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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

|dw:1433715559882:dw|
i've simplified it to 2x-y+z=0 how do i make that into an eigenvector?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I'm having trouble seeing how got an egivenvalue of 3
Okay, so let's check and see if 3 works then
the original matrix was |dw:1433716009320:dw|
I found that one eigenvalue was 3 and the other two were complex numbers
I'm just trying to find the eigenvector for 3 right now
I plugged 3 into the matrix to get the matrix that i originally posted
I just realized the other two eigenvalues aren't complex btw
for this other matrix you have I see how you got 3 now
for the first matrix all the egienvalues were 0
sorry i wasn't clear. The first matrix is the second matrix with the eigenvalue 3 plugged in
|dw:1433716555159:dw|
which produced
|dw:1433716643997:dw|
after i simplified it a bit
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
oops forgot to put my one column vector in
after simplifying i got 2x-y+z=0
the second and third row should cancel
\[(A- \lambda I)x=0 \\ \left[\begin{matrix}2 & -1 & 1 \\ 2 & -1 & 1 \\-2 & 1 & -1 \end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)=\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)\] so we have only need to look at \[2z-y+z=0\] as you said
err 2x-y+z=0
\[2x-y+z=0 \\ \text{ one solution is } (1,1,-1) \\ \text{ now also say we replace } x=t \text{ then } 2t=y-z \\ \text{ so we could replace } y=3t \text{ and } z=t \\ \text{ so we have } \left(\begin{matrix}1 \\ 1 \\ -1 \end{matrix}\right)+t \left(\begin{matrix}1 \\ 2 \\ 1\end{matrix}\right)\] I think I remember something like this
that 2 should be 3
so can i use ( 1 , 1 , -1) as the eigenvector?
as one egienvector
3 is repeated twice
ok so we have to find a second egienvector
for lambda=3
we can use the (1,1,-1) as the first
ok I get it now, thanks a lot!
\[\left[\begin{matrix}4 & -2 & 2 \\ 8 & -4 & 4 \\ -4 & 2 & -2\end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) =\left(\begin{matrix}1 \\ 1 \\ -1\end{matrix}\right)\] to find the second one you will have to solve this for (x,y,z)
notice I replace that zero vector with the other egienvector we found for lambda=3
I think i just need one eigenvector, i need it to solve for a system of differential equations problem
Thanks for the help!
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
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
sorry the first link might not work
Was 3 a repeated eigenvalue?
yah yah!
I think he is going to need that other egienvector
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question