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.