Hi,
I have got the following matrix. I have found the eigen values but in some eq x, y & z terms are vanishing, so how to find the value of eigen vector??? Also why we have to do normalization??
A__=__[1__1__0]
______[1__1__0]
______[0__0__1]

A-λI=0
[1__1__0]_-_[λ__0__0]_____=0____________________
[1__1__0]___[0__λ__0]___________________________
[0__0__1]___[0__0__λ]___________________________
______________________________________________
[1-λ__1__0]=0__________________________________
[1___1-λ__0]____________________________________
[0___0____1-λ]__________________________________
_______________________________________________
1-λ|1-λ__0|____-1|1___0__| +0 =0__________________
___|0__1-λ|______|0___1-λ_|________________________
_______________________________________________
(1-λ)(1-λ)^2____-(1-λ) =0
Taking_(1-λ) common
(1-λ)[(1-λ)^2__-1]=0
First eigen value λ1 = 1
Now consider:
[(1-λ)^2_____-1]=0
1-2λ+λ^2-1=0
-2λ+λ^2=0
λ(λ-2)=0
λ2=0
& λ3=2
______________________________________________________
______________________________________________________
For λ1 = 1
Ax = λ*x*I
_____________________________________________________
[1__1__0]_______[x]___= 1 * [x]___________________________
[1__1__0]_______[y]_______[y]___________________________
[0__0__1]_______[z]_______[z]___________________________
______________________________________________________
x+y=x------eq(1)
x+y=y------eq(2)
z=z---------eq(3)
______________________________________________________
Above x vanishes in eq(1), y vanishes in eq(2) & z vanishes in eq(3).
What is the value of μ1?????
______________________________________________________
For λ2=0_______________________________________________
[1__1__0]_______[x]___= 0 * [x]____________________________
[1__1__0]_______[y]_______[y]____________________________
[0__0__1]_______[z]_______[z]____________________________
______________________________________________________
x+y=0----------eq(4)_______________________________________
x+y=0----------eq(5)_______________________________________
z=0-------------eq(6)_______________________________________
______________________________________________________
In my view it should be μ=[-1]_______________________________
_____________________[-1]______________________________
_____________________[0]_______________________________
but teacher has got different answer. In addition to this he has done normalization, please guide me what is the need for normalization_______________________________
______________________________________________________
For λ=2________________________________________________
_____________________________________________________
[1__1__0]_______[x]___=_2_*_ [x]___________________________
[1__1__0]_______[y]_______[y]___________________________
[0__0__1]_______[z]_______[z]___________________________
x+y=2x------eq(7)-----------------------------------------------------------------
x=-y----------------------------------------------------------------------------------
x+y=2y-------eq(8)----------------------------------------------------------------
y=x----------------------------------------------------------------------------------
z=2z----------eq(9)---------------------------------------------------------------
Therefor, μ=[-1]_________________________________________
__________[1]_________________________________________
__________[0]___________________________________________
Why we have to normalize the vector?
Somebody please guide me.
Zulfi.

 

______________________________________________________
x+y=x------eq(1)
x+y=y------eq(2)
z=z---------eq(3)
______________________________________________________
Above x vanishes in eq(1), y vanishes in eq(2) & z vanishes in eq(3).
What is the value of μ1?????

Hello,

Interesting notation

The way you approached this, arent you supposed to solve that set of three equations simultaneously?
In other words, from the first equation;
______________________________________________________
x+y=x------eq(1)
______________________________________________________

we could get y=0,

then inserting that into the second equation, we could get x=0.

The third equation z=z means any value of z satisfies it and in keeping with the intent of a scalable vector z=1 would be the best choice.

If this doesnt seem to make sense, then just try to solve that set of three equations simultaneously.

Once you have all the eigenvalues the simpler approach i think is to solve the system:
A-L*i=0

where A is your matrix, L is ONE particular eigenvalue, and 'i' is the identity matrix of the same order as A.

Of course this approach is a consequence of your original approach but it may be easier to see how it works.