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Easy math

  • Thread starter tuanle007
  • Start date
  • #1
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can someone look at the image...
can someone teach me/ help me remember how we can get
e11 and e12
thank you so much
tuan
 

Answers and Replies

  • #2
36
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here is the picture
 

Attachments

  • #3
cristo
Staff Emeritus
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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?
 
  • #4
CompuChip
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Homework Helper
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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.
 
  • #5
HallsofIvy
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Homework Helper
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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?
 
  • #6
CompuChip
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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].
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]
 
  • #7
36
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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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