- #1

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can someone teach me/ help me remember how we can get

e11 and e12

thank you so much

tuan

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- Thread starter tuanle007
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In summary, someone needs to learn how to get eigenvalues and eigenvectors from a matrix. They found a way to do it yesterday.f

- #1

- 36

- 0

can someone teach me/ help me remember how we can get

e11 and e12

thank you so much

tuan

- #2

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- #3

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- #5

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[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].

The other eigenvalue, by the way, is 2. What are the eigenvectors corresponding to eigenvalue 2?

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Actually, they choose [itex]e_{11}, e_{12}[/itex] such that the eigenvector has unit lengthIn 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].

[tex]|| (e_{11}, e_{12}) || = \sqrt{ e_{11}^2 + e_{12}^2 } [/tex]

- #7

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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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