Example 1: Diagonalization of a matrix
The matrix
 A = 2 -2 1 -1 3 -1 2 -4 3
has eigenvalues 1,2 = 1 and 3 = 6. The corresponding eigenvalues are 1,2 = 1, 1,2 = t ( 2, 1, 0 ) + s ( -1, 0, 1 ), s, t C, s, t 0 3 = 6, 3 = r ( 1, -1, 2 ), r C r 0.
We choose t = s = r = 1 and thus have 1 = ( 2, 1, 0 ) 2 = ( -1, 0, 1 ) 3 = ( 1, -1, 2 )
From eigenvectors i we form the matrix T.
 det T = 2 -1 1 1 0 -1 0 1 2
 = 1 -1 1 1 2
 - 1 2 -1 0 1
-Compute across the 2nd row

= -2 - 1 - 2 + 0 = -5 0
=> { 1, 2, 3 } linearly independent. => A is diagonalizable

The matrix D is a diagonal matrix, which has the eigenvalues of A as diagonal elements:

 D = 1 0 0 0 1 0 0 0 6
We compute T-1:
 (T | I ) = 2 -1 1 1 0 0 1 0 -1 0 1 0 0 1 2 0 0 1
 ~ 1 0 -1 0 1 0 0 -1 3 1 -2 0 0 1 2 0 0 1
 -Add the 2nd row multiplied by -2 to the 1st row -Interchange the 1st and the 2nd row
 -Add the 2nd row to the 3rd row -Multiply the 2nd row by -1
 ~ 1 0 -1 0 1 0 0 1 -3 -1 2 0 0 0 5 1 -2 1
 ~ 1 0 0 1/5 3/5 1/5 0 1 0 -2/5 4/5 3/5 0 0 1 1/5 -2/5 1/5
 -Add the 3rd row multiplied by 1/5 to the 1st row -Add the 3rd row multiplied by 3/5 to the 2nd row/td> -Multiply the 3rd row by 1/5
 -Add the 2nd row to the 3rd row -Multiply the 2nd row by -1
We thus have
 T-1 = 1/5 3/5 1/5 -2/5 4/5 3/5 1/5 -2/5 1/5
Now A has been diagonalized A = T D T -1
 = 2 -1 1 1 0 -1 0 1 2 1 0 0 0 1 0 0 0 6 1/5 3/5 1/5 -2/5 4/5 3/5 1/5 -2/5 1/5
(The result can be verified by multiplying the matrices.)
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