
#1
Nov307, 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(AsI) = 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 (AsI) 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 



#2
Nov307, 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?




#3
Nov307, 06:25 PM

P: 208

im not sure how i get the eigenvector from this? 



#4
Nov307, 10:02 PM

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3 x 3 matrix eigenvectors
[za,2zabz,z]=z*[a,2ab,1]. Remember eigenvectors are only defined up to an multiplicative constant.




#5
Nov407, 02:02 PM

P: 208

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(1sqrt(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 



#6
Nov407, 05:45 PM

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P: 25,170





#7
Nov507, 05:35 AM

P: 208





#8
Nov507, 05:21 PM

P: 208

I substitute it into the matrix and get (1 sqrt(2))x + y + az = 0 x + y(1sqrt(2)) + bz (1sqrt(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? 



#9
Nov507, 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.




#10
Nov507, 05:41 PM

P: 208

so the eigenvector is x*(1, 1sqrt(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! 



#11
Nov507, 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.




#12
Nov507, 05:51 PM

P: 208

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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