1. Let A be a square matrix of order 2. Prove that A satisfies the equation A² - trace(A)A + det(A)I = 0, where trace(A) = a + d, det(A) = ad - bc, and I is the 2x2 identity matrix.

If we define A to be general matrix of order 2:

> with(linalg):A:=matrix(2,2,[[a,b],[c,d]]);

A :=	[ a	b ]
	[ c	d ]

then we may simply substitute this and the given values for trace(A) and det(A) into the above equation. Define the left hand side:

> LHS:=evalm(A^2-(a+d)*A+(a*d-b*c));

A :=	[ a2 + (- a - d) a + a d	a b + b d + (- a - d) b ]
	[ c a + d c + (- a - d) c	d2 + (- a - d) d + a d ]

We need to simplify this matrix. Here's how:

> for i from 1 to 2 do for j from 1 to 2 do LHS[i,j]:=simplify(LHS[i,j]) od od;
> evalm(LHS);

[ 0	0 ]
[ 0	0 ]

Since we may take 0 to mean the zero matrix, the two sides of the equation are equivalent:

0 = 0
A² - (a + d)A + (ad - bc)I = 0
A² - trace(A)A + det(A)I = 0.

2. Prove that, if A is a 2x2 matrix and Aⁿ = 0 for some natural number n, then A² = 0.

From the above problem we can state that A² = bA + cI if we use the equality A² = trace(A)A - det(A)I and let b = trace(A) and c = (-det(A)).

Now, we need to show that c = 0. To do this we assume that A is invertible. Since A^-1 exists, we may multiply the equation Aⁿ = 0 by (A^-1)ⁿ = A^-n:

(Aⁿ)(A^-n) = 0(A^-n) => A^n-n = 0 => A⁰ = 0 => I = 0

This equation clearly indicates a contradiction. So we conclude that A cannot be invertible. Now assume that c is not 0. The equation A² - bA = cI can be rearranged divided by c. The result is the following: (1/c) * (A² - bA) = I. Utilizing the definition of the identity matrix and associativity, we know A = AI => bA = bAI => bA = bAI. With this new knowledge, we can rewrite this equation as such: (1/c) * A(A - bI) = I. By the definition of an inverse, A - bI has to be equal to c * (A^-1). For this to be true, the matrix A would have to invertible which we proved is not the case. Therefore our assumption is wrong and c = 0, so now we can say that A² = bA.

If we multiply both sides of A² = bA by A^n-2 we achieve the new equation

Aⁿ = bA^n-1. (1)

By substituting n-1 into equation (1) for n we get the equation: A^n-1 = bA^n-1-1 => A^n-1 = bA^n-2. By substituting this new value of A^n-1 into equation (1), we acquire

Aⁿ = b(bA^n-2) =(b²)*(A^n-2). (2)

By substituting n-2 for n into equation (1), we get A^n-2 = b(A^n-3). Substituting this value of A^n-2 into equation (2), we get

Aⁿ = (b²)*b(A^n-3) = Aⁿ = (b³)*(A^n-3). (3)

A pattern is evident:

Aⁿ = b(A^n-1) = (b²)(A^n-2) = (b³)(A^n-3) = . . . = (b^n-2)(A²).

We are given that Aⁿ = 0, hence (b^n-2)(A²) = 0. We conclude that either b^n-2 = 0 or A² = 0. If A² = 0 then we have concluded our proof. If b^n-2 = 0 and we multiply both sides of the equation by b^3-n we get (b^n-2) * (b^3-n) = (b^3-n) * 0 => b = 0. Then A² = bA = 0 in this case also.

3. Find a square 3 x 3 matrix such that A³ is zero but A² is not zero.

If A is the following matrix:

> A:=matrix(3,3,[[0,1,0],[1,0,1],[0,-1,0]]);

A :=	[ 0	1	0 ]
	[ 1	0	0 ]
	[ 0	-1	0 ]

Then A³ is the zero matrix as shown below:

> evalm(A^3);

[ 0	0	0 ]
[ 0	0	0 ]
[ 0	0	0 ]

A² is not the zero matrix though:

> evalm(A^2);

[ 1	0	1 ]
[ 0	0	0 ]
[ -1	0	-1 ]