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Homework Help: Easy math

  1. Apr 29, 2008 #1
    can someone look at the image...
    can someone teach me/ help me remember how we can get
    e11 and e12
    thank you so much
  2. jcsd
  3. Apr 29, 2008 #2
    here is the picture

    Attached Files:

  4. Apr 30, 2008 #3


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    What is this; a homework question or what? I don't really understand what you want explaining to you. Is there a specific question that the attachment is a solution to?
  5. Apr 30, 2008 #4


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    You can get e11 and e12 as in the picture. Just write down the eigenvalue equation, multiply it out and solve the two given equations for the two unknowns. That's all that is happening. If you want more help, you need to be more specific.
  6. Apr 30, 2008 #5


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    Apparently this is to find eigenvectors of the matrix
    [tex]\left(\begin{array}{cc} 3 & \sqrt{2} \\ \sqrt{2} & 4\end{array}\right)[/tex]
    corresponding to eigenvalue 5.

    Since you want, by definition of "eigenvalue" and "eigenvector"
    [tex]\left(\begin{array}{cc} 3 & \sqrt{2} \\ \sqrt{2} & 4\end{array}\right)\left(\begin{array}{c}e_{11} \\ e_{12}\end{array}\right)= \left(\begin{array}{c} 5e_{11} \\ 5e_{12}\end{array}\right)[/tex]

    Which leads to the equation in your post
    [tex]\left(\begin{array}{cc}-2e_{11}+ \sqrt{2}e_{12}\\ \sqrt{2}e_{11}- e_{12}\end{array}\right)= \left(\begin{array}{c}0 \\ 0\end{array}\right)[/tex]

    That gives you the two equations [itex]-2e_{11}+ \sqrt{2}e_{12}= 0[/itex] and [itex]\sqrt{2}e_{11}- e_{12}= 0[/itex]. Because 5 is an eigenvalue, if you try to solve those two simultaneous equations you will find they are "dependent"- they reduce to a single equation: [itex]\sqrt{2}e_{11}= e_{12}[/itex]. Choosing either of those to be any number you want, you can solve for the other. In your attachment, they choose [itex]e_{11}= 1[/itex] so [itex]e_{12}= \sqrt{2}[/itex]. Any eigenvector of the equation, corresponding to eigenvalue 5, is a multiple of [itex](1, \sqrt{2})[/itex].

    The other eigenvalue, by the way, is 2. What are the eigenvectors corresponding to eigenvalue 2?
  7. Apr 30, 2008 #6


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    Actually, they choose [itex]e_{11}, e_{12}[/itex] such that the eigenvector has unit length
    [tex]|| (e_{11}, e_{12}) || = \sqrt{ e_{11}^2 + e_{12}^2 } [/tex]
  8. Apr 30, 2008 #7
    i think compuchip is correct..
    i found out how to do it yesterday..
    u have to set (e11)^2 + (e12)^2 =1
    and solve for e11, and e12...
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