>	with(linalg);

Warning, the protected names norm and trace have been redefined and unprotected

>	A:=matrix([[1, -1, 0, 0, 0, 0], [4, 5, 0, 0, 0, 0], [0, 0, 0, -1, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, -1, 0]]);

>	charpoly(A, t);

>	factor(%);

>	factor(minpoly(A, t));

>	#A is not diagonalizable because its minimal polynomial has repeated roots.

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

>	factor(charpoly(B,t));

>	factor(minpoly(B, t));

>	#B is diagonalizable over the complex numbers. Lets verify that this is indeed the minimal polynomial.

>	multiply(B-3, B^2 + 1);

>	#Lets diagonalize B. We calculate the eigenspaces

>	ESP(3) := kernel(B-3);

>

#Check:

>	multiply(B, ESP(3)[1]);

>	ESP(I):=kernel(B-I);

>

#Check:

>	multiply(B, ESP(I)[1]);

>	evalm(I*ESP(I)[1]);

>	ESP(minusI):=kernel(B+I);

>

#Check:

>	multiply(B, ESP(minusI)[1]);

>	evalm(-I*ESP(minusI)[1]);

>	#By putting the eigenvectors together we get a matrix M that diagonalizes B

>	M:=transpose(matrix([ESP(3)[1], ESP(3)[2], ESP(I)[1], ESP(I)[2], ESP(minusI)[1], ESP(minusI)[2]]));

>	multiply(inverse(M), B, M);

>	#Later we'll learn about the Jordan form. If a matrix is diagonalizable then its Jordan form is diagonal. Else, it's not.

>	jordan(A);

>	jordan(B);

>	#In fact, Maple can givs us the transition matrix too.

>	jordan(B, 'N');

>

print(N);

>	multiply(inverse(N), B, N);

>