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

- anonymous

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

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- anonymous

|dw:1433715559882:dw|

- anonymous

i've simplified it to 2x-y+z=0
how do i make that into an eigenvector?

- anonymous

@iambatman

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## More answers

- freckles

I'm having trouble seeing how got an egivenvalue of 3

- anonymous

Okay, so let's check and see if 3 works then

- anonymous

the original matrix was |dw:1433716009320:dw|

- anonymous

I found that one eigenvalue was 3 and the other two were complex numbers

- anonymous

I'm just trying to find the eigenvector for 3 right now

- anonymous

I plugged 3 into the matrix to get the matrix that i originally posted

- anonymous

I just realized the other two eigenvalues aren't complex btw

- freckles

for this other matrix you have I see how you got 3 now

- freckles

for the first matrix all the egienvalues were 0

- anonymous

sorry i wasn't clear. The first matrix is the second matrix with the eigenvalue 3 plugged in

- anonymous

|dw:1433716555159:dw|

- anonymous

which produced

- anonymous

|dw:1433716643997:dw|

- anonymous

after i simplified it a bit

- 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

oops forgot to put my one column vector in

- anonymous

after simplifying i got
2x-y+z=0

- anonymous

the second and third row should cancel

- 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

err 2x-y+z=0

- 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

that 2 should be 3

- anonymous

so can i use ( 1 , 1 , -1) as the eigenvector?

- freckles

as one egienvector

- freckles

3 is repeated twice

- freckles

ok so we have to find a second egienvector

- freckles

for lambda=3

- freckles

we can use the (1,1,-1) as the first

- anonymous

ok I get it now, thanks a lot!

- 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

notice I replace that zero vector with the other egienvector we found for lambda=3

- anonymous

I think i just need one eigenvector, i need it to solve for a system of differential equations problem

- anonymous

Thanks for the help!

- 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

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

sorry the first link might not work

- anonymous

Was 3 a repeated eigenvalue?

- freckles

yah yah!

- freckles

I think he is going to need that other egienvector

- 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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