Results 1 to 6 of 6

Math Help - Basis for eigenspace!

  1. #1
    Member
    Joined
    Apr 2008
    Posts
    132

    Basis for eigenspace!

    Find a basis for the eigenspace corresponding to the eigenvalue λ=2 of the matrix

    A =
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by matty888 View Post
    Find a basis for the eigenspace corresponding to the eigenvalue λ=2 of the matrix

    A =
    Well assume a vector [x, y, z]^{T} to be the eigenvector corresponding. By definition the eigenspace is the span of the eigenvector.
    So
    |A - 2I| = 0 \Rightarrow 12x + y - 9z=0

    So the if x = a and y= b, we have (a,b,\frac{12a+b}9)

    So (1,0,\frac43) and (0,1,\frac19) form the basis
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Quote Originally Posted by matty888 View Post
    Find a basis for the eigenspace corresponding to the eigenvalue λ=2 of the matrix

    A =
    A=\begin{pmatrix} 26 & 2 & -18 \\ 0 & 2 & 0 \\ 36 & 3 & -25 \end{pmatrix}

    X_2=\begin{pmatrix} x \\ y \\ z \end{pmatrix}

    Solve for X_2 in AX_2=2X_2

    This yields the following system :

    \left\{ \begin{array}{ccccccc} 26x &+ &2y&-&18z&=&2x \\ & & 2y & & & =& 2y \\ 36x&+&3y&-&25z&=&2z \end{array} \right.

    \implies \left\{\begin{array}{ccccccc} 24x&+ &2y&-&18z&=&0 \\ 36x&+&3y&-&27z&=&0 \end{array} \right.

    (we can see that these two equations are actually the same).

    There are two possibilities. Either y=1, either y=0.

    If y=0, then 24x-18z=0 \implies x=\frac 34 z. For example, z=4 and x=3

    So \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix} is an eigenvector of the eigenspace.


    If y=1, then 24x+2-18z=0 \implies x=-\frac{1}{12}+\frac 34 z. For example, z=\frac 13 and x=\frac 16

    So \begin{pmatrix} \frac 16 \\ 1 \\ \frac 13 \end{pmatrix} is the second eigenvector of the eigenspace.


    Does it look clear ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2008
    Posts
    132

    Thanks(????)

    Thanks guys just wondering are both of your answers legitimate,I am getting (3,0,4) and (-1,12,0) also, so not really sure????
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by matty888 View Post
    Thanks guys just wondering are both of your answers legitimate,I am getting (3,0,4) and (-1,12,0) also, so not really sure????
    Your answer is right too because from my basis I can represent

    (3,0,4) = 3(1,0,\frac43)

    (-1,12,0) = -(1,0,\frac43) + 12(0,1,\frac19)

    So both of your vectors are a linear combination of my basis and your vectors need both of my basis vectors.So your set is a basis

    You can try getting your vectors as a linear combination of Moo's basis
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    For my basis, it'll be :

    12(\frac 16,1,\frac 13)-(3,0,4)=(2,12,4)-(3,0,4)=(-1,12,0)

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Generalized eigenspace
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: September 4th 2011, 05:49 PM
  2. Find a basis for the eigenspace
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: January 14th 2011, 05:59 PM
  3. [SOLVED] eigenspace
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 28th 2010, 04:41 PM
  4. Basis of Eigenspace using Eigenvalues
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 17th 2009, 11:58 PM
  5. Eigenspace
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: November 4th 2007, 08:06 PM

Search Tags


/mathhelpforum @mathhelpforum