
#1
Dec808, 06:57 AM

P: 132

revising for a text and got stuck mid way through a question
Find the eigenvalues and vectors of A (in matrix form i will state colum then column then column) A((3,0,0),(0,2,0),(0,0,2) B=((3,0,0),(0,2,1),(0,0,2) for A i got x(lamda)=(lamda3)(lamda2)^2 lamda=2or3 then i got lamda=3 solved to get a non zero multiple of (1,0,0) its the second bit that im stuck on my notes tell me that lamda=2 gives eigenvectors (a,a,b) where a and b are any reals, i dont get why this is a,a,b) and not (a,b,b,) usualy i would just think this is a mistake in my copying down but i remember the teacher explaining why it was 11b and not abb just cant remember why, if someone could shed some light this would be much apprichiated, thanks 



#2
Dec808, 07:03 AM

P: 132

theres also a C matrix((2,1,0),(0,2,1)(0,0,2) the anwer given here states that obviously lamda=2 is the only eigenvalue and that we can check that the eigenspace is only 1 dimensional, is this enough of an answer or would we need to state/find the eigenvector, if so how is this done




#3
Dec808, 07:22 AM

HW Helper
P: 2,149

you want
(0,a,b) 



#4
Dec808, 07:28 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

eigenvectors question
You said the problem is to find the eigenvalues and eigenvectors of A. How are B and C at all relevant?
Isn't it obvious that the eigenvalues of A are 3 and 2 with eigenvectors corresponding to 3 multiples of (1, 0, 0) and eigenvectors of 2 linear combinations of (0, 1, 0) and (0, 0, 1)? That is, the eigenvectors corresponding to eigenvalue 3 are of the form (a, 0, 0) and eigenvectors corresponding to eigenvalue 2 are (0, b, c). Are finding the eigenvalues and eigenvectors of B and C separate problems? Yes, the eigenvalues of B are also 2 and 3 again. The simplest way to find eigenvectors corresponding to them is to write out the equation that says they are eigenvalues: [tex]\left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & 2 & 1\\ 0 & 0 & 2\end{array}\right]\left[\begin{array}{c}x \\ y \\ z\end{array}\right]= \left[\begin{array}{c}2x \\ 2y\\ 2z\end{array}\right][/tex] must have a nontrivial solution (x, y, z) in order that 2 be an eigenvalue. Multiplying that out and setting components equal gives 3x= 2x, 2y+ z= 2y, and 2z= 2z. What values of x, y, z (not all 0) satisfy that? The equations for eigenvalue 3 are 3x= 3x, 2y+ z= 3y, and 2z= 3z which should be easy to solve. For C, the corresponding equations are 2x= 2x, 2y+ z= 2y, and 2z= 2z. What values of x, y, z (not all 0) satisfy that? A is, by the way, a "diagonal" matrix while B and C are in "Jordan Normal Form". If it were possible to "diagonalize" every matrix, the world would be much simpler. It is, however, always possible to either diagonalize or convert any matrix to Jordan Normal Form so these are the "typical" kinds of problems. 



#5
Dec808, 09:38 AM

HW Helper
P: 2,618

I don't know if this would detract from the OP's original question, but is it possible to determine the exact Jordan form of a matrix? So far in the course in intermediate linear algebra we were taught only how to determine all the possible Jordan forms of a matrix given it's characteristic equations (Jordan forms similar to each other are not counted) and minimal polynomial. What must we know in advance so we can determine the exact Jordan form of a matrix?




#6
Dec808, 05:36 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Yes, of course it is. If A has eigenvalues e_{1} of multiplicity m_{1}, e_{2} of multiplicity m_{2}, up to e_{n} of multiplicity m_{n} and and each e_{k} has eigenspace (the space of eigenvectors corresponding to e[sup]k[/sub]) of dimension d_{k}, then the Jordan Normal form is an m_{1}+ m_{2}+ ...+ m_{n} by m_{1}+ m_{2}+ ...+ m_{n} matrix consisting of n "blocks" along the diagonal. The kth block is an m_{k} by m_{k} matrix having the eigenvalue e_{k} on the diagonal with "1" directly above the last m_{k} d_{k}1 eigenvalues, "0" elsewhere. The rest of the matrix is "0"s.
In order to determin the Jordan form of a matrix you need to know the eigenvalues, their multiplicities as roots of the characteristic equation, and the eigenvectors of each eigenvalue. For example, if the characteristic equation of a matrix has eigenvalue [itex]\lambda_1[/itex] of multiplicity 3 and there are are two independent eigenvectors corresponding to eigenvalue [itex]\lambda_1[/itex] and eigenvalue [itex]\lambda_2[/itex] of multiplicity two with only one eigenvalue, then its Jordan Normal Form is [tex]\left[\begin{array}{ccccc} 3 & 0 & 0 & 0& 0 \\ 0 & 3 & 1 & 0& 0\\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{array}\right][/tex] 



#7
Dec808, 08:31 PM

HW Helper
P: 2,149





#8
Dec908, 07:36 AM

HW Helper
P: 2,618





#9
Dec908, 05:33 PM

HW Helper
P: 2,149

One way to go is to consider the numbers rank((AaI)^k) where A is a squre matrix, I is the identitiy matrix the same size as A, k=1,2,3, and a is an eigenvalue. rank((AaI)^1) tells you the number of eigen blocks (at least dim 1) rank((AaI)^2)rank((AaI)^1) tells you the number of eigen blocks at least dim 2 rank((AaI)^3) rank((AaI)^2) tells you the number of eigen blocks at least dim 3 and so on. Av=av or (AaI)v There may not be enough eigenvectors to form a basis. We define a generalized eigenvector (associated to the eigenvalue a) of order k=1,2,3,... as a vector v such that [(AaI)^k]v=0 but [(AaI)^(k1)]v!=0 where != means not equal. With generalized eigenvectors we can always form a basis. One way of putting A in Jordan form is J=[S^1]AS where J is the jordan form of A and S is a matrix whose columbs are generalized eigenvectors. 


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