Normalization & value of eigen vectors

Discussion in 'Homework Help' started by zulfi100, Sep 18, 2018.

  1. zulfi100

    Thread Starter Active Member

    Jun 7, 2012
    495
    1
    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.
     
  2. MrAl

    AAC Fanatic!

    Jun 17, 2014
    6,382
    1,382
    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.
     
    Last edited: Sep 20, 2018
Loading...