Question

eigenvector help needed
I have the eigenvalues(-1,1,2,4) but I need help with the eigenvector for 2

I Will post the matrix

Answer 1

post the matrix!

Answer 2

\[\left[\begin{matrix}1 & 0 & 0&0 \\ 1&2 &0&0 \\ 1&2&-1&1 \\ 1&1&0&4\end{matrix}\right]\]

Answer 3

sorry that took so long

Answer 4

ok

Answer 5

Whn I put it into wolfram alpha, it says I should get (0,-2,-1,1) but I can't figure out how to get that. I got all the other ones, but this one isn't coming out.

Answer 6

so that thing you posted is the matrix you are working with right?

Answer 7

yep

Answer 8

and you got the values of: 4,2,-1,1 correct?

Answer 9

correct

Answer 10

I just can't figure out how to get the correct eigenvector for 2. I got the other ones but this one isn't coming out correctly

Answer 11

the answer (i made it from my hand) is (0, 2, 1, -1) and the wólfram alpha is right. Simply find A-2I where I is the identity matrix. Row-reduce it (applying elementary operations in the rows). You get (1,0,0,0//0,1,0,2//0,0,1,1//0,0,0,0)=B. Solving BX=O produces the eigenvector.

Answer 12

wait how did you get B? I'm not quite seeing where you got that from

Answer 13

Row-reducing A-2I

Answer 14

it may just be a different name but what do you mean by row reducing?

Answer 15

apply one of the 3 elementary operations in rows, to obtain the ECHELON form of the matrix

Answer 16

(A-L)x = 0, should produce the eigenvectors for x, right?

Answer 17

1. Interchange rows
2. Multiply a row by a no zero number
3. Add a row, multiplied by a number, to another row

Answer 18

amistre64 is right. It's A-I, not A-L (what is L???)

Answer 19

L is lamda :) a given eigenvalue

Answer 20

right. a-2I gets (-1,0,0,0//1,0,0,0//1,0,-3,1//1,1,0,0), right?

Answer 21

Noooo A-L means a matrix (A) MINUS a number (L), you can't do this

Answer 22

lambda times the identity matrix, yes?

Answer 23

im assuming that since we are talking about eigen stuff, that the I is implied ...

Answer 24

better: (A-LI)x=o, where I = identity matrix

Answer 25

a- 2 (lambda) times the identity matrix

Answer 26

please do not "imply".

Answer 27

excuse my english

Answer 28

anyway... what I put earlier is what you would get, right?

Answer 29

goalie is right

Answer 30

so then why can I not seem to get the correct vector from that matrix?

Answer 31

i have to go... nice to meet you... Greetings from La Paz, Bolivia!!

Answer 32

... I still haven't figured out why I can't seem to get the right answer from the matrix I obtain by taking A - 2(lambda)*I

Answer 33

send me your e-mail, or please write to: diegoesdiego@hotmail.com. I will return the complete procedure. bye.

Answer 34

\[\left[\begin{matrix}1-2 & 0 & 0&0 \\ 1&2-2 &0&0 \\ 1&2&-1-2&1 \\ 1&1&0&4-2\end{matrix}\right]\]

\[\left[\begin{matrix}-1 & 0 & 0&0 \\ 1&0 &0&0 \\ 1&2&-3&1 \\ 1&1&0&2\end{matrix}\right]\]

and row reduce

Answer 35

what is row reduce though?

Answer 36

getting it into echelon form ... how did you do the others?

Answer 37

set up systems of equations by multiplying with variables for the other part

Answer 38

it row reduces to

{1, 0, 0, 0}
{0, 1, 0, 2}
{0, 0, 1, 1}
{0, 0, 0, 0}

and negate the last column since there is only 1 free variable

Answer 39

well, the last one will not be a zero, itll be a 1

Answer 40

x = w{ 0}
y = w{-2}
z = w{-1}
w = w{ 1}

Answer 41

I'll trust you on that one. I'll have to look into that some more. honestly, I've never seen that. That does make it come out right. Thanks

Answer 42

{-1, 0, 0, 0} *-1
{ 1, 0, 0, 0}
{ 1, 2, -3, 1}
{ 1, 1, 0, 2}

{ 1, 0, 0, 0}
{ 1, 0, 0, 0} add the rows
{ 1, 2, -3, 1} using the top row
{ 1, 1, 0, 2}

{ 1, 0, 0, 0}
{ 0, 0, 0, 0} rearrange in a more echelon form
{ 0, 2, -3, 1}
{ 0, 1, 0, 2}


{ 1, 0, 0, 0}
{ 0, 1, 0, 2} *-2 and add to the 3rd row
{ 0, 2, -3, 1}
{ 0, 0, 0, 0}


{ 1, 0, 0, 0}
{ 0, 1, 0, 2}
{ 0, 0, -3, -3} /-3
{ 0, 0, 0, 0}


{ 1, 0, 0, 0}
{ 0, 1, 0, 2} finished results
{ 0, 0, 1, 1}
{ 0, 0, 0, 0}
^^
free variable

Answer 43

define the rest of them in terms of the free variable. which i did above