# Eigenvectors question

1. Dec 8, 2008

### franky2727

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)=(lamda-3)(lamda-2)^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. Dec 8, 2008

### franky2727

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. Dec 8, 2008

you want
(0,a,b)

4. Dec 8, 2008

### HallsofIvy

Staff Emeritus
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:
$$\left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & 2 & 1\\ 0 & 0 & 2\end{array}\right]\left[\begin{array}{c}x \\ y \\ z\end{array}\right]= \left[\begin{array}{c}2x \\ 2y\\ 2z\end{array}\right]$$
must have a non-trivial 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. Dec 8, 2008

### Defennder

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. Dec 8, 2008

### HallsofIvy

Staff Emeritus
Yes, of course it is. If A has eigenvalues e1 of multiplicity m1, e2 of multiplicity m2, up to en of multiplicity mn and and each ek has eigenspace (the space of eigenvectors corresponding to ek[/sub]) of dimension dk, then the Jordan Normal form is an m1+ m2+ ...+ mn by m1+ m2+ ...+ mn matrix consisting of n "blocks" along the diagonal. The kth block is an mk by mk matrix having the eigenvalue ek on the diagonal with "1" directly above the last mk- dk-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 $\lambda_1$ of multiplicity 3 and there are are two independent eigenvectors corresponding to eigenvalue $\lambda_1$ and eigenvalue $\lambda_2$ of multiplicity two with only one eigenvalue, then its Jordan Normal Form is
$$\left[\begin{array}{ccccc} 3 & 0 & 0 & 0& 0 \\ 0 & 3 & 1 & 0& 0\\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{array}\right]$$

Last edited: Dec 8, 2008
7. Dec 8, 2008

### lurflurf

Of course you can common methods include finding the generalized eigen vectors of using elementary divisor theory. However numerical calculations of Jordan forms tend to be unstable, so an approximate Jordan form can be useless.

8. Dec 9, 2008

### Defennder

So this means that the no. of eigenvectors associated with an eigenvalue tells you the size of the largest Jordan block of that eigenvalue? Which is the same thing as the power k of the $$(x-\lambda_1)^k$$ of the minimal polynomial?

Hmm I haven't learnt anything yet about generalised eigenvectors.

9. Dec 9, 2008

### lurflurf

That is not quite right. The number of eigenvectors associated with an eigenvalue tells you the number of Jordan blocks of that eigenvalue.
One way to go is to consider the numbers
rank((A-aI)^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((A-aI)^1)
tells you the number of eigen blocks (at least dim 1)
rank((A-aI)^2)-rank((A-aI)^1)
tells you the number of eigen blocks at least dim 2
rank((A-aI)^3) -rank((A-aI)^2)
tells you the number of eigen blocks at least dim 3
and so on.

eigenvectors have the property

Av=av
or
(A-aI)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

[(A-aI)^k]v=0
but
[(A-aI)^(k-1)]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.