsinthiscase:letfthelinearapplicationrelatedtoAA3−I=0⇔f3−id=(f−id)∘(f2+f+id)=0∼thenIm(f2+f+id)⊆ker(f−id)ifwewanta2−squarematrix,asA≠Iwewillhaveker(f−id)={0}causeifnotπ(f)(x)=(x−1)(x−a)withaanotherreal(thatcanbeequalto1)butasπ(f)(x)shoulddivide(x3−1)thatshowaisnotreal:absurdSoIm(f2+f+id)={0}thenf2+f+id=0∼usingx2+x+1=34[(2x+13)2+1]wegot(2f+id3)2+id=0∼(nowwejusthavetofind2−squarematrixsuchasB2=−I)andwealreadyhavei≅antidiag(1,−1)and−i≅antidiag(−1,1)sofcanbeA={diag(−12,−12)antidiag(32,−32)ifwewantAa3−squarematrix(sodegree(π(f))=3anditdividex3−1thenπ(f)(x)=(x−1)(x2+x+1)nowtheoremofdecompositionallowthatker(f−id)⊕ker(f2+f+id)=Eandtheareeachofthemsteadyletassumedthatker(f−id)=<u>andker(f2+f+id)=<v,w>B=(u,v,w)isabaseofEsuchasf(ker(f2+f+id))⊆ker(f2+f+id)thereexist(a,b,c,d)∈R∗4suchas{f(v)=av+bwf(w)=cv+dwthatshowthematrixMoftherestrictionoffonker(f2+f+id)andweknowthatitdistinctivepolynomialcanbeπ′(f)(x)=x2−trMx+detMsecondlythetheoremofdecompositiontellusthatπ′(f)=x2+x+1sobyunicitytrM=det=−1then{a+d=−1ad−bc=−1nowjustchooseyourvaluewecantakea=d=−12,thenbc=54,b=5andc=14finalyinthebaseByouwillgotamatrixAsuchasf(u)=u;f(v)=−v2+5w;f(w)=v4−w2 Commented by jagoll last updated on 07/Feb/20 thankyoumister Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-146366Next Next post: calculate-0-arctan-x-2-x-2-4-dx- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![M=diag(1,j,j^− ) with j=[1,−((2π)/3)] the problem will be a little difficult if we want a matrix with reals factors in this case : let f the linear application related to A A^3 −I=0 ⇔ f^3 −id=(f−id)○(f^2 +f+id)=0^∼ then Im(f^2 +f+id)⊆ker(f−id) if we want a 2−square matrix , as A≠I we will have ker(f−id)={0} cause if not π(f)(x)=(x−1)(x−a) with a another real (that can be equal to 1) but as π(f)(x) should divide (x^3 −1) that show a is not real : absurd So Im(f^2 +f+id)={0} then f^2 +f+id=0^∼ using x^2 +x+1=(3/4)[(((2x+1)/( (√3))))^2 +1] we got (((2f+id)/( (√3))))^2 +id=0^∼ ( now we just have to find 2−square matrix such as B^2 =−I ) and we already have i≅anti diag(1,−1) and −i≅anti diag(−1,1) so f can be A= { ((diag(((−1)/2),−(1/2)))),((anti diag(((√3)/2) ,−((√3)/2)))) :} if we want A a 3−square matrix ( so degree(π(f))=3 and it divide x^3 −1 then π(f)(x)=(x−1)(x^2 +x+1) now theorem of decomposition allow that ker(f−id)⊕ker(f^2 +f+id)=E and the are each of them steady let assumed that ker(f−id)=<u> and ker(f^2 +f+id)=<v,w> B=(u,v,w) is a base of E such as f(ker(f^2 +f+id))⊆ker(f^2 +f+id) there exist (a,b,c,d)∈R_∗ ^4 such as { ((f(v)=av+bw)),((f(w)=cv+dw)) :} that show the matrix M of the restriction of f on ker(f^2 +f+id) and we know that it distinctive polynomial can be π′(f)(x)=x^2 −trMx+detM secondly the theorem of decomposition tell us that π′(f)=x^2 +x+1 so by unicity trM=det=−1 then { ((a+d=−1)),((ad−bc=−1)) :} now just choose your value we can take a=d=−(1/2) , then bc=(5/4) , b=5 and c=(1/4) finaly in the base B you will got a matrix A such as f(u)=u ; f(v)=−(v/2)+5w ;f(w)=(v/4)−(w/2)](https://www.tinkutara.com/question/Q80839.png)
