3 x 3 matrix eigenvectors


by karnten07
Tags: eigenvectors, matrix
karnten07
karnten07 is offline
#1
Nov3-07, 05:23 PM
P: 208
1. The problem statement, all variables and given/known data

Find the eigenvalues and eigenvectors of A

1 1 a
1 -1 b
0 0 1

you can assume a and b are not equal to zero
2. Relevant equations




3. The attempt at a solution

Using det(A-sI) = 0 where s is the eigenvalue

i get (s^2 - 2)(1 - s) = 0

therefore giving s1 = 1 and s2/3 = +/- 2

Id be very grateful if someone could check this is correct because im a little uncertain.

From here i struggle to find the eigenvectors, i substitute the eigenvalues to (A-sI) to get

0 1 a
1 -2 b
0 0 0

Then i saw a method where i equated the sum of this matrix with (x, y, z) - a vector written vertically, to (0,0,0) also written vertically.

From that i get y + za = 0, x - 2y + bz = 0

But then i get lost in the method from here because im unsure i can apply it to this case. In fact i think i may have gone about finding the vectors in the complete wrong way, or ive missed a trick or something. Ive searched on the internet for examples and methods but their seems to be so many and i cant seem to decipher which ones i can use or even how to use them. So any help at all would be greatly appreciated, preferably before tuesday morning as that is when it is due. Many thanks
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Dick
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#2
Nov3-07, 05:35 PM
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I think the roots of that polynomial are 1 and +/-sqrt(2), right? You are on the right track for the eigenvector of 1. The first equation tells you y=za. Put that into the second equation and get x in terms of z also. Now write the [x,y,z] vector all in terms of z. Do you see the eigenvector?
karnten07
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#3
Nov3-07, 06:25 PM
P: 208
Quote Quote by Dick View Post
I think the roots of that polynomial are 1 and +/-sqrt(2), right? You are on the right track for the eigenvector of 1. The first equation tells you y=za. Put that into the second equation and get x in terms of z also. Now write the [x,y,z] vector all in terms of z. Do you see the eigenvector?
I get y = -za, and writing the x,y,z all in z i get -za, -2za -bz, z

im not sure how i get the eigenvector from this?

Dick
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#4
Nov3-07, 10:02 PM
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3 x 3 matrix eigenvectors


[-za,-2za-bz,z]=z*[-a,-2a-b,1]. Remember eigenvectors are only defined up to an multiplicative constant.
karnten07
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#5
Nov4-07, 02:02 PM
P: 208
Quote Quote by Dick View Post
[-za,-2za-bz,z]=z*[-a,-2a-b,1]. Remember eigenvectors are only defined up to an multiplicative constant.
Ok i see that now, but for s = +/-sqrt(2) im finding even harder to do it in this way. It seems very lengthy and im wondering if there is an easier way to do it for this case or even if the result i am getting is right.

When i use the eigenvalue of sqrt(2) i again try to write it all in terms of z because i thought this looked easiest as one of them is already written in terms of z.

So i get from subsitution

x(1 - sqrt(2)) + y +az = 0 (1)
x + y(1 - sqrt(2)) + bz = 0 (2)
z(1- sqrt(2)) = 0 (3)

Originally i did think that this meant z = 0 but then i substituted that into 1 and 2 but then ended up getting 2 different equations for x and y eg. x = -y(1-sqrt(2)) and
x = -y/(sqrt(2)), so instead i got x from 1, substituted it back into 1 then rearranged to get y in terms of z. Using y i got x in terms of z from 1.

x = bz((-1 - sqrt(2))/2)
y = -bz/2

Subsituting all into 1, 2 and 3 in terms of z i get:

az = 0
0 = 0
z(1 - sqrt(2)) = 0

so i get eigenvector of z*(a, 0, 1 - sqrt(2))

Is there a way i can check this is right? Many thanks
Dick
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#6
Nov4-07, 05:45 PM
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Quote Quote by Dick View Post
[-za,-2za-bz,z]=z*[-a,-2a-b,1]. Remember eigenvectors are only defined up to an multiplicative constant.
BTW, the x and y components are reversed here, you caught that right? There is an easy way to check an eigenvector. Multiply it by the original matrix and see if you get the eigenvalue times the eigenvector. If you do that with the sqrt(2) you'll see something has gone awry. I can see one thing already. The coefficient of y in the second equation should be -1-sqrt(2).
karnten07
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#7
Nov5-07, 05:35 AM
P: 208
Quote Quote by Dick View Post
BTW, the x and y components are reversed here, you caught that right? There is an easy way to check an eigenvector. Multiply it by the original matrix and see if you get the eigenvalue times the eigenvector. If you do that with the sqrt(2) you'll see something has gone awry. I can see one thing already. The coefficient of y in the second equation should be -1-sqrt(2).
oh no, i didnt see id put them the wrong way around, thanks for pointing it out. I will work on it today and see if i can work it out and report back here. Thanks again for all your help.
karnten07
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#8
Nov5-07, 05:21 PM
P: 208
Quote Quote by Dick View Post
BTW, the x and y components are reversed here, you caught that right? There is an easy way to check an eigenvector. Multiply it by the original matrix and see if you get the eigenvalue times the eigenvector. If you do that with the sqrt(2) you'll see something has gone awry. I can see one thing already. The coefficient of y in the second equation should be -1-sqrt(2).
ok so im back on the eigenvalue of sqrt(2)

I substitute it into the matrix and get

(1- sqrt(2))x + y + az = 0
x + y(-1-sqrt(2)) + bz
(1-sqrt(2))z = 0

I dont see how to get the eigenvector from this stage, do i need to write x, y and z all in terms of just one term?
Dick
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#9
Nov5-07, 05:26 PM
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Well, you do have z=0. So yes, now you just need to eliminate either x or y to get all of the vector components in terms of one variable.
karnten07
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#10
Nov5-07, 05:41 PM
P: 208
Quote Quote by Dick View Post
Well, you do have z=0. So yes, now you just need to eliminate either x or y to get all of the vector components in terms of one variable.
so by eliminating az and bz from 1 and 2, i get y = -x(1-sqrt(2))

so the eigenvector is x*(1, 1-sqrt(2), 0)

cool. th enext part of the question asks me to diagonalise the matrix by finding matrix B so that B = R^-1 A R is a diagonal matrix. Im just looking through my notes to see if i have anything on this, but any hints you can give me, thanks for all the help!
Dick
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#11
Nov5-07, 05:45 PM
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Looks more to me like x*(1,sqrt(2)-1,0), but I think you have the general idea.
karnten07
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#12
Nov5-07, 05:51 PM
P: 208
Quote Quote by Dick View Post
Looks more to me like x*(1,sqrt(2)-1,0), but I think you have the general idea.

oh yes youre right again, well spotted. So for the diagonalization, it says that R is formed by the 3 eigenvectors as the columns of this new matrix. So i will make my first eigenvector corresponding to eigenvalue 1, in terms of y, so i can write out the diagonalized matrix all in terms of y.


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