Question

How do you diagonalise the matrix `M = ((4,-1,2),(-4,1,-4),(-5,1,-3))` ?

Answer 1

`((1, 1, 0), (0, 1, 2), (-1, -1, 1))^(-1) ((4, -1, 2), (-4, 1, -4), (-5, 1, -3)) ((1, 1, 0), (0, 1, 2), (-1, -1, 1)) = ((2, 0, 0), (0, 1, 0), (0, 0, -1))`

Explanation:

Here are some steps to diagonalise the matrix:

`M = ((4,-1,2),(-4,1,-4),(-5,1,-3))`

The characteristic polynomial of `M` is:

`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 `t=2`, `t=1` and `t=-1`

Hence a suitable diagonalised matrix is:

`((2, 0, 0), (0, 1, 0), (0, 0, -1))`

Let us now solve for the diagonalising matrix `S` satisfying:

`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 `S` is `((a),(b),(c))` then:

`{ (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 `a=1`, `b=0`, `c = -1`

If the second column of `S` is `((d), (e), (f))` then:

`{ (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 `d=1`, `e=1`, `f=-1`

If the third column of `S` is `((g), (h), (i))` then:

`{ (-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 `g=0`, `h=2`, `i=1`

So we can put:

`S = ((1, 1, 0), (0, 1, 2), (-1, -1, 1))`