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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