anonymous
  • anonymous
what is the eigenvector of this matrix. The eigenvalue of 3 has already been plugged in
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1433715559882:dw|
anonymous
  • anonymous
i've simplified it to 2x-y+z=0 how do i make that into an eigenvector?
anonymous
  • anonymous
@iambatman

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freckles
  • freckles
I'm having trouble seeing how got an egivenvalue of 3
anonymous
  • anonymous
Okay, so let's check and see if 3 works then
anonymous
  • anonymous
the original matrix was |dw:1433716009320:dw|
anonymous
  • anonymous
I found that one eigenvalue was 3 and the other two were complex numbers
anonymous
  • anonymous
I'm just trying to find the eigenvector for 3 right now
anonymous
  • anonymous
I plugged 3 into the matrix to get the matrix that i originally posted
anonymous
  • anonymous
I just realized the other two eigenvalues aren't complex btw
freckles
  • freckles
for this other matrix you have I see how you got 3 now
freckles
  • freckles
for the first matrix all the egienvalues were 0
anonymous
  • anonymous
sorry i wasn't clear. The first matrix is the second matrix with the eigenvalue 3 plugged in
anonymous
  • anonymous
|dw:1433716555159:dw|
anonymous
  • anonymous
which produced
anonymous
  • anonymous
|dw:1433716643997:dw|
anonymous
  • anonymous
after i simplified it a bit
freckles
  • freckles
ok so lambda=3 I'm going to do A-Lambda*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
  • freckles
oops forgot to put my one column vector in
anonymous
  • anonymous
after simplifying i got 2x-y+z=0
anonymous
  • anonymous
the second and third row should cancel
freckles
  • freckles
\[(A- \lambda I)x=0 \\ \left[\begin{matrix}2 & -1 & 1 \\ 2 & -1 & 1 \\-2 & 1 & -1 \end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)=\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)\] so we have only need to look at \[2z-y+z=0\] as you said
freckles
  • freckles
err 2x-y+z=0
freckles
  • freckles
\[2x-y+z=0 \\ \text{ one solution is } (1,1,-1) \\ \text{ now also say we replace } x=t \text{ then } 2t=y-z \\ \text{ so we could replace } y=3t \text{ and } z=t \\ \text{ so we have } \left(\begin{matrix}1 \\ 1 \\ -1 \end{matrix}\right)+t \left(\begin{matrix}1 \\ 2 \\ 1\end{matrix}\right)\] I think I remember something like this
freckles
  • freckles
that 2 should be 3
anonymous
  • anonymous
so can i use ( 1 , 1 , -1) as the eigenvector?
freckles
  • freckles
as one egienvector
freckles
  • freckles
3 is repeated twice
freckles
  • freckles
ok so we have to find a second egienvector
freckles
  • freckles
for lambda=3
freckles
  • freckles
we can use the (1,1,-1) as the first
anonymous
  • anonymous
ok I get it now, thanks a lot!
freckles
  • freckles
\[\left[\begin{matrix}4 & -2 & 2 \\ 8 & -4 & 4 \\ -4 & 2 & -2\end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) =\left(\begin{matrix}1 \\ 1 \\ -1\end{matrix}\right)\] to find the second one you will have to solve this for (x,y,z)
freckles
  • freckles
notice I replace that zero vector with the other egienvector we found for lambda=3
anonymous
  • anonymous
I think i just need one eigenvector, i need it to solve for a system of differential equations problem
anonymous
  • anonymous
Thanks for the help!
freckles
  • freckles
http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-iv-first-order-systems/matrix-methods-eigenvalues-and-normal-modes/MIT18_03SCF11_s33_8text.pdf could be wrong I think you will need to other egienvector
freckles
  • freckles
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0CCUQFjAB&url=http%3A%2F%2Focw.mit.edu%2Fcourses%2Fmathematics%2F18-03sc-differential-equations-fall-2011%2Funit-iv-first-order-systems%2Fmatrix-methods-eigenvalues-and-normal-modes%2FMIT18_03SCF11_s33_8text.pdf&ei=-st0VdyBGYLusAXO_oOIDg&usg=AFQjCNGPx5I__YKjZsx_PZa86olry6fb7w&sig2=GYhXtaffJuV9ssjfj3nC5w
freckles
  • freckles
sorry the first link might not work
anonymous
  • anonymous
Was 3 a repeated eigenvalue?
freckles
  • freckles
yah yah!
freckles
  • freckles
I think he is going to need that other egienvector
anonymous
  • anonymous
Then I would think youd definitely need another eigenvector. Well, not technically an eigenvector, but a generalized one. If I recall, we'd need this: \[(A-3I)\eta = \xi \] Where \(\xi\) is the first eigenvector. I have no idea how to type up matrices on here though, lol.

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