How do you diagonalise the matrix #M = ((4,-1,2),(-4,1,-4),(-5,1,-3))# ?
1 Answer
Explanation:
Here are some steps to diagonalise the matrix:
#M = ((4,-1,2),(-4,1,-4),(-5,1,-3))#
The characteristic polynomial of
#p(t) = det(M - tI)#
#color(white)(p(t)) = abs((4-t, -1, 2), (-4, 1-t, -4), (-5, 1, -3-t))#
#color(white)(p(t)) = (4-t)abs((1-t, -4), (1, -3-t))-1abs((-4, -4), (-3-t, -5))+2abs((-4, 1-t), (-5, 1))#
#color(white)(p(t)) = (4-t)(t^2+2t+1)-(-4t+8)+2(-5t+1)#
#color(white)(p(t)) = -t^3+2t^2+t-2#
#color(white)(p(t)) = -(t-2)(t-1)(t+1)#
The eigenvalues are the zeros
Hence a suitable diagonalised matrix is:
#((2, 0, 0), (0, 1, 0), (0, 0, -1))#
Let us now solve for the diagonalising matrix
#S^(-1)((4, -1, 2),(-4, 1, -4),(-5, 1, -3))S = ((2, 0, 0),(0, 1, 0),(0, 0, -1))#
If the first column of
#{ (2a = 4a-b+2c " " rarr " " 0 = 2a-b+2c), (2b = -4a+b-4c " " rarr " " 0 = -4a-b-4c), (2c = -5a+b-3c " " rarr " " 0 = -5a+b-5c) :}#
...one solution of which is
If the second column of
#{ (d = 4d-e+2f " " rarr " " 0 = 3d-e+2f), (e = -4d+e-4f " " rarr " " 0 = -4d-4f), (f = -5d+e-3f " " rarr " " 0 = -5d+e-4f) :}#
...one solution of which is
If the third column of
#{ (-g=4g-h+2i " " rarr " " 0=5g-h+2i), (-h=-4g+h-4i " " rarr " " 0=-4g+2h-4i), (-i=-5g+h-3i " " rarr " " 0=-5g+h-2i) :}#
...one solution of which is
So we can put:
#S = ((1, 1, 0), (0, 1, 2), (-1, -1, 1))#