Page 429 - 35Linear Algebra

G.14 Diagonalizing Symmetric Matrices 429

by finding the eigenvalues of this

 
1 − λ 2 0
1 − λ 0
det  2 1 − λ 0  = (1 − λ)
0 5 − λ
0 0 5 − λ

2 0 2 1 − λ
− (2) + 0
0 5 − λ 0 0
= (1 − λ)(1 − λ)(5 − λ) + (−2)(2)(5 − λ) + 0
2
= (1 − 2λ + λ )(5 − λ) + (−2)(2)(5 − λ)
2
= ((1 − 4) − 2λ + λ )(5 − λ)
2
= (−3 − 2λ + λ )(5 − λ)
= (1 + λ)(3 − λ)(5 − λ)

So we get λ = −1, 3, 5 as eigenvectors. First find v 1 for λ 1 = −1
       
x 2 2 0 x 0
(M + I)  y  =  2 2 0   y  =  0  ,
z 0 0 6 z 0
 
1
implies that 2x + 2y = 0 and 6z = 0,which means any multiple of v 1 =  −1  is
0
an eigenvector with eigenvalue λ 1 = −1. Now for v 2 with λ 2 = 3

       
x −2 2 0 x 0
(M − 3I)  y  =  2 −2 0   y  =  0  ,
z 0 0 4 z 0
 
1
and we can find that that v 2 =  1  would satisfy −2x + 2y = 0, 2x − 2y = 0 and
0
4z = 0.
Now for v 3 with λ 3 = 5
       
x −4 2 0 x 0
(M − 5I)  y  =  2 −4 0   y  =  0  ,
z 0 0 0 z 0

Now we want v 3 to satisfy −4x + 2y = 0 and 2x − 4y = 0, which imply x = y = 0,
 
0
but since there are no restrictions on the z coordinate we have v 3 =  0  .
1
Notice that the eigenvectors form an orthogonal basis. We can create an
orthonormal basis by rescaling to make them unit vectors. This will help us

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