given A a square matrix non singular A≠ I. find A such that A^3 = I


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Question Number 80830 by jagoll last updated on 07/Feb/20

$g i v e n A a s q u a r e m a t r i x n o n$ $s i n g u l a r A \neq I .$ $f i n d A s u c h t h a t A^{3} = I$
	Answered by ~blr237~ last updated on 07/Feb/20
	$M=diag(1,j,j^− ) with j=[1,−((2π)/3)] the problem will be a little difficult if we want a matrix with reals factors in this case : let f the linear application related to A A^3 −I=0 ⇔ f^3 −id=(f−id)○(f^2 +f+id)=0^∼ then Im(f^2 +f+id)⊆ker(f−id) if we want a 2−square matrix , as A≠I we will have ker(f−id)={0} cause if not π(f)(x)=(x−1)(x−a) with a another real (that can be equal to 1) but as π(f)(x) should divide (x^3 −1) that show a is not real : absurd So Im(f^2 +f+id)={0} then f^2 +f+id=0^∼ using x^2 +x+1=(3/4)[(((2x+1)/(√3)))^2 +1] we got (((2f+id)/(√3)))^2 +id=0^∼ ( now we just have to find 2−square matrix such as B^2 =−I ) and we already have i≅anti diag(1,−1) and −i≅anti diag(−1,1) so f can be A= { ((diag(((−1)/2),−(1/2)))),((anti diag(((√3)/2) ,−((√3)/2)))) :} if we want A a 3−square matrix ( so degree(π(f))=3 and it divide x^3 −1 then π(f)(x)=(x−1)(x^2 +x+1) now theorem of decomposition allow that ker(f−id)⊕ker(f^2 +f+id)=E and the are each of them steady let assumed that ker(f−id)=<u> and ker(f^2 +f+id)=<v,w> B=(u,v,w) is a base of E such as f(ker(f^2 +f+id))⊆ker(f^2 +f+id) there exist (a,b,c,d)∈R_∗ ^4 such as { ((f(v)=av+bw)),((f(w)=cv+dw)) :} that show the matrix M of the restriction of f on ker(f^2 +f+id) and we know that it distinctive polynomial can be π′(f)(x)=x^2 −trMx+detM secondly the theorem of decomposition tell us that π′(f)=x^2 +x+1 so by unicity trM=det=−1 then { ((a+d=−1)),((ad−bc=−1)) :} now just choose your value we can take a=d=−(1/2) , then bc=(5/4) , b=5 and c=(1/4) finaly in the base B you will got a matrix A such as f(u)=u ; f(v)=−(v/2)+5w ;f(w)=(v/4)−(w/2)$
	$M = d i a g (1, j, \bar{j}) w i t h j = [1, - \frac{2 π}{3}]$ $t h e p r o b l e m w i l l b e a l i t t l e d i f f i c u l t i f w e$ $w a n t a m a t r i x w i t h r e a l s f a c t o r s$ $i n t h i s c a s e : l e t f t h e l i n e a r a p p l i c a t i o n r e l a t e d t o A$ $A^{3} - I = 0 \Leftrightarrow f^{3} - i d = (f - i d) \circ (f^{2} + f + i d) = \tilde{0}$ $t h e n I m (f^{2} + f + i d) \subseteq k e r (f - i d)$ $i f w e w a n t a 2 - s q u a r e m a t r i x, a s A \neq I w e w i l l h a v e k e r (f - i d) = {0}$ $c a u s e i f n o t π (f) (x) = (x - 1) (x - a) w i t h a a n o t h e r r e a l (t h a t c a n b e e q u a l t o 1)$ $b u t a s π (f) (x) s h o u l d d i v i d e (x^{3} - 1) t h a t s h o w a i s n o t r e a l : a b s u r d$ $S o I m (f^{2} + f + i d) = {0} t h e n f^{2} + f + i d = \tilde{0}$ $u s i n g x^{2} + x + 1 = \frac{3}{4} [{(\frac{2 x + 1}{\sqrt{3}})}^{2} + 1]$ $w e g o t {(\frac{2 f + i d}{\sqrt{3}})}^{2} + i d = \tilde{0} (n o w w e j u s t h a v e t o f i n d 2 - s q u a r e m a t r i x s u c h a s B^{2} = - I)$ $a n d w e a l r e a d y h a v e i ≅ a n t i d i a g (1, - 1) a n d - i ≅ a n t i d i a g (- 1, 1)$ $s o f c a n b e A = {\begin{cases} d i a g (\frac{- 1}{2}, - \frac{1}{2}) \\ a n t i d i a g (\frac{\sqrt{3}}{2}, - \frac{\sqrt{3}}{2}) \end{cases}$ $i f w e w a n t A a 3 - s q u a r e m a t r i x (s o d e g r e e (π (f)) = 3 a n d i t d i v i d e x^{3} - 1$ $t h e n π (f) (x) = (x - 1) (x^{2} + x + 1)$ $n o w t h e o r e m o f d e c o m p o s i t i o n a l l o w t h a t$ $k e r (f - i d) \oplus k e r (f^{2} + f + i d) = E a n d t h e a r e e a c h o f t h e m s t e a d y$ $l e t a s s u m e d t h a t k e r (f - i d) =< u > a n d k e r (f^{2} + f + i d) =< v, w >$ $B = (u, v, w) i s a b a s e o f E$ $s u c h a s f (k e r (f^{2} + f + i d)) \subseteq k e r (f^{2} + f + i d)$ $t h e r e e x i s t (a, b, c, d) \in R_{*}^{4} s u c h a s {\begin{cases} f (v) = a v + b w \\ f (w) = c v + d w \end{cases}$ $t h a t s h o w t h e m a t r i x M o f t h e r e s t r i c t i o n o f f o n k e r (f^{2} + f + i d)$ $a n d w e k n o w t h a t i t d i s t i n c t i v e p o l y n o m i a l c a n b e π^{'} (f) (x) = x^{2} - t r M x + d e t M$ $s e c o n d l y t h e t h e o r e m o f d e c o m p o s i t i o n t e l l u s t h a t π^{'} (f) = x^{2} + x + 1$ $s o b y u n i c i t y t r M = d e t = - 1$ $t h e n {\begin{cases} a + d = - 1 \\ a d - b c = - 1 \end{cases}$ $n o w j u s t c h o o s e y o u r v a l u e$ $w e c a n t a k e a = d = - \frac{1}{2}, t h e n b c = \frac{5}{4}, b = 5 a n d c = \frac{1}{4}$ $f i n a l y i n t h e b a s e B y o u w i l l g o t a m a t r i x A s u c h a s$ $f (u) = u; f (v) = - \frac{v}{2} + 5 w; f (w) = \frac{v}{4} - \frac{w}{2}$
	Commented by jagoll last updated on 07/Feb/20

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