To obtain the eigen values, subtract the diagonals with a value . Obtain the determinant of the new matrix obtained and equate to 0 to calculate the eigen value. There are many finite solutions for an eigen vector. Find the eigen vector where x=1. Calculate the eigen vectors by solving the simultaneous equations by substitution technique.
The 3×3 matrix is obtained by:
A=[a b c d e f g h i]
The determinant is:
|A|= |a b c d e f g h i| = aei+ bfg + cdh - afh- bdi - ceg
[5 -18 -32 0 5 4 2 -5 -11]=[5-λ -18 -32 0 5-λ 4 2 -5 -11-λ]
=|5-λ -18 -32 0 5-λ 4 2 -5 -11-λ| =0
=(5-λ)(5-λ)(-11-λ)+(-18)(4)(2)+(-32)(0)(-5)-(5-λ)(4)(-5)-(-18)(0)( -11-λ)-(-32)(5-λ)(2)=0
= (25-5λ-5λ+λ^2)( -11-λ) - 144+0 - (5-λ)(-20)-0-(-64)(5-λ)=0
= (25-10λ+λ^2 )(-11-λ)-144-(-100+20λ)-(-320+64λ)=0
= -275-25λ+110λ+10λ^2-11λ^2-λ^3-144+100-20λ+320-64λ = 0
= -λ^3+10λ^2-11λ^2-25λ+110λ-64λ-20λ-275-144+100+320 = 0
=-λ^3-λ^2+λ+1 = 0
=-λ^3-λ^2+λ=-1
= - λ(λ^2+ λ-1)= -1
- λ=-1,thus λ=1
λ^2+ λ-1=-1
λ^2+ λ=-1+1
λ^2+λ=0
λ(λ+ 1)=0
λ=0
λ+1=0
λ=-1
The eigen values are λ= -1,0 and 1
When λ= -1,let x be 1
[5-λ -18 -32 0 5-λ 4 2 -5 -11-λ]= [ 5--1 -18 -32 0 5--1 4 2 -5 -11--1] = [6 -18 -32 0 6 4 2 -5 -10]
[6 -18 -32 0 6 4 2 -5 -10][ x y z]=[0 0 0]
6x-18y-32z=0
0x+6y+4z=0
Let x =1: 18y + 32z = 6 and 6y + 4z =0
6y=4z ; y =4/6 z
18× 4/6 z – 32z = 6
12z – 32z = 6. -20 z = 6.
Z= (-3)⁄(10 )
y= 4⁄6 ×(-3)⁄(10 ) = -1⁄5
[1 ,-1⁄5,-3⁄10]
When λ= 0,let x be 1
[5-λ -18 -32 0 5-λ 4 2 -5 -11-λ]= [5-0 -18 -32 0 5-0 4 2 -5 -11-0] = [5 -18 -32 0 5 4 2 -5 -11]
[5 -18 -32 0 5 4 2 -5 -11][x y z]=[0 0 0)]
5x-18y-32z=0
0x+5y+4z=0
Let x =1; 18y+32z=5 and 5y+4z=0
5y = - 4z, y = (-4)/5 z
18×(-4)⁄5 z+32z=5
-14.4z+32z=5
17.6z=5; z=25⁄88
y= (-4)/5×25⁄88 y = -5⁄22
[1 ,-5⁄22,25⁄88]
When λ= 1,let x be 1
[5-λ -18 -32 0 5-λ 4 2 -5 -11-λ]= [5-1 -18 -32 0 5-1 4 2 -5 -11-1] = 4 -18 -32 0 4 4 2 -5 -12]
[4 -18 -32 0 4 4 2 -5 -12][x y z)]=[0 0 0)]
4x-18y-32z=0
0x+4y+4z=0
Let x =1: 18y+32z=4, 4y+4z=0
4y=-4z, y=-z
18(-z)+32z=4
-18z+32z=4
14z=4, z=2⁄7
y= -2/7
[1 ,-2⁄7,2⁄7]
The eigen vectors are: [1 , -1⁄5, -3⁄10], [1 , -5⁄22, 25⁄88] and [1 , -2⁄7, 2⁄7]
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