Find an invertible matrix P and a diagonal matrix D such that P^(-1)AP=D for eac

Question

Find an invertible matrix P and a diagonal matrix D such that P^(-1)AP=D for each of the following symmetric matrices.
A = [2 1 0]
[1 2 0]
[0 0 1],

B = [-2 -2 -1]
[-2 2 1]
[-1 1 5]

Explanation / Answer

det(A - ?I) = (2 - ?)^2(1 - ?) - (1 - ?) = (4 - 4? + ?^2)(1 - ?) - (1 - ?) = (3 - 4? + ?^2)(1 - ?) = (1 - ?)^2(3 - ?) so we have two eigenvalues 1, and 3. if A(x1 x2,x3) = (x1,x2,x3) then (2x1 + x2,x1 + 2x2,x3) = (x1,x2,x3) then x1 = -x2 and we can write the solution space as: c1(1,-1,0) + c2(0,0,1) if A(x1 x2,x3) = 3(x1,x2,x3) so that (2x1 + x2,x1 + 2x2,x3) = (3x1,3x2,3x3) we must have x3 = 0, and x1 = x2 so the solution space is c(1,1,0) {(1,-1,0),(0,0,1),(1,1,0)} is a linearly independent set: c1(1,-1,0) + c2(0,0,1) + c3(1,1,0) = 0 implies (c1 + c3,-c1 + c3,c2) = (0,0,0) so c2 = 0 c1 + c3 = 0 implies c3 = -c1, but then this means -c1 + c3 = 0 so -c1 - c1 = 0 --> -2c1 = 0 --> c1 = 0, so c3 = 0. thus the matrix P = [ 1 0 1] [-1 0 1] [ 0 1 0] is invertible, and P^(-1)AP = [1 0 0] [0 1 0] [0 0 3] B is a LOT harder. det(B - ?I) = -?^3 + 5?^2 + 10? -36 this cubic does not factor over the rationals. it does have 3 real roots x1 ~ -2.67813 x2 ~ 2.22569 x3 ~ 5.64144 and so B is similar to D = [x1 0 0] [0 x2 0] [0 0 x3] but finding P explicitly is a bit messy, for example an eigenvector corresponding to x1 is: (1,-(4 + x1),(x1 - 5)/(x1 - 1)) so P should be: [1.............. ...............1........ ..............1........] [-(4 - x1)...............-(4 - x2).............-(4 - x3).....] [(x1 - 5)/(x1 - 1) (x2 - 5)/(x1 - 1) (x3 - 5)/(x3 - 1)]

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