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anonymous
 one year ago
what is the eigenvector of this matrix. The eigenvalue of 3 has already been plugged in
anonymous
 one year ago
what is the eigenvector of this matrix. The eigenvalue of 3 has already been plugged in

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433715559882:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i've simplified it to 2xy+z=0 how do i make that into an eigenvector?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I'm having trouble seeing how got an egivenvalue of 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so let's check and see if 3 works then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the original matrix was dw:1433716009320:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I found that one eigenvalue was 3 and the other two were complex numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm just trying to find the eigenvector for 3 right now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I plugged 3 into the matrix to get the matrix that i originally posted

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just realized the other two eigenvalues aren't complex btw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1for this other matrix you have I see how you got 3 now

freckles
 one year ago
Best ResponseYou've already chosen the best response.1for the first matrix all the egienvalues were 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry i wasn't clear. The first matrix is the second matrix with the eigenvalue 3 plugged in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433716555159:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433716643997:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0after i simplified it a bit

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok so lambda=3 I'm going to do ALambda*I=0 \[\left[\begin{matrix}7 & 2 & 2 \\ 8 &1 & 4 \\ 4 & 2 & 1\end{matrix}\right] \left[\begin{matrix}3 & 0 & 0 \\ 0 & 3 &0 \\ 0 & 0 & 3\end{matrix}\right] \\ =\left[\begin{matrix}4 & 2 & 2 \\ 8 & 4 & 4 \\ 4 & 2 & 2\end{matrix}\right] =\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)\] \[\left[\begin{matrix}2 & 1 & 1 \\ 2 & 1 & 1 \\ 2 & 1 & 1 \end{matrix}\right]=\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)\] ok I'm cool with that

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oops forgot to put my one column vector in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0after simplifying i got 2xy+z=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the second and third row should cancel

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[(A \lambda I)x=0 \\ \left[\begin{matrix}2 & 1 & 1 \\ 2 & 1 & 1 \\2 & 1 & 1 \end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)=\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)\] so we have only need to look at \[2zy+z=0\] as you said

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[2xy+z=0 \\ \text{ one solution is } (1,1,1) \\ \text{ now also say we replace } x=t \text{ then } 2t=yz \\ \text{ so we could replace } y=3t \text{ and } z=t \\ \text{ so we have } \left(\begin{matrix}1 \\ 1 \\ 1 \end{matrix}\right)+t \left(\begin{matrix}1 \\ 2 \\ 1\end{matrix}\right)\] I think I remember something like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so can i use ( 1 , 1 , 1) as the eigenvector?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok so we have to find a second egienvector

freckles
 one year ago
Best ResponseYou've already chosen the best response.1we can use the (1,1,1) as the first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok I get it now, thanks a lot!

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\left[\begin{matrix}4 & 2 & 2 \\ 8 & 4 & 4 \\ 4 & 2 & 2\end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) =\left(\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right)\] to find the second one you will have to solve this for (x,y,z)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1notice I replace that zero vector with the other egienvector we found for lambda=3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think i just need one eigenvector, i need it to solve for a system of differential equations problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the help!

freckles
 one year ago
Best ResponseYou've already chosen the best response.1http://ocw.mit.edu/courses/mathematics/1803scdifferentialequationsfall2011/unitivfirstordersystems/matrixmethodseigenvaluesandnormalmodes/MIT18_03SCF11_s33_8text.pdf could be wrong I think you will need to other egienvector

freckles
 one year ago
Best ResponseYou've already chosen the best response.1sorry the first link might not work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Was 3 a repeated eigenvalue?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I think he is going to need that other egienvector

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then I would think youd definitely need another eigenvector. Well, not technically an eigenvector, but a generalized one. If I recall, we'd need this: \[(A3I)\eta = \xi \] Where \(\xi\) is the first eigenvector. I have no idea how to type up matrices on here though, lol.
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