Problem. Prove that if A is a 3 by 3 matrix and An=0 for some n then A3=0.
Problem 1 from the homework makes us guess that A must satisfy an equation of
the form
A3-tA2+qA-d
where t=trace(A), d=det(A). Using Maple, we found the q.
> with(linalg);
> A:=matrix(3,3,[[a,b,c],[d,e,f],[g,h,i]]);
| A := |
[ a |
b |
c ] |
| [ d |
e |
f ] |
| [ g |
h |
i ] |
> B:=evalm(A3-trace(A)*A2-det(A));
B :=
[a3 + 2 a b d + 2 a c g + d b e + g c i + (- a - e - i) (a2 + b d + c g) - a e i
+ a f h + d b i + g c e, b a2 + b2 d + b c g + e a b + b e2 + e c h + h a c
+ h b f + h c i + (- a - e - i) (a b + b e + c h), c a2 + c b d + c2 g
+ f a b + f b e + f c h + i a c + i b f + c i2
+ (- a - e - i) (a c + b f + c i)]
[d a2 + a e d + a f g + b d2 + d e2 + d f h + g d c + g e f + g f i
+ (- a - e - i) (d a + e d + f g), a b d + 2 d b e + e3 + 2 e f h + h f i
+ (- a - e - i) (b d + e2 + f h) - a e i + a f h + d b i + g c e, c d a
+ c e d + c f g + f b d + f e2 + f2 h + i d c + i e f + f i2
+ (- a - e - i) (d c + e f + f i)]
[g a2 + a h d + a i g + d g b + d h e + d i h + c g2 + g f h + g i2
+ (- a - e - i) (g a + h d + i g), b g a + b h d + b i g + e g b + h e2
+ e i h + h c g + f h2 + h i2 + (- a - e - i) (g b + h e + i h), a c g
+ 2 g c i + e f h + 2 h f i + i3 + (- a - e - i) (c g + f h + i2 ) - a e i
+ a f h + d b i + g c e]
> for m from 1 to 3 do for n from 1 to 3 do B[m,n]:=simplify(B[m,n]) od od;
> > evalm(B);
[a b d + a c g - e a2 - i a2 - a e i + a f h,
b2 d + b c g - e a b + h b f - i a b - i b e,
c b d + c2 g + f c h - i a c - e a c - e c i]
[- a e d + b d2 + d f h + g d c - i d a - i e d,
d b e + e f h - a e2 - i e2 - a e i + g c e,
c f g + f b d + f2 h - i e f - a e f - a f i]
[- a i g + d g b + c g2 + g f h - e g a - e i g,
b h d - e i h + h c g + f h2 - a h e - a i h,
g c i + h f i - a i2 - e i2 - a e i + d b i]
> evalm("/A);
[b d + c g - a e - a i - e i + f h, 0, 0]
[0, b d + c g - a e - a i - e i + f h, 0]
[0, 0, b d + c g - a e - a i - e i + f h]
Thus q=bd+cg-ae-ai-ei+fh.
We can state that A3 = tA2 -qA +dI.
We know that An is 0 for some n>0. If n<=3 then A3 is zero (zero multiplied by A or A2 is zero) and we are done.
So suppose that n>3. We can also suppose that n is the smallest number such that An=0.
Multiply our equation by An-1. We get
An+2=tAn+1-qAn+dAn-1
Since An=An+1=An+2=0, we have that dAn-1=0. But An-1
is not zero by our assumption (since n-1<n), so d=0. Thus
A3=tA2-qA
Multiply by An-2:
An+1=tAn-qAn-1
Again An+1=An=0, so qAn-1=0. Thus q=0. So
A3=tA2
Multiply by An-3 (notice that n-3>1 since n>3):
An=tAn-1
And we get t=0. So A3=0. The proof is complete.
Problem. Prove that if A is a 3 by 3 matrix and An=0 for some n then A3=0.
Problem 1 from the homework makes us guess that A must satisfy an equation of
the form
A3-tA2+qA-d
where t=trace(A), d=det(A). Using Maple, we found the q.
> with(linalg);
> A:=matrix(3,3,[[a,b,c],[d,e,f],[g,h,i]]);
| A := |
[ a |
b |
c ] |
| [ d |
e |
f ] |
| [ g |
h |
i ] |
> B:=evalm(A^3-trace(A)*A^2-det(A));
If our guess is correct, B must be proportional to A. Using Maple,
we can find the coefficient, q.
> q:=B[1,1]/A[1,1];
This gives q=bd+cg-ae-ai-ei+fh.
Check that qA=B
>evalm(B-q*A);
The result is 0, so our guess was correct and q=bd+cg-ae-ai-ei+fh.
We can state that A3 = tA2 -qA +dI.
We know that An is 0 for some n>0. If n<=3 then A3 is zero (zero multiplied by A or A2 is zero) and we are done.
So suppose that n>3. We can also suppose that n is the smallest number such that An=0.
Multiply our equation by An-1. We get
An+2=t An+1-qAn+dAn-1
Since An=An+1=An+2=0, we have that dAn-1=0. But An-1
is not zero by our assumption (since n-1*<n), so d=0. Thus
A3=tA2-qA
Multiply by An-2:
An+1=tAn-qAn-1
Again An+1=An=0, so qAn-1=0. Thus q=0. So
A3=tA2
Multiply by An-3 (notice that n-3>1 since n>3):
An=tAn-1
And we get t=0. So A3=0. The proof is complete.